Matter and its properties

Throughout this section, you will explore chemistry and its branches, focusing on essential topics such as the storage of chemicals and laboratory safety. You will delve into the scientific method of inquiry, as well as key historical developments in atomic theory including Bohr’s model of the atom, Dalton’s atomic theory, J.J Thomson’s cathode ray tube experiment, and the Rutherford model of the atom.

Do you wonder how an X-Ray machine works? Well, you are in luck, in this section you will study the rules for filling electrons in orbitals, as well as explore the concepts of radioactivity and properties of radioactive radiations.

At the end of this section, you should be able to:

1. Describe chemical processes around us and their applications in everyday life.

2. Discuss and explain safety rules and hazard symbols in the laboratory.

3. Explain why chemicals should be stored by compatibility and not alphabetically in the laboratory.

4. Investigate the scientific method of inquiry.

5. Identify the main postulates of Dalton’s atomic theory and explain the weaknesses of the theory.

6. Describe the cathode ray experiment and alpha particle scattering experiment and identify the weaknesses of J. J. Thompson and Rutherford’s models of the atom.

7. State the main postulates of Bohr’s planetary theory and explain the importance of the quantum numbers to the electron structure of the atom.

8. Apply Aufbau’s principle, Pauli’s exclusion principle and Hund’s rule of maximum multiplicity to write the electron configuration of the first thirty elements of the periodic table.

9. Describe radioactivity, and the properties of radiations and compare isotopes based on their stability as well as their applications in everyday life.

Key Ideas

• A chemical reaction is a process where substances change into new substances.

• Fermentation is a process where microorganisms convert sugars into alcohol or acids.

• Photosynthesis is the process by which green plants convert light energy (sunlight) into chemical energy.

• Respiration is the process by which cells convert glucose into energy.

• Combustion is a reaction involving the burning of fuel to release energy.

• Personal Protective Equipment (PPE) is a gear worn to protect oneself from hazards.

• A chemical is a substance that consists of atoms or molecules with specific properties and characteristics.

• A fire blanket is a safety device designed to extinguish incipient (starting) fires.

• A fire extinguisher is a handheld active fire protection device usually filled with a dry or wet chemical used to extinguish or control small fires.

• A hypothesis is a testable explanation or guesswork designed to guide experimentation and checking information.

• A scientific theory is a well-established explanation for experimental data.

• Scientific method is a way of learning that emphasises observation and experimentation.

• Scientific law A relationship between physical observables, often represented by a mathematical formula, tested and developed with numerous and diverse experimental observations.

• Alpha particle is a essentially a helium nucleus.

• Gold foil experiment is an experiment where alpha particles were directed at a thin sheet of gold foil to study atomic structure.

• Deflection is the change in direction of a particle.

• Scattering is the process by which particles are deflected or spread out in different directions after colliding with another particle or barrier.

• Plum pudding model is an early model of atomic structure proposed by J.J. Thomson.

• Nuclear model is the atomic model proposed by Rutherford.

• Atomic spectra are series of coloured bands which are formed when white light passes through a prism.

• Continuous spectrum is a spectrum that has no breaks or gaps between the wavelength range.

• Line spectrum is a spectrum that has discrete lines that can be categorised as excited atoms.

• Radioactivity is the emission of energy from an unstable nucleus, either in terms of particles or electromagnetic radiation.

• Radioisotopes are unstable elements.

• Half-life is the time taken for a radioisotope to reduce the number of unstable nuclei to half of the original value OR the time taken for the

Activity of a radioactive substance to reduce to half of the original value.

• Isotopes are atoms with the same number of protons and different numbers of neutrons, e.g. a carbon atom will have 6 protons but can have 6, 7 or 8 neutrons.

• Unstable atom is one that has too much energy or an imbalance of protons and neutrons in its nucleus.

Chemistry is a scientific discipline that focuses on the study of matter, its composition, structure, and properties as well as the principles governing its behaviour. It intersects with fields like physics, biology, environmental science, and engineering and is crucial in understanding and explaining the natural world.

Chemistry plays a vital role in developing new technologies, materials, and drugs for various applications.

Activity 1.1: The chemistry of burning wool Steps:

1. Use the link below to watch the video on burning of stell wool:

https://thewonderofscience.com/phenomenon/2018/7/8/burning-steel- wool

2. Now! Use the questions below to explore what role you think chemistry may play in explaining burning.

a. What was the mass of the steel wool before burning?

b. While the wool was burning, did you see any sparks? If so, why do you think the were sparks (Hint: consider the components of combustion)

c. Did you notice a change in mass when the steel wool stopped burning? If you did notice a change, why do you think this is so.

Activity 1.2: The chemistry of photosynthesis Materials needed: 2 green plants (potted) Steps:

1. Label your plants A and B.

2. Place plant A in a dark room and plant B in sunlight.

3. Observe any changes to plants A and B after three (3) days.

4. Record your observations and discuss them with your colleagues.

Note: Plants convert sunlight, water, and carbon dioxide into oxygen and glucose.

Activity 1.3: The chemistry of an acid-base reaction Materials needed: a beaker, spatula, 250 cm³of vinegar and baking soda reacting.

Note: For access to all apparatus, this activity should be done in the laboratory.

Steps:

1. Using a measuring cylinder, measure about 50 cm³of vinegar into a beaker.

2. Add 2-3 spatula full of baking soda to the vinegar solution in the beaker.

3. Observe any changes in the reactions.

Fun fact: Did you know that the solution of vinegar and baking soda is used in homes to clean stains?!

Note: Mixing acids like vinegar with bases like baking soda produces salt, water, and carbon dioxide gas.

Discussion:

Chemistry is the study of the substances that make up the world around us and how they interact, transform, and affect our lives. By examining processes like photosynthesis, chemical reactions (for example combustion observed in

activity 1.1), and rusting we can see chemistry’s fundamental role in both natural phenomena and practical applications, emphasising its importance in science and everyday life.

Activity 1.4: Distinguish among the traditional branches of chemistry Materials needed: Internet access for research or textbooks Steps:

1. Use the internet to find differences between the branches of chemistry.

https://www.slideshare.net/slideshow/different-branches-of- chemistry/263492854

2. Record and discuss your observations with your friends. Let the following pointers guide your discussion.

a. What are some of the branches you read about?

b. Were there any similarities and or differences between any of the branches?

1. Pure Chemistry

Pure Chemistry is the study of basic principles and theories of chemistry without considering practical use or application. It involves exploring the properties, structure, and behaviour of matter at a molecular and atomic level, analysing the interactions and transformation of substances, understanding the behaviour of atoms and molecules, discovering new compounds, and improving technologies.

Branches of Pure Chemistry

The main branches of pure chemistry are as follows.

a. Physical Chemistry is a branch of chemistry that combines principles from physics and chemistry to study the relationship between the physical properties of matter and its chemical composition and behaviour.

b. Organic Chemistry is the branch of chemistry that studies carbon-based molecules and their properties, composition, and reactions.

c. Inorganic chemistry is the branch of chemistry that studies non-carbon- based compounds and their properties, composition, and reactions. It includes the study of the properties of elements, their compounds and their behaviours in different conditions.

2. Applied Chemistry

Applied Chemistry is the branch of chemistry that studies the practical applications of chemical knowledge in various fields. It focuses on applying chemistry and its principles to solve real-world problems using scientific methods. It has diverse applications in food science, medicine, pharmaceuticals, material sciences, agriculture and environmental science.

Figure 1.1: Chemistry is central to understanding your world.

Chemistry is a central scientific discipline that plays a key role in various aspects of our daily lives, from health and well-being to the environment around us. It is often called the “central science” because it is connected to other disciplines such as physics and biology. It is critical in the development of new materials in various industries, such as electronics, textiles, and construction.

Chemistry is a crucial discipline that provides a fundamental understanding of our world. Its applications are vast and include technology, medicine, industry, and environmental management, making it central to scientific progress and human development.

Chemistry thus has close relationships with various other subjects, including physics, biology, and environmental science, due to the fact that it overlaps with them in terms of content and techniques.

Activity 1.5: A flowchart to demonstrate how chemistry affects daily lives Materials needed: Poster paper, markers, or coloured pencils, internet access, chemistry textbooks Steps:

1. Find out different ways in which chemistry affects the following aspects of life:

• Food and cooking

• Health and medicine

• Cleaning and hygiene

• Environment and sustainability

• Technology and industry

2. Create the flowchart:

• In the centre of the poster paper write “Chemistry.”

• Draw branches from the central idea to the main categories identified.

• Use markers or coloured pencils to draw arrows from “Chemistry” to the various categories.

Chemistry has an enormous impact on daily life, as it is essential for various aspects of modern life. Here are some ways in which chemistry affects daily life.

a. Food and nutrition: Chemistry has a significant impact on food and nutrition by improving food quality, safety, and preservation (e.g. Treatment of water at Kpong and Weija; Standardisation of products at Ghana Standards Authority). It helps us understand the composition of different foods, develops various food processing techniques, and uses chemicals as food additives to improve taste and prevent spoilage. It also provides tools and techniques for analysing food components, contaminants, and nutrients, contributing to research aimed at improving health and disease prevention.

b. Agriculture: Chemistry is crucial in agriculture to maximise crop yield and quality while minimising costs and environmental impact. It impacts agriculture through the development of animal feed (for example, Koudijs Gh Ltd.), fertilisers (for example, Glofert- fertiliser company in Tema, Ghana), chemical pesticides to control pests, understanding soil chemistry, genetic modifications (Ghana Atomic Energy Commission), and water management with chemicals. Chemistry has revolutionised agriculture, providing valuable insights, technologies, and solutions to enhance crop yields, control pests and diseases and improve soil and water quality.

c. Medicine: Chemistry has a significant impact on medicine as it contributes to the development of drugs (for example, Tobinco Pharmaceutical Ltd and Ernest Chemist.) and medical devices (for example, Intravenous Infusions PLC, Koforidua, Ghana), their production, and analysis. Chemistry plays a role in discovering new compounds and synthesising them to optimise their therapeutic use.

d. Transportation: Chemistry greatly affects transportation in various ways, which include fuel production (for example, Tema Oil Refinery, TOR, Ghana), vehicle material designs, lubricants and additives (for example, Ghana Oil, Goil), emissions control, and battery technologies. These chemical advancements enhance fuel efficiency, decrease emissions, and improve the transition to eco-friendly transportation methods.

e. Energy: Chemistry affects energy through its involvement in the production of traditional and renewable energy, energy storage solutions, the development of energy-efficient technologies, and technologies that reduce emissions from energy production. Through chemical principles, researchers can identify solutions that promote more sustainable and environmentally friendly energy production and consumption. Chemistry plays a crucial role in the production of traditional energy sources such as coal, oil (TOR), and natural gas through processes such as extraction, refining and combustion.

Chemistry is also involved in the production of renewable energy sources such as solar panels (for example, Global Engineering and Drilling Ghana Ltd., East Legon) and wind turbines through the development of new materials and processes. Battery technology relies on electrochemistry.

Use the link below to observe the impacts of chemistry in daily life.

https://www.youtube.com/watch?v=L2Q2q20KaEk

There are many career opportunities in the field of chemistry and chemistry- related fields. Below are just a few examples:

1. Pharmacist

2. Medical doctor

3. Biochemist

4. Chemical engineer

5. Chemistry teacher

6. Nurse

7. Laboratory technician

Activity 1.6: Investigating careers available in chemistry and related fields Material needed: Internet access, textbooks, any available learning resources on chemistry Steps:

1. Research: Research into the following areas of chemistry and related fields:

Pharmaceuticals and medicine, environmental science, industrial chemistry, forensic science, academic and research institutions, chemical engineering, food and agriculture

2. Activity sheet - Create an activity sheet with answers to the following guidelines:

a. What are the main responsibilities and tasks associated with careers in each of these fields?

b. What are the educational requirements and skills needed?

c. What are some specific job titles within this field?

d. What impact do these careers have on society and the environment?

Education and Training Required for Careers in

Chemistry The education and training required for careers in chemistry and related fields vary depending on the specific job and employer.

Some jobs in chemistry or chemistry-related fields require a minimum of a bachelor’s degree, while some specialised positions may require an advanced degree. Employers often require laboratory or research experience, relevant work experience and problem-solving skills. Some routine laboratory jobs may not require a degree but would need school qualifications in chemistry.

Activity 1.7: Educational pathways and training required for various careers in chemistry and related fields Materials needed: Internet access for research Steps:

1. In small groups, find out the educational pathways and training required for the following careers in Chemistry:

• Pharmacist

• Chemical Engineer

• Environmental Scientist

• Forensic Scientist

• Toxicologist

• Biochemist

• Food Scientist

• Materials Scientist

2. Gather information on the following points:

• Required high school subjects and skills

• Necessary degrees (e.g., Bachelor’s, Master’s, PhD)

• Specialised training or certifications

• Internship or work experience opportunities

• Continuing education and professional development

3. Discuss your findings and prepare a brief presentation.

4. Compare the educational pathways and discuss any similarities or differences.

Activity 1.8: Exploring the importance of chemistry to the Ghanaian society.

1. Work with a friend or in groups.

2. Find out on the internet, textbooks, resource persons, and chemistry- related outfits about the importance of chemistry to Ghanaian society in the field of agriculture and present your findings using PowerPoint or flipcharts.

3. Record your findings in a PowerPoint format and present them to the class.

The school chemistry laboratory must be a safe place for effective learning. Given this, the following rules and regulations must be strictly observed.

1. Do not eat or drink anything in the laboratory.

2. Never taste chemicals in the laboratory.

3. Any water spilt on the floor must be wiped off immediately.

4. Do not add water to acid but rather acid to water.

5. Never walk barefoot in the laboratory.

6. Keep the laboratory clean and organised.

7. Report all accidents and spills immediately.

8. Follow proper handling and disposal procedures for chemicals.

9. Follow instructions for conducting experiments and using equipment.

10. Wear appropriate protective equipment such as a lab coat, safety goggles and gloves.

11. Know the location and use of emergency equipment such as fire extinguishers, eye wash stations, and safety showers.

12. Be aware of the potential hazards in the laboratory and take precautions to prevent accidents.

Chemical hazards are solids, liquids, gases, and solutions can pose potential hazards and dangers.

Depending on the chemical, the dose, the exposure route, and the duration of exposure, these hazards can have various effects on human health and the environment.

Some common chemical hazards include:

1. Explosives These are chemicals that can rapidly release energy in the form of heat, light, gas and sound, causing physical damage and injury. Examples are dynamite, nitro- glycerine, ammonium nitrate and nitrocellulose.

2. Flammable Liquids and Gases

These are chemicals that can ignite (catch fire) or explode when exposed to heat, sparks, or flames, causing burns, fires, and explosions.

Examples are gasoline, propane, butane, ethanol, diesel, acetone, paint thinners, aerosol sprays, lubricating oils, cooking oils, and fats.

3. Corrosive Substances

These are chemicals that can destroy or damage materials such as metals, plastics, or human tissue, causing severe burns and tissue damage.

Examples are hydrochloric acid, sulphuric acid, nitric acid, sodium hydroxide, potassium hydroxide, bleach and ammonia solution.

4. Toxic Substances

These are chemicals that can harm or kill living organisms such as humans, animals, or plants by interfering with biological functions or disrupting vital organ systems.

Examples are arsenic, lead, mercury, carbon monoxide, pesticides, cyanide, benzene, chlorine gas and ammonia.

5. Oxidising Substances

These are chemicals that can accelerate and promote combustion (burning) in other materials by providing oxygen or other oxidising agents. They cause severe burns, respiratory damage, and explosions.

Examples are hydrogen peroxide, potassium permanganate, oxygen gas, chlorine gas, bleach, nitric acid, and potassium nitrate.

6. Radioactive Substances

These materials spontaneously emit radiation as a result of the decay of their atomic nuclei. Proper handling and disposal are essential, as they can pose significant hazards due to their potential for radiation exposure and contamination.

Examples of radioactive substances are uranium, radon, iodine-131 and cobalt-60.

7. Irritant Substances

These are materials that can cause irritation or inflammation when they come into contact with the skin, eyes, respiratory system, or other organs.

Irritant substances can have a range of adverse effects on humans, such as itching, pain, redness, swelling, and blistering of the skin.

Examples are ammonia, bleach, hydrochloric acid, detergents, insecticides, sodium hydroxide and gasoline.

8. Harmful Substances

These are materials that can pose a risk to the health and safety of humans or the environment. They can cause acute or chronic effects on exposure, depending on the dose, duration and mode of exposure.

Examples are lead, asbestos, pesticides, carbon monoxide, tobacco smoke, mercury, and arsenic.

9. Biohazard Substances

These are materials that can pose a threat to the health and safety of living organisms, including humans, plants, and animals. They may contain living and non-living biological agents that can cause harm, such as bacteria, viruses, toxins and biological waste.

Examples of biohazard substances include blood, bodily fluids, tissues, organs, and microorganisms.

Hazard symbols are visual signs or markings used to indicate the potential danger or risks associated with a particular substance or product. These pictograms are usually displayed on containers or packing to help users identify and handle hazardous materials safely.

Table 1.1: Hazard Symbols

Explanation Symbols

1 Harmful symbol: The harmful symbol is used to indicate that a substance is harmful if ingested, inhaled, or absorbed through the skin. The harmful symbol as shown.

Figure 1.5: Harmful Symbol.

2 Irritant symbol: The irritant symbol is used to indicate that a substance may irritate the skin, eyes or respiratory system. The irritant symbol as shown.

Figure 1.6: Irritant Symbol

3 Corrosive symbol: The corrosive symbol is used to indicate that a substance is capable of causing irreversible damage to living tissues or corroding materials including metals, plastics, and other substances. The corrosive symbol as shown. Figure 1.7: Corrosive Symbol.

4 Toxic symbol: The toxic symbol is used to indicate that a substance is highly poisonous and can cause serious harm to human health or the environment.

The toxic symbol shown.

Figure 1.8: Toxic Symbol

Explanation Symbols

5 Oxidising symbol: The oxidising symbol is used to indicate that a substance is capable of promoting the combustion (burning) or ignition of other materials. The oxidising symbol as shown.

Figure 1.9: Oxidising Symbol

6 Flammable symbol: The flammable symbol is used to indicate that a substance is combustible and can catch fi re easily. The symbol for flammable as shown Figure 1.10: Flammable Symbol.

7 Explosive symbol: The explosive symbol is a sign that warns people about the presence of explosives or other hazardous substances. The explosive symbol as shown. Figure 1.11: Explosive Symbol.

8 Radioactive symbol: The radioactive symbol, also known as radiation hazard symbol, is a warning symbol that is used to indicate the presence of radioactive materials or areas that emit radiation. The radiation hazard symbol is shown below. Figure 1.12: Radioactive Symbol 9 Biohazard symbol: The biohazard symbol is used to indicate the presence of biological hazards such as infectious agents, toxins, and other biohazardous materials that can cause harm to human health or the environment. The biohazard symbol is shown below.

Figure 1.13: biohazard Symbol

Prohibition signs are warning signs that indicate that certain activities, actions, or objects are not allowed (prohibited) in a particular area.

Table 1.2: Prohibition Signs

Explanation Symbol

1 No naked flame: It is a prohibition sign that indicates that open flames and any activity involving unprotected flames are restricted or prohibited in a certain area. The sign is intended to improve safety, guard against fire hazards, and ensure adherence to applicable legislation.

Common locations include industrial settings, laboratories, fuel storage areas, construction sites, and places with flammable materials. A ‘no naked flame’ sign as shown.

Figure 1.2: No Naked

Flame Symbol.

2 Danger: ‘Danger’ is a term used to describe a specific situation, activity, or condition that poses a significant risk of harm, injury or damage to individuals, property or the environment. Any situation that has the potential to cause harm, injury or damage can be considered as dangerous, and it is important to respond to such situations promptly to prevent harm. A ‘danger’ sign as shown.

Figure 1.3: Danger

Symbol.

3 No smoking: ‘No smoking’ is a common sign that is seen in public places, workplaces, and other areas where smoking is prohibited. A ‘no smoking’ sign as shown.

Figure 1.4: No

smoking Symbol.

First aid signs serve as visual markers intended to designate the whereabouts of first aid facilities, equipment, or stations within a given area. Their essential function lies in enhancing workplace safety and furnishing explicit directions to individuals during emergency situations. These signs commonly employ universally acknowledged symbols, colours, and text to communicate details regarding the accessibility and position of first aid resources.

Table 1.3: First Aid Signs

Explanation Symbols

1 First aid: First aid is the immediate assistance provided to a person who has been injured or has suddenly taken ill. It involves a series of simple, life-saving techniques and procedures that can be performed by anyone with basic training. The primary objective of first aid is to preserve life, prevent the condition from worsening, and promote recovery while waiting for professional medical attention. A first aid sign is shown.

Figure 1.14: First aid Symbol 2 Safety Shower: This sign is a visual cue that shows where first aid supplies and safety showers are located in a building.

This sign is usually located in areas where there is a possibility of exposure to hazardous compounds that may require quick decontamination or first aid, is essential to emergency response. This sign is placed strategically in places where there is a greater chance of coming into contact with potentially harmful products, such as industrial facilities, laboratories, chemical storage areas, and other workspaces with potentially harmful materials.

A safety shower sign is shown.

Figure 1.15: 1. Safety

Shower Symbol

Explanation Symbols

3 Eye Wash: This sign is a visual indicator designed to identify the location of emergency eye wash stations. These stations are crucial in environments where there is a risk of exposure to hazardous substances that can cause eye injuries or irritation. The sign helps individuals quickly locate the nearest eye wash station, promoting prompt action in case of an emergency. It is strategically placed in areas where there is a risk of eye exposure to chemicals, dust, or other hazardous materials.

Common locations include laboratories, manufacturing facilities, chemical storage areas, and places where workers handle potentially harmful substances. An eye wash sign is shown.

Figure 1.16: Eye Wash

Symbol.

Personal protective equipment (PPE) is any equipment or clothing worn by individuals to protect against the specific hazards in the workplace or other environments. PPE is designed to protect the wearer from potential hazards that could cause injury, illness, or death. Some of the common types of PPE include:

1. Respirators/Gas masks: These are devices designed to reduce the inhalation of hazardous substances such as dust, fumes, and gases.

2. Hand gloves: These are specialised hard-worn protective coverings that protect the skin against harmful substances or injuries.

3. Eye protectors: These devices include safety glasses, chemical goggles, or face shields that protect the eyes from flying particles, dust, or splashes of hazardous substances.

4. Protective clothing: This includes specialised clothing such as lab coats, aprons, and full-body suits that protect against chemicals, heat, and other hazardous materials.

Example of PPEs

Safety earmuff Safety helmet Respiratory mask Hair net Dust mask Safety goggle Safety boots Disposable Overall Hand gloves Lab Coat Apron

Safety equipment refers to devices or clothing that are designed to protect individuals from injuries or hazards while performing activities or tasks. This includes:

1. Eye shower station: An eye shower station is a safety device found in workplaces where hazardous exposure is possible. It provides immediate treatment to individuals who have contact with hazardous materials or chemicals in the eye. The device consists of a basin attached to a water supply, and an injured person is instructed to flush the eyes with water directed at the eyes and not the face.

2. Fume chamber: A fume chamber is an enclosed space or hood designed to contain and capture hazardous fumes, dust, or vapours that may be produced during laboratory experiments or industrial processes. It is often used to protect workers from dangerous substances that may be emitted during experiments, testing, or production and to prevent contamination of the external environment.

Activity 1.9: Various practices in a chemistry laboratory.

Material needed: Internet access Look at the images and discuss the various practices in a chemistry laboratory

a. Describe what you observed in each image.

b. Mention some of the good practices and wrong practices in the images.

c. Identify some of the personal protective equipment in the images.

d. Identify some of the laboratory glassware in the images.

e. Identify any potential hazards in a laboratory environment as seen in any of the images.

Activity 1.10: ‘Dos’ and ‘Don’ts’ in the Chemistry Laboratory Materials Needed: Short clips demonstrating proper and improper lab behaviour.

Use the links below to watch a video on the ‘Dos’ and ‘Don’ts in the lab.

https://www.youtube.com/watch?v=saXFQR86ziM https://youtu.be/MEIXRLcC6RA After watching the videos:

a. Discuss and make a list of ‘dos’ and ‘don’ts’ in the chemistry laboratory.

b. Discuss the general rules and regulations in the chemistry laboratory with a colleague. Note down the rules and regulations and let your teacher have a look at it.

Activity 1.11: Examining the assay of chemical containers or reagents, electrical gadgets, and other materials and identifying the hazard symbols on them.

Materials needed: A variety of empty chemical containers, reagent bottles, and electrical gadgets with hazard symbols printed on them; a printed sheet of common hazard symbols and their meanings.

Steps:

1. In a small group carefully observe and identify the hazard symbols on the chemical containers or reagents, and electrical gadgets. (If real items are not available, use pictures or printed images.)

2. Take note of the hazard symbols identified (e.g., flammable, corrosive, toxic)

3. Refer to the printed sheet to determine the meaning of each symbol.

4. State the precautions that should be taken when handling the item.

5. State any personal protective equipment (PPE) that might be required.

6. Present your findings to the class, explaining the symbols identified and the associated hazards and safety precautions.

Activity 1.12: The essential safety rules and regulations in the chemistry laboratory Materials needed: A printed list of common laboratory rules and regulations Steps:

1. Discuss the provided rules and regulations.

2. Analyse why each rule is important and what potential risks it mitigates.

3. Give examples or scenarios where each rule would be applicable.

4. Create a poster that highlights key laboratory rules and regulations.

5. Present your poster to the class.

Activity 1.13: Various prohibition signs related to laboratory safety Materials needed: Printed examples of prohibition signs (e.g., First Aid, Danger, No Smoking, High Voltage, etc.)

Procedure:

1. Printed examples of various prohibition signs is assigned to the group.

2. Focus on specific prohibition signs (e.g., First Aid, Danger, No Smoking, High Voltage).

3. Discuss the prohibition signs related to their assigned heading.

4. Analyse the meaning of each sign and discuss why it is important in a laboratory setting.

5. Give examples or scenarios where each sign would be applicable.

6. Create a poster that highlights the prohibition signs under your assigned heading. Use drawings, symbols, and brief descriptions to make your poster informative and engaging.

7. Present your poster to the class.

Activity 1.14: Handling hazardous chemicals safely using personal protective equipment (PPE) and safety equipment.

Materials needed: Printed sheets with different types of PPE and safety equipment.

Steps:

1. Think about your experiences or knowledge of handling hazardous chemicals safely.

2. Write down your thoughts on how to use specific PPE (chemical goggles, hand gloves, aprons/laboratory coats, and respirators/gas masks) and safety equipment (eye shower station, fume hood).

3. In pairs share your ideas and discuss the following points:

4. The importance of each type of PPE and safety equipment.

5. Specific scenarios where each item would be necessary.

6. Each pair to share your ideas and discussion points with the class.

7. In a small group, discuss at least five laboratory rules and at least five hazard symbols.

8. Present your findings to the class.

Chemicals in the laboratory should be stored safely and organised to prevent accidents and ensure the safety of laboratory workers. It is better to store these chemicals by compatibility rather than alphabetically.

Here are some reasons why chemicals should be stored by compatibility and not alphabetically:

1. Chemicals should be stored by compatibility in the laboratory because some chemicals can react explosively or dangerously when they come into contact with other chemicals.

2. Storing chemicals alphabetically can result in incompatible substances being stored next to each other. This can cause chemical reactions that can result in fires, explosions, toxic fumes, and other hazardous situations.

3. Storing chemicals by compatibility reduces the risk of accidents and ensures the safety of laboratory workers.

4. Storing chemicals by compatibility also helps organise their storage, making it easier to locate specific chemicals when needed.

Therefore, it is important to follow the guidelines for the storage of chemicals by compatibility to prevent harmful incidents or accidents in the laboratory.

The following guidelines can be used to store chemicals in the laboratory.

Chemical Incompatibility Chart

Table 1.4: Chemical Copatibility Chart.

How to put out a Small Fire Using Fire Blanket and Fire Extinguisher If a small fire breaks out in the laboratory, it is important to act quickly and appropriately to minimise any potential damage or injuries.

Here is how to use a fire blanket and fire extinguisher to put out a small fire in a laboratory setting:

Using a Fire Blanket

1. If there is a small fire on or near a person, the person should stop, drop, and roll to smother the flames. If the fire is caused by a flammable liquid or in a pan, turn off the heat source.

2. Pull the blanket out of its bag or storage container.

3. Hold the corner of the blanket and, if possible, cover the fire starting from the base. If the fire is on a person’s clothing, wrap the blanket around him or her to smother the flames.

4. Make sure the edges of the blanket create a seal with the surface around the fire to prevent the spread of flames.

5. Leave the blanket in place until the fire has completely stopped or until help arrives.

Using Fire Extinguisher

1. Before using the fire extinguisher, pull the fire alarm and make sure everyone in the area is aware of the fire.

2. Identify the type of fire extinguisher you are using to ensure it is appropriate for the fire you are dealing with.

3. Hold the fire extinguisher in an upright position and aim it at the base of the flames.

4. Squeeze the handle or trigger to release the extinguishing agent.

5. Sweep the nozzle from side to side while aiming at the base of the flames until the fire is completely extinguished.

6. Stay alert for any re-ignition of the flames and keep the fire extinguisher aimed at the base of the flames until it is safe to put away.

Understanding the type of fire extinguisher, you have and the type of fire you are dealing with is crucial to ensure the right type of action.

Activity 1.15: Principles and practices of chemical storage and compatibility Materials needed: Laboratory coat, gloves, safety goggles, printed checklist of storage practices and safety guidelines, notebook, pen, Fire blanket Fire extinguisher Steps

1. In your groups, or with a partner, take a trip to the chemistry laboratory or chemical store near you.

2. Meet the laboratory technician or staff in charge.

3. Ask the laboratory technician or staff in charge to give you a brief overview of the storage area, highlighting key safety features and practices.

4. In small groups, explore the storage area, observing how chemicals are organised and stored.

5. Use your checklists to note specific storage practices, such as:

a. Are flammable chemicals stored away from heat sources?

b. Are corrosive substances stored in corrosion-resistant containers?

c. Are toxic chemicals clearly labelled and stored securely?

d. Is there proper segregation of incompatible chemicals?

e. Is there adequate ventilation in the storage area?

f. Are safety signs and labels clearly visible?

g. Is safety equipment such as eye wash stations and fume hoods easily accessible?

6. Discuss what you observed with the laboratory technician.

7. Find out why it is essential to store chemicals by compatibility in a laboratory setting.

8. Write a brief summary in your notebook. The write-up should include:

a. The storage practices you observed.

b. Why it is important to separate different types of chemicals?

c. How proper labelling contributes to safety.

d. What you learned about the role of safety equipment in the laboratory?

Activity 1.16: Why it is important to store chemicals based on compatibility rather than alphabetically in the laboratory.

Materials needed: Internet access, access to chemical storage guidelines, chemical compatibility charts, worksheet with questions.

Steps 1.

a. Use the link A below to watch the video on why chemicals should be stored on compatibility not alphabetically:

A: https://youtu.be/6EYxVqLj7NI

b. Click on link B below to watch a video of what happens when chemicals are not stored properly:

B: https://www.youtube.com/watch?v=ZIAyNFLRFuw

2. In small group discuss and answer the following questions:

a. What is compatible or incompatible chemicals?

b. Why might storing chemicals alphabetically be dangerous?

c. Give examples of chemical reactions that could occur if incompatible chemicals are stored together.

d. How do chemical compatibility charts help in organising a laboratory?

3. Present your findings to the class.

Activity 1.17: Putting out a small fire using a fire blanket and a fire extinguisher.

Steps

1. Discuss the following questions in your groups or with a friend:

a. What type of fire is it (paper, electrical, chemical)?

b. Which extinguishing method is appropriate (fire blanket or extinguisher)?

c. What are the steps to use each method correctly?

2. Share your pair’s discussion with the larger group.

Summarise the key points, including:

a. Identifying the type of fire

b. Choosing the appropriate extinguishing method

c. Using a fire blanket to smother the fire

d. Using a fire extinguisher (e.g., PASS method: Pull, Aim, Squeeze, Sweep)

The Scientific method of inquiry is a systematic approach used by scientists to investigate and learn about the natural world around us. The scientific method involves a series of steps that are followed to ensure that scientific investigations are made in a logical, objective, and repeatable manner.

Here are the steps involved in the scientific method:

1. Make Observations: Scientists begin by carefully observing and recording information about a phenomenon or problem they wish to investigate, or gather prior knowledge about a certain topic or concept

2. Formulate a Question: Based on their observations and prior knowledge, scientists create a question or issue they want to investigate.

3. Develop a Hypothesis: A hypothesis is an educated guess about the answer to the question or issue that has been formulated.

4. Conduct Experiments: In this step, scientists design and conduct experiments to test the hypothesis.

5. Collect and analyse Data: Scientists record their observations and collect data from their experiments.

6. Draw Conclusions: Based on the results of their experiments and observations, scientists use logic to draw conclusions about their hypothesis.

7. Communicate Results: Finally, scientists communicate their findings through scientific papers, presentations or other means.

The scientific method allows scientists to avoid bias and to ensure that their results are valid, reliable and replicable. Through this methodical approach, scientists can discover new knowledge, solve problems and explore and understand the natural world.

Activity 1.18: Using the scientific methods of inquiry to solve a problem in the school environment or nearby community.

Materials needed: Materials needed: Notebooks, pens, and pencils, measuring tapes, rulers, or other instruments to measure physical dimensions, sample containers, Lab equipment (e.g., beakers, test tubes, microscopes) for scientific experiments, chemicals (if necessary), graph paper, calculators, Safety gear such as gloves, goggles, and lab coats, poster paper and computer.

Cameras (optional, to capture visual evidence), internet (optional), Steps

1. State the problem: For example, the decline in students’ performance in integrated science.

2. State your observation: Example –

• Data collected on recent test scores and grades showed a decline in students’ performance in integrated science.

• Feedback from parents, administrators and teachers confirmed the decline.

• Most students could not satisfactorily answer chemistry questions in integrated science.

3. State hypothesis: Example – ‘The decline in integrated science performance is due to a lack of understanding of fundamental concepts.

4. Carry out experimentation:

a. Conduct surveys and interviews with students to understand their perception of the difficulty in integrated science.

b. Review teaching methods and curriculum to identify any potential gaps or areas for improvement.

c. Implement interventions, such as additional instruction sessions, interactive workshops, and changes in teaching strategies.

5. Analysis: Analyse data collected from the experiments conducted. This includes:

a. Analysing survey responses to identify common challenges or misconceptions among students.

b. Comparing the performance of students who received interventions with those who did not.

c. Assessing changes in student attitudes and engagement with integrated science after implementing interventions.

6. Conclude: Based on the analysis of data, write your conclusion regarding the effectiveness of the interventions in addressing the decline in integrated science performance.

a. If the interventions were successful, consider implementing them on a larger scale or adjusting based on feedback.

b. If the interventions were not effective, revisit the hypothesis and consider alternative explanation for the decline in performance.

This might involve further experimentation or research.

Activity 1.19: Step by step application of scientific method of enquiry in the localities.

1. Design a poster outlining the method used and share with your class for discussion.

2. Outline at least five steps involved in the scientific method of enquiry.

3. Identify at least a problem in the school environment that can be solved using the scientific method of enquiry.

4. Formulate a hypothesis to drive the investigation for at least one of the problems identified.

5. Design an experiment that can be used to solve the problem(s) identified.

Quite a few scientists contributed to the modern model structure of the atom.

We now look at a few of the most important scientific discoveries that led to the modern atomic theory. Amongst them are Dalton’s Atomic Theory, J.J. Thomson’s Cathode ray experiment, and Rutherford’ alpha scattering experiment.

The idea that elements are made up of atoms is called the atomic theory.

The postulates of the Dalton’s atomic theory are as follows:

a. All elements are made up of small indivisible particles called atoms.

b. Atoms cannot be created or destroyed.

c. Atoms of the same element are identical, that is, they have the same mass and size, but atoms of different elements have different masses and sizes.

d. Atoms of different elements combine in simple whole number ratios to form compounds.

The theory in its broad outline is still valid, however, some of the postulates have been modified in the light of subsequent discoveries.

Modification of the Dalton’s Atomic Theory

1. All elements are made up of small indivisible particles called atoms:

Atoms are not indivisible. This is because it was later discovered that atoms are not the smallest particles and are further broken down into subatomic particles such as protons, neutrons and electrons.

2. Atoms cannot be created or destroyed: This postulate is still acceptable for ordinary chemical reactions. In nuclear reactions, however, atoms of the same element are destroyed, and new ones are created.

3. Atoms of the same element are identical, that is, they have the same mass and size, but atoms of different elements have different mass and size: The discovery of isotopes (atoms of the same element having the same number of protons but different number of neutrons) contradicts this postulate.

4. Atoms of different elements combine in simple whole-number ratios to form compounds: This postulate is still acceptable for inorganic compounds, which usually contain few atoms per molecule. Carbon, however, forms very large organic compounds such as polymers, proteins, and starch, which can contain thousands of atoms. Silicon, which is inorganic, also forms very complex silicates involving a large number of atoms.

Activity 1.20: Review of the atom and its sub-atomic particles

1. What is an atom?

2. Draw a model of the atom.

3. Name the three sub-atomic particles.

4. State where each sub-atomic particle can be located.

5. What is the nucleus, and what does it contain?

Activity 1.21: To discuss the main postulates of Dalton’s atomic theory.

Materials needed: Internet access, printed postulates of Dalton’s atomic theory Steps

1. In small groups, introduce the four main postulates of Dalton’s atomic theory.

2. Discuss the postulates using the following questions as guides:

a. What does each postulate mean?

b. Give an example to illustrate each postulate.

3. Share your understanding of the postulates with the class.

4. Research and explore the limitations and contributions of Dalton’s atomic theory to modern chemistry.

Activity 1.23: Constructing the atomic model.

1. Construct a model to represent the atom as a simple sphere with no internal structure.

2. Draw a diagram of the atom modelled.

3. Display the model and diagram for class discussion.

Activity 1.24: The Dalton’s Atomic theory

1. In your own words, state the main postulates of Dalton’s atomic theory.

2. State at least one strength and one weakness of Dalton’s atomic theory.

3. Explain at least one strength and one weakness of Dalton’s atomic theory.

4. How relevant is Dalton’s atomic theory to the evolution of modern chemistry?

J.J. Thomson’s cathode ray experiment was a series of experiments that laid the foundation for the discovery of the electron.

Thomson passed an electric current through a vacuum tube and observed a stream of negatively charged particles that travelled from the negatively charged electrode, known as the cathode, to the positively charged electrode, known as the anode. These particles were called cathode rays. The cathode ray tube used by Thomson is shown below.

Figure 1.2: A cathode ray tube To further study the cathode rays, Thomson conducted experiments using electrical and magnetic fields. He found that the rays were deflected by both fields, indicating that the particles that made up the ray had a negative charge.

He also measured the charge-to-mass ratio of the particles and found that it was much smaller than any known atom, leading him to conclude that the cathode rays were made of particles smaller than the atoms. These particles were named as electrons.

Thomson’s discovery of the electron was a major breakthrough in the understanding of the structure of matter.

J.J. Thomson’s Model of the Atom

J.J. Thomson’s model of the atom, also known as the ‘plum pudding’ model was developed on the basis of his discovery of the electron. According to this model, the atom is composed of a positively charged sphere, like a pudding, in which negatively charged electrons are embedded, like plums.

The diagram below shows Thomson’s model of the atom.

Figure 1.3: Plum Pudding Model of the Atom.

While this model was a significant step in the understanding of atomic structure, it had some weaknesses.

Some main weaknesses of J.J. Thomson’s model of the atom are:

1. The model assumed that the positive and negative charges were spread evenly throughout the atom, which would produce a neutral electric charge.

However, the model failed to explain the presence of a nucleus in atoms.

2. The model also did not provide any information about the number or arrangement of the electrons in an atom. Thomson’s model just suggested that the electrons were dispersed throughout the atom but not in a pattern or orbit.

3. The model did not explain the atom’s overall mass. The electrons are much lighter than the protons and neutrons that make up the nucleus, and it was not clear how the atom’s overall mass was distributed.

4. Rutherford’s discovery of the atomic nucleus disproved Thomson’s model by showing that it was inaccurate in describing the atomic structure.

Despite the limitations, Thomson’s model was significant in showing that atoms are not indivisible but could be further broken down into constituent particles.

It also led to the development of further models for the structure of the atom, including Rutherford’s model, which corrected some of the shortcomings of Thomson’s model.

Rutherford’s alpha scattering experiment was a landmark experiment aimed to investigate the structure of the atom and the nature of its constituent particles.

The experiment involved firing positively charged alpha particles at a thin gold foil and observing their trajectory as they passed through the foil. The expectation was that the alpha particles would pass straight through the foil or be slightly deflected by the atomic structure of the atoms within the foil. However, some alpha particles were scattered at very large angles, and some even scattered backwards.

The diagram below shows Rutherford’s alpha scattering experiment.

Figure 1.4: Rutherford’s alpha scattering experiment.

Rutherford analysed the results of the experiment and proposed a new atomic model that showed that the atom has a small, dense, positively charged nucleus at its centre, surrounded by negatively charged electrons. He concluded that the deflected alpha particles were deflected by a strongly charged nucleus, while others passed straight through the atom’s empty space.

This experiment provided evidence for the existence of the atomic nucleus and paved the way for further discoveries about the structure and behaviour of atoms.

Rutherford’s Model of the Atom

Rutherford’s atomic model, also known as the nuclear model, was proposed based on his famous Alpha particle scattering experiment. The model introduced the concept of a small, positively charged. nucleus in the centre of the atom, surrounded by negatively charged electrons. The electrons would orbit the nucleus in specific energy levels and paths, similar to the planets orbiting the sun.

Rutherford’s model of the atom is shown below.

Figure 1.5: Rutherford’s model of the atom.

Despite its contribution to the understanding of the structure of the atom, Rutherford’s atomic model had several weaknesses, which are as follows:

1. The model is unable to explain the stability of atoms. In the model, negatively charged electrons move around the positively charged nucleus and should eventually lose energy and spiral into the nucleus, causing the atom to collapse. However, this does not happen in reality, and the model failed to explain why.

2. The model cannot account for the high energy emission spectra of atoms.

According to the model, electrons must travel in specific paths and can only transition between certain energy levels, which would result in a limited spectrum of radiation. But experimental observations showed that the atoms emitted a much larger range of radiation than predicted by the model.

3. Rutherford’s atomic model could not explain the existence of isotopes, atoms of elements with the same atomic number but different mass numbers. The model proposes that the number of electrons in an element is equal to its atomic number, which would also determine the number of protons in the element. However, isotopes of the same element have a different number of neutrons even though they have the same number of protons.

Despite these limitations, Rutherford’s atomic model was a fundamental stepping stone towards the modern understanding of atomic structure and formed the basis of further models, including Bohr’s atomic model, which built upon Rutherford’s concept of a nucleus and attempted to address some of the model’s weaknesses.

The Structure of the Atom

Based on the results of J.J. Thomson’s cathode ray and Rutherford’s alpha scattering experiments, the following structure of the atom was proposed:

The atom consists of a positively charged nucleus, which contains most of the mass of the atom. The protons are positively charged. The electrons, which have a negative charge, are located outside the nucleus.

The number of electrons in an atom is equal to the number of protons, giving an atom a neutral overall charge. The electrons are held in the shells by the electrostatic attraction to the positively charged nucleus.

The table below shows the location, charges and relative masses (amu = atomic mass units) of the subatomic particles of the atom.

Table 1.5: Subatomic Particles

Subatomic Particles

Particle Charge Relative Mass (AMU) Location

Proton Positive 1 Proton

Neutron Neutral 1 Neutron

Electron Negative 1/1840 Outside Nucleus

Activity 1.25: Investigating the properties of cathode rays through interactive simulations.

Materials needed: A computer or tablet with internet access, simulation videos or interactive charts demonstrating cathode ray properties.

Steps:

1. Use the link below to watch the video to explore the properties of cathode rays.

https://www.youtube.com/watch?v=vXOeehVTcRA

2. Observe and discuss the direction of deflection relative to the orientation of the magnetic field.

3. Observe and discuss the direction of deflection relative to the orientation of the electric field.

4. Discuss the concept of fluorescence when cathode rays strike certain materials, causing them to emit light or leave a visible trace.

5. Share your findings with a colleague.

Activity 1.26: Findings of Rutherford’s alpha scattering experiment Materials needed: Access to internet and computer or tablets, video of Rutherford’s alpha scattering experiment Steps

1. Watch the video (use the link below) that demonstrates Rutherford’s alpha scattering experiment.

https://www.youtube.com/watch?v=XBqHkraf8iE

2. Take notes on key aspects such as:

a. How alpha particles were emitted towards a thin gold foil.

b. What Rutherford expected to happen to the alpha particles.

c. What actually happened to the alpha particles as observed on the detector screen.

3. Describe what you observed in the video.

Activity 1.27: Description of the model of the atom by J. J Thompson and Rutherford experiments.

Describe the structure of the atom based on analysis of the evidence gathered from both experiments.

Activity 1.28: Constructing J.J Thompson and Rutherford atomic models.

1. Construct a model to represent the atom using the evidence gathered from of JJ Thomson and Rutherford experiments.

2. Draw a diagram of the modelled atom.

This lesson looks at the theories that explain how electrons behave in the atom and the effects their behaviours have on the physical and chemical nature of the atom.

Bohr’s planetary theory explained the structure of an atom and the behaviour of electrons.

According to this theory, electrons orbit the nucleus in a fixed, circular orbits at specific energy levels, similar to the planets in the solar system. These energy levels were quantised. This means the electrons could only occupy specific energy states and could only transition between those levels by either absorbing or emitting a discrete amount of energy in the form of a photon. This theory also introduced the concept of ground state, where the electron orbits the nucleus at its lowest energy level, excited states, where the electron absorbs energy and jumps to higher energy levels.

Bohr’s theory provided a foundation for modern atomic theory and led to further developments in the understanding of quantum mechanics.

Main Postulates of Bohr’s Planetary Theory

Bohr’s planetary theory of the atom was based on the following postulates:

1. Electrons move around the nucleus of an atom in fixed, circular orbits.

2. The electrons can exist only in certain allowed orbits, which correspond to specific energy levels.

3. While an electron is in a particular energy level, it does not radiate energy.

This energy is only emitted when the electron jumps from one energy level to another.

4. The energy of the emitted radiation corresponds to the difference in energy between the initial and the final energy levels.

5. The size of the orbit and the energy of the electron are related. Electrons in larger orbits have more energy than those in smaller orbits.

6. Electrons can only make transitions between energy levels that correspond to a specific amount of energy, known as a quantum. These transitions produce or absorb photons that have a frequency proportional to the difference in energy.

These postulates provided a framework for understanding the behaviour of electrons within atoms.

The diagrams below show Bohr’s model of the atom.

Figure 1.6: Bohr’s model of the atom Continuous and Line Spectra Continuous spectrum A continuous spectrum is a spectrum of electromagnetic radiation containing photons of all energy levels within a specific range. Unlike atomic spectra, which consists of only specific energy levels, a continuous spectrum contains a radiation of all energies resulting in a smooth display of colours or wavelengths.

Examples of sources of continuous spectra include a hot solid object, such as a light bulb filament, a glowing gas or plasma.

Continuous spectra are important in both astronomy and laboratory experiments, as they can be used to help identify the composition and temperature of objects emitting radiation.

Line spectrum A line spectrum is a spectrum produced by an excited atom or molecule that contains only discrete wavelengths or colours of electromagnetic radiation. These wavelengths correspond to specific energy level transitions within the atom or molecule.

The line spectrum appears as a series of coloured lines or bands rather than a continuous spectrum, which contains radiation at all wavelengths. For example, the line spectrum of hydrogen consists of several discrete lines of colour in the visible spectrum. Each line corresponds to a specific transition between energy levels in the hydrogen atom. Similarly, each chemical element has a unique line spectrum, which can be used to identify the element based on the wavelengths of the lines observed.

Line spectra are important in many areas of science, including astronomy, chemistry and physics. Chemists use line spectra to help identify unknown substances or verify the purity of a sample.

Physicists use line spectra to study the behaviour of electrons within atoms and molecules, providing insights to the nature of matter and energy.

The diagrams below show examples of continuous and line spectra.

Figure 1.6: Diagram showing continuous spectrum and emission spectrum Differences between a continuous spectrum and a line spectrum There are several differences between a continuous spectrum and a line spectrum.

These are summarised in the table below.

Table 1.6: Differences between a continuous spectrum and a line spectrum Definition A continuous spectrum contains radiation of all energies within a certain range, whereas an emission line spectrum contains only specific wavelengths of radiation.

Source A (nearly) continuous spectrum is emitted by a hot, dense object such as a light bulb filament or a star, whereas an emission line spectrum is produced when an excited atom or molecule emits light.

Appearance A continuous spectrum appears as a smooth display of colours or wavelengths while an emission line spectrum appears as a series of discrete lines or bands.

Composition A continuous spectrum contains radiation of all energies, while an emission line spectrum contains only specific energies that correspond to the energy level transitions of the emitting atom or molecule.

Usage Continuous spectra are used to identify the temperature and composition of a source emitting light while emission line spectra are used to identify the chemical composition of a source emitting light.

Examples Examples of sources for continuous spectra include the sun (almost continuous), light bulbs, and blackbody radiators, while emission line spectra are typically emitted by excited atoms or molecules, such as hydrogen, helium or neon.

Relationship of the lines in the emission spectrum of hy- drogen to electron energy levels The lines in the emission spectrum of hydrogen are directly related to the electron energy levels of the hydrogen atom. When an electron in a hydrogen atom is excited, it moves from the ground state (lowest energy level) to a higher energy level. This higher energy state is not stable, and the electron will eventually return to its ground state by releasing energy in the form of a photon of light.

The energy of this photon is directly proportional to the difference between the higher and lower energy levels of the electron. Since each energy level of the electron in a hydrogen atom is fixed, the energy of the released photon is also fixed. It corresponds to specific wavelength or colour of light. Therefore, each emission line in the hydrogen spectrum corresponds to a specific energy level transition for the electron in the hydrogen atom.

The lines in the spectrum represent wavelengths of the photons emitted as the electron transitions back to a lower energy level.

The diagram below shows the relationship between the line spectrum of hydrogen and energy levels.

Figure 1.7: Line Spectrum of Hydrogen and Energy Levels

Quantum theory has made significant contributions toward the development of the atomic structure. Some of these contributions are:

Wave-particle duality: Quantum theory introduced the concept of wave-particle duality, which suggests that particles could exhibit both wave-like and particle- like behaviours. This theory helped scientists to understand the behaviour of electrons in atoms, as electrons exhibit wave-like behaviour in their movement around the nucleus.

Discrete energy levels: Quantum theory introduced the concept of energy levels in atoms. This means that electrons can only exist at certain energy levels around the nucleus and cannot exist anywhere in between.

Uncertainty principle: Quantum theory introduced the uncertainty principle, which states that it is impossible to know both the position and the momentum of an electron at the same time.

Quantum numbers: Quantum theory introduced the concept of quantum numbers, which describe the energy levels and positions of electrons in atoms.

These quantum numbers help predict the properties of atoms and their behaviour during chemical reactions.

Electron spin: Quantum theory also introduced the concept of electron spin, which explains why two electrons in the same electron orbital have opposite spins. This concept helps scientists understand the behaviour of electrons in atoms and their contribution to the magnetic properties of materials. These concepts introduced by quantum theory have helped scientists understand the fundamental behaviour of electrons and predict their properties and behaviour during chemical reactions.

Quantum numbers are integers or half-integers that describe the properties of electrons in an atom. There are four types: principal, momentum/azimuthal, magnetic and spin quantum numbers.

1. Principal quantum number (n): This quantum number determines the energy level of an electron and describes the size of the electron cloud. It can have any positive value starting from 1.

2. Angular Momentum/Azimuthal quantum number (l): This quantum number indicates the shape of the electron cloud or the subshell in which an electron is present. It can have value from 0 to (n-1).

3. Magnetic quantum number (m): This quantum number specifies the orientation of the electron cloud in space. It can have values from (-l) to (+l).

4. Spin quantum number (s): This quantum number describes the spin of an electron, which is a fundamental property of all particles. It can be either (+½) or (-½).

These quantum numbers play a crucial role in determining the electron configuration of an atom and understanding its behaviour. They also help us in predicting the position of electrons in an atom by providing a framework for describing energy states and orbitals.

The Importance of Quantum Numbers to the

Electron Structure of the Atom

Quantum numbers play an important role in understanding the electron structure of an atom. The electron structure refers to the arrangement of electrons in an atom’s different energy levels and subshells.

Here are some reasons why quantum numbers are important in this regard:

1. Describing the electron energy levels: The principal quantum number (n) allows us to determine the energy levels available to electrons in an atom. Each energy level corresponds to an electron shell, and the value of n determines the number of subshells and electrons that can reside in each shell.

2. Specifying the subshells: The angular momentum/azimuthal quantum number (l) provides information about the subshells within each shell. It determines the shape of the subshell. This helps us to predict the electron distribution more accurately.

3. Determining the electron orientation: The magnetic quantum number (m), indicates the orientation of the electron cloud in space. It helps us to understand spatial arrangement of electrons within subshells.

4. Predicting electron spin: The spin quantum number (s), describes the spin of each electron, which is important for understanding the electron configuration of an atom. Two electrons with opposite spins can occupy the same orbital, which has important consequences for the chemical and physical properties of different elements.

ORBITALS An orbital is defined by a set of quantum numbers (n, l, m). Here are a few examples of orbitals:

a. The s orbital is the lowest orbital, with a spherical shape and can hold up to two electrons. The s orbital is shown below.

Figure 1.8: Shape of the S-orbital

b. The p orbitals (px, py and p z ), have a dumb-bell shape and can hold up to six electrons (two in each orbital). These orbitals are found in the second and higher energy levels. The p orbitals are shown below:

The three p orbitals are aligned along perpendicular axes.

Figure 1.9: The three P orbitals aligned along perpendicular axes

c. The d orbitals have more complex shapes and can hold up to ten electrons in total (5 orbitals with 2 electrons in each orbital). These orbitals are found in the third energy level and higher.

Activity 1.29: The Bohr’s planetary atomic model.

1. State Bohr’s planetary model of the atom.

Activity 1.30: Illustrating the Bohr model of the atom.

Materials needed: Poster paper or large sheet of paper or cardboard, markers, coloured pencils, access to textbooks, or internet.

Steps:

1. Research and find out basic information about the Bohr model.

2. Create a large diagram of the Bohr model on paper.

3. Indicate the following in the diagram:

a. The nucleus at the centre (use different colours to draw the nucleus and the electron orbits

b. Multiple electron orbits around the nucleus

c. Electrons in the orbits

d. Labels for the energy levels

Activity 1.31: Understanding Bohr’s theory and the stability of the atom through research and structured writing tasks.

Materials needed: Textbooks covering atomic theory, access to the internet for research, notebook, and pen.

Steps:

1. Use textbook and online articles to gather information about Bohr’s theory. Keywords for online research: “Bohr model”, “Bohr theory of the atom”, “stability of the atom”, “postulates of Bohr’s atomic theory”.

2. Take notes of the key postulates of Bohr’s theory and how it explains the stability of the atom.

3. Write a brief summary of each of Bohr’s postulates. The summary should cover the following postulates:

a. Electrons orbit the nucleus in specific, fixed orbits or energy levels.

b. Each orbit corresponds to a specific energy level.

c. Electrons can move from one orbit to another by absorbing or emitting a quantum of energy.

d. The angular momentum of an electron in an orbit is quantized.

3. Compose explanations of Bohr’s theory and the stability of the atom in your own words.

Learner to address the following points:

a. How Bohr’s model differs from previous atomic models.

b. How the fixed orbits prevent electrons from spiralling into the nucleus.

c. The significance of quantized energy levels.

4. Discuss the key points and common themes found in the summaries and explanations.

Activity 1.32: The difference and similarities between continues and line spectra.

Analyse the differences and similarities between continuous and line spectra and carry out the following in your analysis:

a. delve into the fundamental principles that govern each of the spectra.

b. with the fundamentals principles identified present the differences and similarities between the two spectra in a tabular form.

Activity 1.33: Description of the energy levels in Bohr’s planetary structures of the atom.

1. Carefully study the following diagrams illustrating the planetary structure of the atom.

a. Identify the fixed orbits and energy levels as proposed by Bohr.

Figure 1.20 Figure 1.21

(b) Describe the meaning of the numbers in Fig. 1.21

Activity 1.34: Learning about the s, p, d and f orbitals.

Visit the following websites and given links to learn more about the s, p, d and f orbitals.

• Chemistry Steps

• Wikipedia

• https://byjus.com/chemistry/bohrs-model/

• https://byjus.com/physics/bohr-model-of-the-hydrogen-atom/

• https://www.space.com/bohr-model-atom-structure

Activity 1.35: Using physical models to illustrate the concept of quantum numbers Materials needed: Different coloured balls or beads to represent electrons, rings of different sizes to represent energy levels (orbits), string or wire to suspend the rings at different heights, labels for principal quantum number (n), angular momentum quantum number (l), magnetic quantum number, and spin quantum number.

Steps

1. Arrange the rings or hoops on the large sheet or poster board to represent different energy levels. Suspend them at different heights using string or wire to signify increasing energy levels.

2. Label each ring with the corresponding principal quantum number (n = 1, 2, 3, ...).

3. Use the coloured balls or beads to represent electrons.

4. Place the balls on the different rings to show electrons in various energy levels.

5. Discuss how the principal quantum number (n) increases with distance from the nucleus and how energy increases with n.

6. Introduce the concept of subshells (s, p, d, f) corresponding to the angular momentum quantum number (l).

7. Use different coloured sections on each ring to represent different subshells (e.g., one section for s, three sections for p, five sections for d, and seven sections for f).

8. Place electrons in these sections to show how they occupy different subshells within an energy level.

Activity 1.36: Illustrate the relationship between quantum numbers and electron orbitals.

Materials needed: Quantum number charts, computer or tablet with internet access.

Steps

1. Identify the four types of quantum numbers (principal, angular momentum, magnetic, and spin).

2. Use the link below to watch the video on the relationship between quantum numbers.

https://www.youtube.com/watch?v=eRIN9CPDrpo&t=234s

3. Discuss the relationship between quantum numbers and electron orbitals

Worked Example

Identify the principal quantum number (n) and the angular momentum (l) in following energy levels:

i. 2p

ii. 3d Answer

i. In the above, the 2 in the 2p is the principal quantum number and the angular momentum is 1(that is l = 2 – 1).

ii. 3 in the 3d is the principal quantum number and the angular momentum is 2 (that is l = 3 – 1)

Activity 1.37: properties of s and p in terms of shape, orientation energy level.

Compare and contrast the properties of s and p in terms of shape, orientation energy level. Justify your reasoning with relevant examples.

Worked Example

Identify the principal quantum number (n) and the angular momentum (l) in the following energy levels:

(i) 2p

(ii) 3d

Aufbau’s principle, also known as the building-up principle, is a fundamental principle in chemistry stating that the atomic orbitals are filled with electrons in order of increasing energy. Specifically, electrons will fill lower energy atomic orbitals before moving on to higher energy levels. This principle is used to determine the electron configuration of atoms and the order in which the orbitals are filled. Electrons fill subshells in the following order: 1s, 2s, 2p, 3s, 3p, 4s, 3d.

Pauli’s exclusion principle is a fundamental principle in quantum mechanics that states that no two electrons in an atom can have the same set of quantum numbers.

In other words, if two electrons are in the same orbital, they must have opposite spins. The principle is essential in determining the structure of atoms, as it limits the number of electrons that can occupy each orbital and organises the electron configuration of atoms in the periodic table.

Hund’s rule of maximum multiplicity is a principle in quantum mechanics that states that within a subshell, electrons will occupy orbitals singly, with their spins parallel’ before they pair up with opposite spins.

In other words, if two or more orbitals having the same amount of energy are unoccupied, then electrons will start occupying them singly, before they fill them in pairs.

This means that electron pairing in p and d orbitals cannot occur until each orbital of a given subshell contains one electron or is singly occupied.

This rule is based on the fact that, electrons in orbitals with parallel spins repel each other less than electrons with opposite spins, leading to a lower potential energy for the system. Therefore, when electrons occupy a subshell, they will make the least energetic configuration by occupying orbitals singly before pairing up.

This rule is important in determining the electronic configuration of atoms.

How to Express Electron Configuration Using s, p, d Notation To express the electronic configuration of an atom using s, p, d notation, you first need to identify the principal quantum number of the highest energy level occupied by electrons, which is equal to the period number of the element in the periodic table. Then you assign the electrons to each subshell in the following order:

1. First, the 1s orbital is filled before any other orbital.

2. Next, the 2s orbital is filled before the 2p orbitals start to fill.

3. Then, the 3s orbital is filled before the 3p orbitals start to fill.

4. After that, the 4s orbital is filled before the 3d orbitals begin to fill.

That is, 1s 2s 2p 3s 3p 4s 3d So, for example, the electron configuration of Carbon (C), which has 6 electrons, can be expressed in s, p and d notation as follows: 1s 2 2s 2 2 p 2.

This indicates that the first energy level (n = 1) is filled with two electrons in the 1s orbital, while the second energy level (n = 2) is filled with four electrons in the 2s and 2p orbitals, (with two electrons in the 2s orbital and two electrons in the 2p orbitals).

Electron configuration of ions and elements can be written as seen in the examples below. Writing these configurations usually follows the Aufbau’s principle.

Example

Write the electron configuration for each of the following ions and elements:

i. ₁₃Al ³⁺ii. ₁₆S²−:

iii. ₂₄Cr

iv. ₂₉Cu

iii. ₂₄Cr

iv. ₂₉Cu Answers

i. ₁₃Al (3+)/13Al+3: 1 s²2 s²2 pₓ ²2 p_(y) ²2 p_(z) ²ii. ₁₆S²−: 1 s²2 s²2 pₓ ²2 p_(y) ²2 p_(z) ²3 s²3 pₓ ²3 p_(y) ²3 p_(z) ²iii. ₂₄Cr: 1 s²2s²2p⁶3s²3p⁶3 d⁵4s¹iv. ₂₉Cu: 1 s²2s²2p⁶3s²3p⁶3 d¹⁰4s¹How to Express Electron Configuration Using ‘Electrons-in-Boxes’ Method The electron configuration of an atom can also be expressed using the ‘electrons- in-boxes’ method.

In this method, each orbital is represented as a box, and the electrons are represented by arrows, with the direction indicating their spin.

For example, the electron configuration of carbon (C) with 6 electrons can be represented as follows:

In this representation, the first energy level (n= 1) has only one orbital, the 1s orbital, with electrons represented by a pair of arrows pointing up and down. The second energy level (n=2) contains four orbitals: the 2s orbital, which has two electrons represented by arrows pointing up and down, and the three 2p orbitals, which has two electrons represented by two arrows pointing up.

Differences in Stability between Fully Filled, Half- Filled and Partially Filled Orbitals The stability of an atom depends on the electron configuration of its orbitals. A fully filled or half-filled subshell is more stable than a partially filled subshell.

Activity 1.38: To assess prior knowledge on electron configuration.

Materials needed: Periodic table charts, paper and pen, handouts with basic questions on electron configuration.

Steps

1. In small groups answer basic questions on electron configuration.

Example questions:

a. What is an electron configuration?

b. How are electrons arranged in an atom?

c. What do the terms ‘orbital’ and ‘energy level’ mean?

2. Share your answers with other groups.

3. Use the periodic table to explain the order of filling orbitals (1s, 2s, 2p, 3s, etc.).

Activity 1.39: To understand the rules for writing electron configurations and how to apply these rules to different elements.

Materials needed: Access to textbooks and the internet for research, periodic

table charts, paper and pen, key questions and tasks.

Steps

1. In small groups Investigate on the following principles stated in 1a and b:

a. Aufbau Principle

b. Pauli Exclusion Principle

c. Hund’s Rule

2. Research on the rule stated in 1c. Focus on its significance, and examples of how it is applied in writing electron configurations.

3. After your investigations and research, prepare for a group presentations.

The information in your presentation should answer the following questions:

a. What is the Aufbau principle?

b. Why is it important in writing electron configurations?

c. Examples of electron configurations using the Aufbau principle.

d. What is the Pauli Exclusion Principle?

e. Why is it important in writing electron configurations?

f. Provide examples of electron configurations using the Pauli Exclusion Principle.

g. What is the Hund’s rule?

h. Why is it important in writing electron configurations?

i. Provide examples of electron configurations using the Hund’s rule.

Activity 1.40: Application Activity

1. Write the electron configurations of the following using the researched rule and principles in Activity 1.38:

a. ₁H, ₃Li, ₆C, ₁₁Na, ₁₇Cl

b. ₂₄Cr, ₂₉Cu

c. ₂He, ₁₀Ne, ₁₈Ar

2. Using the ‘electron in box method’ write the electron configuration of the following elements showing how the electrons occupy the orbital.

13Al, ₂₀Ca, ₂₆Fe

Activity 1.41: To help learner understand the stability associated with fully filled, half-filled, and partially filled orbitals in subshells.

Materials needed: Whiteboard and markers, periodic table charts, handouts or slides with examples and key points Steps

1. Write the electron configuration of noble gases (Neon, Argon)

2. Discuss why noble gases are inert and stable due to their fully filled orbitals.

Note: Fully filled orbitals (e.g., s², p⁶, d¹⁰, f¹⁴) are especially stable due to symmetry and maximised electron pairing. Noble gases are inert and stable due to their fully filled orbitals.

3. Write electron configuration of nitrogen and manganese.

4. Discuss why elements with half-filled orbitals tend to have extra stability

Note: half-filled orbitals (e.g., p³, d⁵, f⁷) have a special stability due to exchange energy and symmetry.

5. Write electron configuration of oxygen and iron.

6. Discuss why these configurations are less stable and often more reactive.

Activity 1.42: Filling the orbitals with electrons in box method Visit https://www.youtube.com/watch?v=9ogq50CBgCg to watch videos or observe demonstrations of the process of filling orbitals using the s, p and d notations and electron-in- box method.

Activity 1.43: Application of Aufbau’s principles in writing electron configuration

1. Write the electron configuration of oxygen using s, p and d notation

2. Explain the Aufbau principle and how it applies to writing of electron configurations.

Relative atomic mass is the average mass of an atom of an element, considering all its naturally occurring isotopes, relative to the mass of an atom of carbon-12, which has been assigned a mass of exactly 12.

This value is expressed in atomic mass units (amu). The relative atomic mass is determined by the abundance of each isotope of an element.

Relative molecular mass is the sum of the relative atomic masses of all the atoms in a molecule, relative to the mass of carbon-12, which has been assigned a mass of exactly 12.

The relative molecular mass is calculated by adding up the atomic masses of all the atoms in a molecule, considering the number of atoms in each element present in a molecule.

A mass spectrometer is an instrument used for measuring the mass-to-charge ratio (m/z) of ions in a sample. It operates by generating ions from a sample and then separating the ions based on their mass-to-charge ratio (m/z) using electric and magnetic fields. The separated ions then hit a detector where they create a signal that is proportional to their abundance.

This signal is then analysed to determine the mass-to-charge ratio (m/z) of the ions and their relative abundance, providing information about the composition of the sample.

The mass spectrometer is shown below.

Figure 1.22: Mass spectrometer Mass spectrometers are used in a wide variety of fields, including chemistry, biochemistry, physics, geology and environmental science, for applications such as identifying the presence of specific compounds in a sample, analysing the composition of proteins and other biomolecules, studying the isotopic composition of elements, and monitoring air and water pollution.

Parts of the Mass Spectrometer and how they Work The five main processes which take place in the mass spectrometer are:

1. A sample of the substance is vaporised in the vaporisation chamber.

2. The vapour is ionised into positive ions in the ionisation chamber.

3. The positive ions are attracted and accelerated by an electric field.

4. The accelerated positive ions are deflected in the magnetic field and focused onto a detector according to their mass-to-charge ratio (m/z).

5. The ions generate a current or signal that is proportional to their abundances.

The detector then draws a mass spectrum of the different ions.

The diagrams below show how the mass spectrometer is used.

Figure 1.23: Diagram of a mass spectrometer process that takes place in it.

Source: K. Bhavyasri et al., IJSRR 2019, 8(2), 3161-3176 Mass Spectrum A mass spectrum is a graphical representation of the ion intensities (relative abundances) as a function of their mass-to-charge ratios (m/z). Simply put, a mass spectrum is a graph of the percentage abundance versus the relative atomic masses of the ions present in the sample.

Generally, the mass spectrum shows a series of peaks, each representing an ion with a specific mass-to-charge ratio (m/z). The height of the peak is proportional to the abundance of the ion in the sample.

The mass spectrum of chlorine is shown below.

Fig 1.24: Mass spectra of Chlorine element How to calculate the relative atomic mass of different elements The relative atomic mass of an element can be calculated by using:

1. The mass spectrum

2. Percentage abundance data.

Generally, the relative atomic mass is calculated by using the following steps:

1. Identify the isotopes of the element and their percentage abundances.

The percentage abundance refers to the amount of each isotope present in a sample, expressed as a percentage of the total number of atoms of the elements.

2. Multiply the mass of each isotope by its percentage abundance.

3. Add up the products obtained in step 2.

4. Divide the sum obtained in step 3 by 100 to obtain the relative atomic mass of the element.

Worked Example

Let us consider the element boron. Boron has two isotopes, boron-10 and boron-11 with natural abundances of approximately 20% and 80% respectively. To calculate the relative atomic mass of boron, we can use the steps as follows:

1. Boron-10 has a percentage abundance of 20% and boron-11 has a percentage abundance of 80%.

2. The relative atomic mass (Aᵣ) of boron-10 is 10 amu and the relative atomic mass (Aᵣ) of boron-11 is 11 amu.

Therefore, 10 × 20 = 200 and 11 × 80 = 880

3. 200 + 880 = 1080

4. 1080/100 = 10.8 Therefore, the relative atomic mass of boron is 10.80 amu.

Worked Example

Chlorine naturally exists as two isotopes, chlorine-35 and chlorine-37. The abundance of chlorine-35 is 75% and the abundance of chlorine-37 is 25%.

Calculate the relative atomic mass of chlorine.

Answer

1. % of chlorine-35 = 75% and % of chlorine-37 = 25%

2. The Aᵣ of chlorine-35 is 35 amu and the Aᵣ of chlorine-37 is 37 amu.

Therefore 75 × 35 = 2625 and 25 × 37 = 925

3. 2625 + 925 = 3550

4. 3550/100 = 35.50 Therefore, the relative atomic mass of chlorine is 35.50 amu.

Activity 1.44: Operation of the mass spectrometry and it’s application

1. Design a flow chart to show the operation of the mass spectrometer.

2. (a) Design a model to show how to calculate the relative atomic mass of different elements.

(b) Use the information provided in table-1 to calculate the relative atomic mass of naturally occurring copper isotopes. Give your answer to 1 decimal place.

Mass number % Abundance 63 69 65 31

3. Bromine has two main isotopes: Br-79 with an atomic mass of 78.91833 amu and a relative abundance of 50.69%, and Br-81 with an atomic mass of 80.91629 amu and a relative abundance of 49.31%. Calculate the average atomic mass of bromine.

RADIOACTIVITY Radioactivity is the process by which unstable atomic nuclei decay, releasing energetic particles or electromagnetic radiation including alpha, beta, and gamma rays. This natural phenomenon occurs in each radioactive isotope (radionuclides) at a fixed rate, characterized by a half-life, which is the time for half of the radioactive atoms to decay into stable ones. This rate of decay is unique to each individual radioactive isotopes, e.g. Carbon-14 will have a different half-life to Uranium-235, but all carbon-14 isotopes will have the same half-life. Radioactivity has practical applications in medicine, industry, and energy production, but also poses health risks and safety concerns that require careful management.

Nuclear reactions and chemical reactions are two different reactions that involve changes in the composition of matter.

The main differences between these two types of reactions are as follows:

1. Nature of particles involved: In a chemical reaction, the atoms of the reacting substances do not change their identities. In contrast, in a nuclear reaction, the atomic nuclei of the reacting substances undergo changes, resulting in the formation of different nuclei and subatomic particles.

Chemical reactions can be summarised in the movement and arrangement of electrons within Chemical bonds, however nuclear reactions involve the nucleus of the atoms themselves.

2. Energy changes: Nuclear reactions involve much larger energy changes than chemical reactions. This is because nuclear reactions involve changes in the binding energies of atomic nuclei, which are much larger than the energies involved in the breaking and forming of chemical bonds.

3. Rate of reaction: Chemical reactions generally occur at a faster rate than nuclear reactions. This is because chemical reactions involve the interaction of electrons, which are much lighter and move much faster than atomic nuclei.

4. Triggering factors: Chemical reactions are triggered by factors such as temperature, pressure, and concentration, while nuclear reactions are triggered by factors such as particle bombardment and radioactive decay.

In other words, chemical reactions are affected by environmental conditions such as temperature and pressure, but nuclear reactions are not.

5. Product stability: The products of a chemical reaction are typically more stable than the products of a nuclear reaction. This means that the products of a nuclear reaction may undergo further decay or transformation into products over time.

6. Emission of radiation: Nuclear reactions emit radiation such as alpha particles, beta particles, and gamma rays while chemical reactions do not.

Alpha, beta, and gamma radiation are types of ionising radiation, which have different properties based on their composition and energy.

Here are the properties of each type of radiation:

Alpha Radiation

1. Consists of a helium nucleus of two protons and two neutrons.

2. Has a charge of +2 and a mass number of 4 amu.

3. Travels only a few centimetres in the air and are blocked by a sheet of paper or the dead cells on the surface of skin.

4. Has a high ionisation potential and can cause significant damage to living tissue.

5. Emitted by heavy nuclei like uranium and radium as they decay into lighter elements.

6. Symbol ₂ ⁴He Beta Radiation

1. Consists of a negatively charged particle identical to a high-energy electron.

2. It has a charge of -1 and has a negligible mass.

3. Are more penetrating than alpha particles and can penetrate several millimetres of aluminium or plywood.

4. Have lower ionisation potential than alpha particles and may cause damage to living tissue.

5. Are emitted by elements with an excess of neutrons, like carbon-14.

6. Symbol − 1 0e Gamma Radiation

1. Consists of high-energy photons of electromagnetic radiation.

2. Has no charge or mass.

3. Is highly penetrating and can pass through several centimetres of lead or concrete.

4. Has a lower ionisation energy potential than alpha and beta radiation but can still cause damage to living tissues by ionising atoms along their path.

5. Emitted by the nucleus of some radioactive atoms, such as Cobalt-60.

6. Symbol γ

The stability of atomic nuclei is determined by the balance between the strong nuclear force, which holds the nucleus together and the electrostatic repulsion between protons in the nucleus.

The neutron-to-proton ratio and the binding energy per nucleon (one of the subatomic particles of the nucleus) are the two most important factors that determine the stability of an atomic nucleus.

Neutron-to-Proton Ratio

The neutron-to-proton ratio is the ratio of the number of neutrons to the number of protons in the nucleus. This ratio affects the stability of the nucleus because it determines the balance between the strong nuclear force and the electrostatic repulsion between protons.

When the ratio is low, meaning there are more protons than neutrons, the electrostatic forces between the protons are stronger than the strong nuclear forces holding the nucleus together, making the nucleus unstable. On the other hand, nuclei with high neutron-to-proton ratios are also unstable as there are more neutrons than protons and the strong nuclear force becomes weaker leading to decay of the nucleus.

For stability, the ratio must be equal to the values shown in the graph below. As you can see this is close to one (1) for low atomic mass isotopes but drifts away from this as the atomic mass increases.

Also shown in this diagram are the types of decay that will occur depending upon the specific isotope (it is worth noting that some of the types of decay shown here are not covered by this course e.g. beta +, neutron or proton).

Figure 1.22: Graphical representation of neutron-to-proton ratio after decay Binding Energy per Nucleon The energy needed to keep the protons, neutrons, and electrons together in an atom is generally called binding energy.

The binding energy per nucleon refers to the average amount of energy required to remove a nucleon from the nucleus of an atom. It is calculated by dividing the total binding energy of a nucleus by the total number of nucleons (protons and neutrons) in the nucleus. The binding energy per nucleon reflects the strength of the strong nuclear force that holds the nucleus together.

When the binding energy per nucleon is high, the nucleus is more stable, meaning the strong nuclear force is strong enough to overcome the electrostatic repulsion between the protons. Nuclei with low binding energy per nucleon are less stable and tend to undergo radioactive decay which increases their binding energy per nucleon.

Carry out following activities (1, 2 and 3) in the laboratory.

Activity 1.45 Demonstrating radioactivity in a safe and engaging way.

Materials needed: naturally-occurring radioactive substances, Geiger counter, radioactive sources, lead, aluminium, paper, spark counter, High voltage power supply.

1. Observe the teacher (or any trained personnel invited by the teacher hold the radioactive sources in turn over the spark counter to show the presence of radiation.

2. For each source in turn, observe the teacher or trained personnel demonstrate the range and penetration power of each source.

3. For learners who may not have access any of the materials or receive a demonstration by a teacher or trained personnel, alternatively, you can use any of the links below to watch the video on radioactivity.

https://www.youtube.com/watch?v=Pc3- y6Omzzw&list=PLEEC940EB121761B3 https://www.youtube.com/ watch?v=xwYSvzojFbo&list=PLEEC940EB121761B3&index=3 https://www.youtube.com/ watch?v=IUp9xa8tYs&list=PLEEC940EB121761B3&index=4 https://www.youtube.com/ watch?v=t9bKVL6hdRQ&list=PLEEC940EB121761B3&index=5 Do well to write down what you observe while watching the video(s) and compare your notes with that of your colleagues in class.

Half-life of a Nuclide

The half-life of a nuclide is the time it takes for half of the undecayed atoms in a sample of the nuclide to decay. This means that after one half-life has passed, half of the original nuclide atoms will have decayed, and the remaining half will remain.

A second half-life will have passed when a further half (i.e. half of a half) has decayed.

The half-life is a characteristic property of each radioactive nuclide and is constant for that nuclide, which means that regardless of the size of the sample, each atom has the same probability of decaying during a certain period.

Some nuclides have very short half-lives, which means they decay rapidly and are highly radioactive for only a short time, while others have very long half- lives, which means they decay slowly and can remain radioactive for thousands or millions of years.

Consider the implications of a substance which has a long half-life; why might this be beneficial and why might this not be beneficial? Discuss your thoughts with a neighbour and feedback to the class.

Consider the implications of a substance with a short half-life; why might this be beneficial and why might this not be beneficial? Discuss your thoughts with a neighbour and feedback to the class.

Half-life is an important concept in the study of nuclear chemistry and is used to calculate the amount of time needed for a given amount of radioactive material to decay to a desired amount, as well as how much of a substance will remain after a certain period of time. It is also used in radioactive dating methods to estimate the age of geological samples or archaeological artefacts.

Activity 1.46 Determination of half-life from experimental data.

Material Needed: A set of 100 dice.

1. Lay out a results table as shown below:

Number of rolls Number of dice 0 100 1

2. Shake all 100 dice.

3. Remove all the dice which show a number 1 or 2.

4. Count the remaining dice

5. Record this result in your table next to “number of rolls = 1”

6. Roll the remaining dice again.

7. Repeat steps 3 to 7 until all the dice have been removed.

8. Plot a graph showing how the number of dice changes as the number of rolls increases.

9. Repeat this experiment but now remove all dice which show a number of 1, 2 or 3.

a. Using observations from the activity. Discuss how this model relates to half life

b. Determine the half-life of both of your experiments (how many rolls did it take to reach 50?).

c. Discuss how the experiment could be developed to show a radionuclide with a longer half-life.

How to calculate the half-life using experimental data and calculation To calculate the half-life of a substance a few things need to be known:

1. The initial activity or number of undecayed nuclei.

2. The final activity or number of undecayed nuclei.

3. The total time that the sample has been left for.

Worked Example

An isotope of Uranium-235 has an initial activity of 32 Bq and a final activity of 2 Bq. This sample was left for 200 years. Calculate the half-life of the sample:

Answer Firstly, begin by halving the initial activity value:

32/2 = 16.

Next, continue the halving process until the final activity value is reached:

16/2 = 8.

8/2 = 4.

4/2 = 2.

Then count how many times the activity has halved: 4 This process has undergone four (4) half-lives.

Finally divide the total time taken by the number of half-lives.

200/4 = 50 years.

The time for one (1) half-life was 50 years.

Alternatively, To calculate the half-life of a radioactive nuclide using experimental data, you need to measure the decay rate or the amount of radioactive material remaining at specific time intervals.

Here is how to calculate half-life using experimental data and calculation:

1. Measure the initial activity (or quantity) of the radioactive sample i.e. No.

2. Measure the activity (or quantity) of the radioactive sample at a specific time interval, Δt i.e. Nt

3. Calculate the fraction of the original material that has decayed during the time interval, Δt Using the formula:

Fraction decayed = Activity at Δt__________ initial activity = Nₜ_ Nₒ

4. Calculate the natural logarithm of the fraction decayed:

i.e. In ( Nₜ__ Nₒ) = λΔt (λ is the decay constant of the nuclide)

5. Solve for the decay constant:

λ = − In ( Nₜ__ Nₒ)_______ Δt

6. Calculate the half-life, t₁/2 :

i.e.

t1/2 = In2___ λ , where (In2 ) is the natural logarithm of 2, which is approximately 0.693.

i.e.

t₁_ ₂ = 0.693____ λ

Note that you can repeat this process with different time intervals, and the resulting half-lives should be consistent if the sample is homogeneous and the decay process is steady.

NB:

The expressions In(^(No)_ _(N)ₜ ) = λΔt and log(No_ Nt ) = λt/2.303 can also be used Now try your hand at using equations to calculate half-life using experimental data.

Activity 1.47

Calculate the required values for each of these questions:

1. A sample of radioactive isotope X starts with an activity of 1000 counts per minute (cpm) and decays to 250 cpm in 60 years. What is the half- life of isotope X?

2. A sample of element Y has an initial activity of 800 counts per minute (cpm) and decays to 100 cpm in 24 years. Determine the half-life of element Y.

3. A sample of isotope Z starts with an activity of 1600 counts per minute (cpm) and decays to 100 cpm in 50 years. What is the half-life of isotope Z?

4. A sample of substance W starts with an activity of 3200 counts per minute (cpm) and decays to 200 cpm over a period of 48 years. Find the half-life of substance W.

5. A sample of material Q starts with an activity of 1600 counts per minute (cpm) and decays to 200 cpm in 21 years. Calculate the half-life of material Q.

Uses of radioisotopes and the principle behind each use Radioisotopes are used in a wide range of applications, from medical diagnosis and treatment to industry and scientific research.

Here are some common uses of radioisotopes and the principle behind each use:

1. Medical diagnosis: Radioisotopes like technetium-99, iodine-131, and gallium-67 are commonly used in medical imaging techniques such as positron emission tomography (PET), single photon emission computed tomography (SPECT,) and computed tomography (CT) scans.

These isotopes emit gamma rays that can be detected by special cameras to create images of internal body structures, allowing doctors to diagnose and monitor diseases such as cancer, heart disease, and neurological disorders.

Figure 1.25: Medical diagnosis

2. Medical treatment: Radioisotopes can be used to treat certain types of cancer, such as thyroid cancer, by selectively targeting and destroying cancerous cells.

Iodine-131 is used to treat thyroid cancer, while strontium-89 is used to relieve pain in bone cancer patients.

Figure 1.26: Medical treatment

3. Industrial applications: Radioisotopes such as cobalt-60 and iridium-192 are used to sterilise medical equipment and food products, as well as measure the thickness of materials in manufacturing processes.

4. Agricultural applications: Radioisotopes can be used to measure plant and soil properties, such as moisture content and nutrient uptake, to improve crop yield and quality. Radioisotopes are also used for insect and pest control.

5. Scientific research: Radioisotopes can be used to study various biological and chemical processes in cells, tissues, and organisms. Carbon-14 is used in dating archaeological artefacts and geological samples, while tritium (hydrogen-3) is used to label molecules and track their movement within biological systems.

6. Source of heat energy: Controlled nuclear fission is used to generate electricity and heat energy.

The principle behind each use of radioisotopes is based on their unique properties, including their half-life, decay mode, and energy spectrum. For

example, in medical imaging, radioisotopes that emit gamma rays are used because they can penetrate the body and be detected by specialised cameras.

In cancer treatment, radioisotopes that emit beta particles or alpha particles are used because they can travel short distances and deliver a high dose of radiation to cancerous cells while sparing healthy tissue.

In industrial and agricultural applications, radioisotopes are used to measure various properties based on their ability to emit radiation or interact with other materials.

In a nutshell, the uses of radioisotopes are based on their ability to provide useful information, diagnose and treat diseases, improve quality control in manufacturing, and advance scientific research.

How to complete and balance simple nuclear reactions Balancing a nuclear reaction involves ensuring that the total number of protons (atomic number, or proton number) and the total mass numbers (nucleon number) are the same on both sides of the equation. Additionally, the sum of the charges on each side must be equal. Example of a balanced nuclear reaction is shown below:

Note that nuclear reactions involve the emission or capture of particles, such as alpha particles (helium nuclei) or beta particles (electrons).

These particles should be accounted for in the equation with the appropriate symbol.

Worked Example

A certain radioactive element, N, disintegrated into another element, K, when an alpha particle was released. The atomic mass and atomic number of N are 289 and 87, respectively.

a. Write the unbalanced equation

b. Balance the equation in (a) above.

c. Determine the values of x and y

d. Discuss how you balanced your nuclear equation with a friend.

Answer

1. Write out the radioactive elements and radiations involved in the form equation N → K + α

2. Attach the mass numbers and atomic numbers 87 289N → _(y) ˣK + ₂ ⁴α NB. The total atomic mass on the left-hand side is equal to that on the right side.

The total atomic number on the left-hand side is equal to that on the right side.

3. Find x and y by equating the mass numbers on the left-hand side to that on the right-hand side and the atomic numbers on the left-hand side to that on the right-hand side.

289 = x + 4 ∴ x = 289 − 4 = 285 87 = y + 2 ∴ y = 85 Hence the nuclear equation is balanced 289 87 N → 285 85 K + 4 2α Self-Assessment Try to balance the following nuclear reactions using the worked example above (ask your teacher for assistance if you need it):

1. 232 690Th → 228 88 Ra + X

2. 14 6 C → 14 7 N + Y

3. 9 4Be + Z → 12 6 c + 1 0n (₀ ¹n = neutron) Risks associated with radioactivity Radioactivity can pose several health and environmental risks.

Here are some of the risks associated with radioactivity:

1. Radiation exposure: Exposure to ionising radiation emitted by radioactive materials can damage the DNA in cells, leading to cellular mutations and cancer. Long-term exposure to low levels of radiation is associated with an increased risk of cancer, while high levels of radiation exposure can cause sickness and death.

Figure 1.27: Modes of exposure

2. Health risk to workers: Individuals who work with or around sources of radiation, such as nuclear power plant workers and medical professionals, are at a higher risk of radiation exposure and associated health risks.

Figure 1.28: Effects of exposure to radiation

3. Environmental risks: Radioactive materials can contaminate soil, water, and air, leading to environmental contamination and increased radiation exposure for humans and other living organisms.

Figure 1.29: Environmental risks

4. Nuclear accidents: Accidents at nuclear power plants can release large amounts of radioactive materials into the environment, leading to increased radiation exposure and environmental contamination.

Figure 1.30: Nuclear accidents

5. Nuclear weapons: The use of nuclear weapons can release large amounts of ionising radiation, causing both immediate and long-term health effects for humans and the environment.

To mitigate the risks associated with radioactivity, it is important to follow proper safety protocols when working with radioactive materials. This includes using protective equipment, maintaining proper monitoring and dosimetry practices, and ensuring that radioactive waste is properly disposed of and secured.

In the event of a nuclear accident or radiation release, evacuation procedures and clean-up exercise should be carried out to minimise exposure and contamination.

Fig 1.31: Nuclear weapons

Activity 1.48 Exploring the applications of radioisotopes in various fields.

Materials needed: Internet access, presentation tools (e.g., posters, PowerPoint)

1. In a group of 5 learners each, explore a different application of radioisotopes (e.g., medical imaging, cancer treatment, food irradiation, energy, carbon dating).

2. Present you findings to the class as a poster or PowerPoint, focusing on how radioisotopes are used and their benefits and risks.

3. Discuss the different applications and the importance of radioisotopes in everyday life.

Activity 1.49 The principal parts of a mass spectrometer and how they work.

Materials needed: Charts or printed images of a mass spectrometer, worksheet with questions and space for notes

1. Use a diagram or a chart of a mass spectrometer,

2. Identify the principal parts of a mass spectrometer:

3. Explain the function of the parts.

4. Explain the step-by-step process of how a mass spectrometer works following the guiding questions below:

a. What is the purpose of the ion source in a mass spectrometer?

b. How are ions accelerated in a mass spectrometer?

c. What role does the magnetic field play in the deflection of ions?

d. How does the detector measure the ions?

Activity 1.50: Identification of peaks on the mass spectrum To learn how to identify peaks on a mass spectrum and use them to calculate the relative abundance and masses of isotopes.

Materials needed: Printed mass spectra charts and calculators Using the given mass spectrum Identify how the x-axis (horizontal axis) and y-axis (vertical axis) represent.

1. Identify the number of peaks on the mass spectrum.

(Each peak corresponds to an isotope, and the height (or intensity) of the peak indicates its relative abundance.)

2. Calculating Relative Abundance:

Aᵣ = (m₁ × a₁) + (m₂ × a₂) + … Where: m₁, m₂,…,… are the masses of the isotopes a₁, a₂,…,… are the abundances of the isotopes

Worked Example

Using the simple mass spectrum of boron isotopes below.

(a) Identify the percentage abundance of the boron isotopes.

(b) Calculate the relative atomic mass of boron.

Answer

a. The peak heights show the relative abundance of the boron isotopes:

boron-10 has a relative abundance of 19.9% and boron-11 has a relative abundance of 80.1%

b. (Aᵣ(B) = (m₁ × a₁) + (m₂ × a₂) = (10 × 0.199) + (11 × 0.801) = 10.8 Therefore, relative atomic mass of boron is 10.8

Review Questions 1.1a

1. State three careers that are related to chemistry.

2. State and explain at least three benefits of chemistry to the Ghanaian society.

3. Conduct research on the importance of chemistry to Ghanaian society in the field of agriculture and present your findings using PowerPoint or flipcharts.

4. Why is the study of chemistry important, and how does it contribute to our understanding of the natural world and address global challenges?

5. Explain why chemistry is considered the “central science”.

6. What are some examples of interdisciplinary research that combine chemistry with other sciences like biology, physics, or environmental science?

7. Explain how chemistry informs and improves our daily lives, from medicine to technology to consumer products.

8. Create a flow chart showing how chemistry affects daily lives, such as in

a. Personal care and hygiene

b. Food and nutrition

c. Health and medicine

d. Technology and gadgets

e. Environment and conservation

Review Questions 1.1b

1. State at least one (1) type of fire that can be extinguished using a carbon dioxide fire extinguisher.

2. Explain why potassium permanganate should not be stored near a bottle containing ethanol.

3. Design a chemical storage compatibility chart using the following chemicals in the laboratory: HN O₃, C H₃ COOH, NaOH, H₂ O₂, C H₃ C H₂ OH

Review Questions 1.1c

1. State at least two (2) examples of personal protective equipment used in the chemistry laboratory.

2. State the precautions to take when handling chemicals with the following hazard labels.

a. Corrosive substance

b. Toxic substance

3. In a certain school, the authorities are concerned about the increasing rate of accidents in the chemistry laboratory during practical sessions.

(a) Suggest three (3) possible causes of these accidents.

(b) Write and explain three (3) ways of curbing the problem identified.

4. State five wrong practices that should be avoided in the chemistry laboratory.

5. What should you do when handling chemicals in the laboratory?

6. Explain what a chemistry student should do in case of a chemical spillage in the laboratory.

7. Discuss how to handle hazardous chemicals safely using protective clothing and safety equipment. Consider the following: gloves, goggles, laboratory coats, respirators and safety showers.

Review Questions 1.2

1. What is the role of curiosity and wonder in driving scientific inquiry?

2. How does the scientific method relate to real-world problems and applications?

Review Questions 1.3a

1. How does Dalton’s atomic theory help us understand the basic structure of matter?

2. Why is it important to learn about the historical development of atomic theory in science?

3. How did Dalton’s theory contribute to the field of chemistry?

4. How did Dalton’s atomic theory differ from earlier concepts of atoms?

Review questions 1.3b

1. Which subatomic particles did Thomson include in the plum-pudding model of the atom?

2. How did the results of Rutherford’s gold foil experiment differ from his expectations?

Review Questions 1.4a

1. What was the name of Bohr’s model of the atom?

2. Briefly describe how Bohr’s model of the atom contributed to our understanding of the stability of the atom.

3. Evaluate the possible values of the magnetic quantum number (m) for an electron in an orbital with l = 2 .

4. Which of the following sets of quantum numbers are not allowed in the hydrogen atom? For the sets of quantum numbers that are incorrect, state what is wrong in each set.

(a) n = 3, l = 2, m = 2

(b) n = 4, l = 3, m = 4

(c) n = 0, l = 0, m = 0

(d) n = 2, l = − 1, m = 1

5. Given the quantum numbers n = 4, l = 3 , determine the number of orbitals and the maximum number of electrons possible in this subshell.

6. If an electron in a hydrogen atom transitions from the n = 3 level to the n = 2 level, what are the possible values of the azimuthal quantum number l for the initial and final states?

7. Explain why an electron orbiting around the nucleus of an atom does not collapse into the nucleus, according to Bohr’s model.

Review Questions 1.4b

1. What is an isotope, and how do isotopes of the same element differ?

2. What happens when an atom becomes unstable and decays?

3. Explain what happens when there more neutrons than protons in the nucleus.

4. List and explain three (3) uses of radioactivity

5. Half-life of a radionuclide is 5 days. How many days will it require for a 160 g sample of this radionuclide to decay to 5 g ?

6. An unknown radioactive isotope, X, decays to one-fourth of its original amount in 60 years. What is the half-life of isotope X?

7. A sample of element Y decays to 12.5% of its initial quantity in 24 years.

Determine the half-life of element Y.

8. Isotope Z decays to 6.25% of its initial amount in 50 years. What is the half- life of isotope Z?

9. A sample of substance W reduces to one-sixteenth of its original amount over a period of 48 years. Find the half-life of substance W.

10. Material Q decays to one-eighth of its original mass in 21 years. Calculate the half-life of material Q.

11. A fossil contains 25% of its original Carbon-14. The half-life of Carbon-14 is 5730 years. How old is the fossil?

12. A sample of Radium-226 has decayed to one-eighth of its original amount.

If the half-life of Radium-226 is 1600 years, how much time has passed since the sample started decaying?

13. A medical treatment involves Iodine-131, which has a half-life of 8 days. If a patient’s body initially contains 200 mg of Iodine-131, how much time will it take for the amount to reduce to 25 mg?

14. Uranium-238 has a half-life of 4.5 billion years. If a rock originally had 60 grams of Uranium-238 and now contains 15 grams, how old is the rock?

15. Use the worksheet provided to answer questions about the types of radiation emitted in each decay process.

Complete the graph by following the sequence of alpha and beta decays that actually occurs in nature from the decay event table.

Start with U-235 and follow the steps (in order) of alpha and beta decays.

Graph the path of decays by drawing an arrow tracing the mass number per atomic number and writing the correct atomic symbol for each mass number per atomic number event.

STEP # DECAY EVENT 1 ALPHA 2 BETA 3 ALPHA 4 ALPHA 5 BETA 6 ALPHA 7 ALPHA 8 ALPHA 9 BETA 10 ALPHA 11 BETA 12 STABLE

16. Fill in the blanks with the correct notation, atomic symbols, mass number, atomic number, decay type, and/or emission particles. You may use a periodic table to identify the element.

(a) (b)

17. Using flow charts, distinguish between nuclear reactions and chemical reactions.

18. What materials can the teacher use to explain the concept of radioactivity to learners?

19. What precautions should a learner take in disposing of materials like mobile phones that can release nuclear radiations that will contaminate the environment?

Review Questions 1.4c

Calculate the relative atomic masses of the following elements:

1. Magnesium has three main isotopes: Mg-24 with an atomic mass of 23.98504 amu and a relative abundance of 78.99%, Mg-25 with an atomic mass of 24.98584 amu and a relative abundance of 10.00%, and Mg-26 with an atomic mass of 25.98259 amu and a relative abundance of 11.01%.

Calculate the average atomic mass of magnesium.

2. Carbon has two main isotopes: C-12 with an atomic mass of 12.00000 amu and a relative abundance of 98.93%, C-13 with an atomic mass of 13.00335 amu and a relative abundance of 1.07%. Calculate the average atomic mass of carbon.

3. Silicon has three main isotopes: Si-28 with an atomic mass of 27.97693 amu and a relative abundance of 92.23%, Si-29 with an atomic mass of 28.97649 amu and a relative abundance of 4.67%, and Si-30 with an atomic mass of 29.97377 amu and a relative abundance of 3.10%. Calculate the average atomic mass of silicon.

Review Questions 1.5a

1. Explain the following terms:

(a) relative atomic mass (b) relative molecular mass

2. Below is a list of elements with their corresponding atomic masses. Use this information to answer the questions that follow.

Element Symbol Element Name Relative Atomic Mass

H Hydrogen 1.01 He Helium 4.00

Li Lithium 6.94

Be Beryllium 9.01

B Boron 10.81 Element Symbol Element Name Relative Atomic Mass C Carbon 12.01 N Nitrogen 14.01 O Oxygen 16.00 F Fluorine 19.00 Ne Neon 20.18 Na Sodium 22.99 Mg Magnesium 24.31 Al Aluminium 26.98 Si Silicon 28.09 P Phosphorus 30.97 S Sulphur 32.07 Cl Chlorine 35.45 K Potassium 39.10 Ca Calcium 40.08

(a) What is the relative atomic mass of nitrogen (N)?

(b) Which element has a relative atomic mass of 12.01?

(c) What is the relative atomic mass of sulphur (S)?

(d) Which element has the highest relative atomic mass in the table?

Review Questions 1.5b

1. Explain why the electron configuration of chromium (Cr) is different from what is expected based on the Aufbau principle

2. Analyse the relationship between electron configuration and the chemical properties of elements within a period.

3. Write the electron configuration for ₁₁Na and ₁₉K. Using the configuration, explain why they are considered reactive elements.

4. Using the electron configurations for (₁₈Ar and ₁₀Ne), explain why they are considered to be non-reactive.

5. Explain why half and fully-filled sub-shells are more stable than those that are neither half nor fully filled.

6. The electron configurations for the elements named are shown here. Each configuration is incorrect in some way. Identify the error in each and write the correct configuration.

(a) ₁₃Al: 1 s²2 s²2 pₓ ²2 p_(y) ²2 p_(z) ²3 s²3 d₁ ¹(b) ₁₇Cl: 1 s²2 s²2 pₓ ²2 p_(y) ²2 p_(z) ²3 s²3 pₓ ²3 p_(y) ²4 s¹(c) ₁₀Ne: 1 s²2 s²2 pₓ ²2 p_(y) ²2 p_(z) ¹3 s¹(d) ₁₉K: 1 s²2 s²2 pₓ ²2 p_(y) ²2 p_(z) ²3 s²3 pₓ ²3 p_(y) ²3 p_(z) ²3 d₁ ¹(e) ₇N: 1 s²2 s²2 pₓ ²2 p_(y) ¹

In this section you will be introduced to fundamental concepts such as relative atomic mass, relative molecular mass, and the mole as a unit of amount of substance. You will also learn how to perform calculations based on the amount of substance and understand the importance of the mole concept in preparing standard solutions.

At the end of this section, you should be able to:

• Explain relative atomic mass and relative molecular mass.

• Describe the atomic mass unit as an average mass.

• Describe the mole as a unit of the amount of substance.

• Calculate different physical quantities (number of entities, mass and volume) based on the amount of substance.

• Explain the mole concept and its relevance in preparation of standard solutions.

Key Ideas

• Atomic mass unit (amu) is equal to 1/12th the mass of a carbon-12 atom.

• Relative atomic mass is the average mass of an element’s atoms, considering isotopes.

• Relative molecular mass: the sum of the relative atomic masses of all atoms in a molecule.

• Mole is a way of measurement in chemistry.

• Avogadro’s constant is number of units in one mole of any substance.

• Entities refers to any distinct atom, molecule, ion.

• Avogadro’s number is the number of particles in one mole.

• Avogadro’s number is 6.022 × 10 ²³particles (atoms, molecules, and ions)

• Mole (mol) is a unit of measurement.

• Molar mass is mass of mole a substance.

Relative Atomic Mass (Aᵣ) is defined as the average mass of one atom of the element compared to 1/12ᵗʰof the mass of one atom of carbon-12.

Mathematically Relative Atomic mass is represented as, Aᵣ = Average mass of one atom of the element_____________________________ ₁_ ₁₂ₜₕ the mass of one atom of carbon − 12 This value is expressed in atomic mass units (amu). The relative atomic mass is determined by the abundance of each isotope of an element.

Worked Example 2.1

One atomic mass unit of Carbon – 12 is 1.6603 × 10⁻²⁴g . If the average mass of an atom of oxygen is 2.65659 × 10 ⁻²³g . Determine its relative atomic mass.

Solution:

Aᵣ(O) = Average mass of one atom of the element_____________________________ ₁_ ₁₂ₜₕ the mass of one atom of carbon − 12 = 2.65659 × 10⁻²³g/1.6603 × 10⁻²⁴g = 16.0 Relative Molecular Mass (Mᵣ) Relative molecular mass is defined as the average mass of one molecule of a substance compared with ¹_ ₁₂ₜₕ of the mass of one atom of carbon-12.

Mᵣ = average mass of one molecule of the substance_________________________________ ₁_ ₁₂ₜₕ the mass of one atom of carbon − 12 It also has no unit. For ionic compounds, the relative molecular mass is called its Relative formula mass (as ionic substances do not exist as molecules).

Relative molecular mass is the sum of the masses of the elements that make up the molecule.

Activity 2.1: Determining the mass of an element or compound using a beam balance Materials needed:

Beam balances, standard carbon-12 samples (represented by 12 beads or any small, identical objects), samples of different elements or compounds (using different numbers of beads/objects), worksheets for recording observations and calculations, calculators Steps:

Carry out this activity in small groups.

1. Set up the beam balance, get a standard carbon-12 sample (12 beads), and samples of other elements or compounds (different numbers of beads).

2. Place the carbon-12 sample in one pan of the beam balance.

3. Place the unknown element or compound sample in the other pan of the beam balance.

4. Adjust the number of beads/objects in the unknown sample until the beam balance is level, indicating that the masses are equal.

5. Record the number of beads/objects used for the unknown sample to balance the carbon-12 standard.

6. Calculate the relative mass of their unknown sample compared to carbon-12.

Example Calculation:

If the unknown sample balanced with 22 beads, the relative mass compared to carbon-12 (12 beads) is:

Aᵣ = Average mass of one atom of the element_____________________________ ₁_ ₁₂ₜₕ the mass of one atom of carbon − 12 So then as a balance is set up:

12 × (mass of 1 carbon atom) = 22 × (mass of 1 unknown atom) So: as we know the mass of a carbon atom is 1 amu we can state that:

Aᵣ = Number of beads in carbon − 12__________________________ Number of beads in unknown sample = 12_ 22 = 0.54 Therefore, the unknown sample has a relative mass of 0.54 times that of carbon-12.

7. Record your calculations and results on the worksheet.

8. Present your findings and explain your calculations.

Worked Example 2.2

Determining mass using a beam balance.

(a) Number of beads in carbon-12 sample: 12

(b) Number of beads in unknown sample: x

(c) Calculate the relative mass of the unknown sample compared to Carbon-12.

Aᵣ = 12__________________________ Number of beads in unknown sample = 12_ x = y Therefore, the relative mass of the unknown sample is y.

Activity 2.2: Understanding relative molecular mass Materials needed: different sets of coloured balls, Steps:

Carry out this activity in small groups.

1. Provide different sets of coloured balls and connectors.

2. Construct different molecules such as H₂O, CO₂, and CH₄.

3. Create a “molecule” by connecting a few balls (e.g., two blue balls and one red ball to represent H₂O).

4. Calculate the relative molecular mass of the constructed molecules using the atomic masses provided.

Calculating the Mᵣ of a compound using Aᵣ:

Worked Example 2.3

Show the steps to calculate the relative molecular mass of water.

Solution:

1. Identify the chemical formula of water i.e. H₂ O .

2. Using the relative atomic mass of each element multiply the relative atomic mass by the number of atoms of each element:

a. In water (H₂ O ), there are 2 hydrogen atoms and 1 oxygen atom.

b. The relative atomic mass of hydrogen = 1.0

c. Number of hydrogen atoms = 2

d. Contribution of hydrogen to the relative molecular mass = 1.0 × 2 = 2.0

e. Relative atomic mass of oxygen = 16.0

f. Number of oxygen atoms = 1

g. Contribution of oxygen to the relative molecular mass = 16.0 × 1 = 16.0

3. Add the contributions from all elements:

Sum the contributions of hydrogen and oxygen to get the relative molecular mass of water:

Mᵣ( H₂ O) = Contribution of hydrogen + Contribution of oxygen Mᵣ(H₂ O)= 2.0 + 16.0 = 18.0

Activity 2.3: Trial Question

Calculate the relative molecular masses (Mᵣ) of the following substances:

For reference, the atomic masses are approximately: H = 1; O = 16; C = 12

1. Water (H ₂O)

2. Carbon Dioxide (CO₂)

3. Methane (CH₄)

4. Glucose (C₆H₁₂O₆).

The relative atomic mass scale is based on an isotope of carbon-12. Carbon-12 contains 6 protons and 6 neutrons and a mass of 12 atomic mass units. The carbon-12 scale is therefore defined as an atomic mass reference scale in which one atom of carbon-12 isotope has 12 units.

Therefore, The mass of one carbon-12 = 12 amu.

1 amu = Mass of one carbon − 12/12 NB: One atomic mass unit of Carbon-12 is the same as 1/12ᵗʰof the mass of one atom of carbon-12.

Recall that most naturally occurring elements have different isotopes with different natural abundance and masses. Therefore, relative atomic mass is an average mass.

Applications of Relative Atomic Mass in everyday life

1. Relative atomic and relative molecular mass is used to calculate the concentration of a stock solution from chemical stores.

2. The idea of relative molecular mass or formula mass and the law of conservation of mass are used to do quantitative calculations in chemistry.

3. The idea of relative atomic mass is used to determine the empirical formula of a substance.

Activity 2.4: How to calculate the atomic mass unit (amu) of an individual particle (atom or molecule) Materials needed:

• Access to a computer or smart device with internet

• Worksheet for notes and questions

• Access to an educational video on atomic mass units (amu) (e.g., a video science education channel on YouTube) How to Calculate Atomic M ass Practice Problems.mp4 https://www.youtube.com/watch?v = ULRsJYhQmlo Steps:

1. Watch the video and take notes on key points, especially on how the amu is defined and measured.

2. Pause the video at key moments to discuss important concepts and ensure understanding.

3. After watching the video, discuss the following questions:

a. Write a brief definition of an atomic mass unit (amu).

b. Explain how the carbon-12 isotope is used to define the amu.

c. Describe how the mass of a single atom or molecule is measured in amu.

d. Why is it important to have a standard unit like the amu in chemistry?

In everyday life, units such as pair and dozen are used to represent a specific number of items. Scientists use the term mole to represent a specific number of elementary entities (atoms, ions or molecules).

One mole of a substance is defined as the amount of substance that contains as many elementary entities as there are atoms in 12 g of the carbon-12 isotope.

The term “elementary entities” refers to the basic units that make up a substance.

These can include atoms, molecules, ions, electrons, protons etc.

1 mole of every substance contains 6.02 × 10²³elementary entities.

For example, 1 mole of magnesium metal contains 6.02 × 10²³atoms inside it.

How do you determine the number of particles (N) of a substance contained in each number of moles (n)?

Number of formula units in 1 mole of any substance = 1 × 6.02 × 10²³Number of formula units in 2 moles of any substance = 2 × 6.02 × 10²³ Number of formula units in n mole of any substance = n × 6.02 × 10²³N = n × 6.02 × 10²³But 6.02 × 10²³is termed Avogadro’s number or constant and it is denoted by N_(A) or L.

Mathematically, n = N_ N_(A) or n = N_ L Where, n = number of moles (amount of substance) measured in mol. The mole is the base unit of the fundamental quantity called the amount of substance.

N = number of entities L = Avogadro’s number expressed as defined particles mol⁻¹. L is a molar quantity, that is, a quantity expressed per mole.

Worked Example 2.4

Calculate the number of moles contained in 9.5 × 10²³molecules of oxygen.

[ L = 6.02 × 10²³]

Solution:

Use the problem-solving approach.

Analyse the question Known Number of molecules Avogadro’s Constant Formula to use: n = N_ L Unknown Amount of substance Solve: Apply the formula N = 9.5 × 10²³ L = 6.02 × 10²³n = ?

n = N_ L Substituting the values, n = 9.5 × 10²³molecules/6.02 × 10²³molecules⁻¹n = 1.58 mol

Activity 2.5: Trial Questions

Find the answers to the following questions:

1. How many moles are there in 1.204 × 10²⁴molecules of water (H₂O)?

2. Calculate the number of moles in 3.011 × 10²²atoms of helium (He).

3. If you have 5.000 × 10²³molecules of carbon dioxide (CO₂), how many moles do you have?

Activity 2.6: Exploring the mole concept Materials needed:

Sample substances (e.g., salt, water, sugar); calculators, 2 small cups or containers, electronic balance, periodic table, table salt (NaCl), water (H₂O) Steps:

1. Use a large poster or chart to show what a mole represents: 6.02 × 10²³elementary entities.

2. Use the periodic table to calculate the molar masses of given compounds:

Sample substance Formula of compound Molar mass in g/mol Actual mass as measured in (g) water H₂ O H₂ O 2 H : 2 × 1 = 2 1 O : 1 × 16 = 16 = 18 g mol⁻¹18 g Sodium chloride NaCl

3. Measure and record the mass of water equal to 1 mol in a container.

4. Determine the molar mass of NaCl.

5. Measure and record the mass of NaCl equal to 1 mol in another container

6. Calculate the number of moles of (H₂ O and NaCl):

n(H₂ O) = m(H₂ O)_ M(H₂ O) = 18 g_ 18 g mol⁻¹= 1.0 mol n(NaCl) = m(NaCl)_ M(NaCl) = g_ g mol⁻¹= mol

7. Count the number of atoms in each molecule:

• Water (H₂ O ) has 2 hydrogen atoms and 1 oxygen atom.

• Sodium chloride (NaCl ) has 1 sodium ion and 1 chloride ion.

Explanation:

1 mol of water contains 6.02 × 10²³molecules of water, which means it contains 2 × 6.02 × 10²³hydrogen atoms and 6.02 × 10²³oxygen atoms.

Discussion Questions:

1. Do 1 mol of H₂O and 1 mol of NaCl have the same mass?

2. Would 1.50 mol of H₂O have the same number of particles as 1.50 mol of NaCl

Hello, learner, you are about to be introduced to how the amount of substance can be used to calculate different quantities such as the number of entities, mass and volume of gases. You will also learn how to relate the mole concept in the preparation of standard solutions.

Recall that, Amount of substance = Number of entities______________ Avogadro′sconstant n = N_ N_(A) or n = N_ L To calculate for the number of entities, multiply both sides of the equation by L N = n × L Number of Entities = number of moles of substance × Avogadro′s Constant

Worked Example 2.5

Calculate the number of atoms contained in 0.25 mol of sodium.

[ L = 6.02 × 10²³]

Solution (Using problem-solving strategy):

i. Analyse the question Known:

n = 0.25 mol L = 6.02 × 10²³Unknown:

N = ?

Use the formula: N = n × L Solve: Apply the formula By definition, N = n × L N = 0.25 mol × 6.02 × 10²³N = 1.51 × 10²³• Evaluate: Check to see if the answer makes sense and if the correct unit is stated.

Moles in Mass of Atoms or Molecules

The Molar mass (M) of a substance, is the relative atomic mass (Aᵣ) or relative molecular mass (Mᵣ) expressed in grams per mole; e.g. the M(H₂)is 2 g mol⁻¹, or the M(CaC O₃)is 100 g mol⁻¹How do you determine the mass of a given number of moles of a substance?

Mass of 1 mole of atom X = 1 × M(X) Mass of 1 mole of O = 1 × 16 = 16g Mass of 2 moles of O = 2 × 16 = 32g Mass of n moles of O = n × M = m g m = n × M Amount of substance (n) = mass of substance (m)________________ Molar mass (M)

Worked Example 2.6

Calculate the number of moles contained in 20 g of Aluminium atoms [Al = 27]

Solution:

Analyse the question Known:

Mass (m)= 20 g Relative atomic mass ( Aᵣ) = 27 Molar mass = 27 g mol⁻¹Formula to use:

Amount of substance (n) = mass of substance (m)________________ Molar mass (M) Unknown:

Number of moles (n) Solve: Apply the formula By definition, Substituting the values, Amount of substance (n) = 20 g_ 27 gmol⁻¹= 0.74 mol Therefore, the number of moles of aluminium atoms is 0.74 mol Calculating for the mass of a given amount of substance Recall that, Amount of substance (n) = mass of substance (m)________________ Molar mass (M) Making m the subject yields, m = n × M

Worked Example 2.7

Calculate the mass of 0.50 mol of water H₂ O. [ H = 1, O = 16 ]

Solution:

Analyse the question Known:

Number of moles (n) = 0.5 mol Relative Atomic masses [ H = 1, O = 16 ] Formula to use:

m = n × M Unknown:

Mass (m) =?

Solve: Apply the strategy Calculate the M M (H₂ O)= 2(1)+ 16 = 18 g mol⁻¹Calculate the mass By definition, m = n × M m = 0.5 × 18 m = 9 g Therefore, the mass of water is 9g.

Activity 2.7: Trial Question

Calculate the number of moles in 36 g of water (H₂O).

Quantity of Substance and Molar Volume of Gases

The volume occupied by a gas depends on:

1. Quantity of substance

2. Temperature

3. Pressure of the gas At standard temperature of 273 K and pressure of 101.3 kPa (known as standard temperature and pressure, or s.t.p.), the volume occupied by one mole of any gas is called molar volume, denoted by Vₘ. Vₘ is a constant and has a value of 22.4 dm³mol⁻¹at s.t.p.

The Molar Volume Vₘ, the number of moles of substance n and volume of gas are related by the formula:

Amount of substance (n) = volume of substance in dm³(V)______________________ Molar volume (Vₘ) Multiplying both sides of the equation by Vₘ gives V = n × Vₘ NB:

a. This equation is used to calculate the volume of a gas at s.t.p., given the quantity of substance or number of moles. The equation (V = n × Vₘ) cannot be used for any other values of temperature or pressure.

b. If the gas volume is measured in cm³, convert to dm³.

c. You can calculate the volume of a named gas, given the formula and relative atomic masses of the elements.

Worked Example 2.8

Calculate the volume occupied by 0.75 mol of ammonia gas (N H₃) at s.t.p.

[Vₘ = 22.4 dm³mol⁻¹]

Solution:

Use the problem-solving approach Known:

Mole (n) = 0.75 mol Vₘ = 22.4 dm³mol⁻¹Formula to use: V = n × Vₘ Unknown:

Volume of gas, V =?

Solve: Apply the problem-solving approach By definition, V = n × Vₘ Substituting the values, V = 0.75 mol × 22.4 dm³mol⁻¹V = 16.8 dm³The concept map of the relationship between mole and other variables are:

Fig. 2.1: A concept map showing the relationship between mole and other variables

Activity 2.8: Trial Question

Calculate the number of moles of oxygen gas (O₂) present in 44.8 dm³ at standard temperature and pressure (stp).

Calculating the number of entities, mass and volume of a gas using mathematical equations You can use a periodic table, calculator and worksheets with specific problems to help you to calculate or determine the number of entities (ion, atoms, molecules, etc), mole and mass of gases.

1. Conversions between Mass and Number of Particles Fig. 2.2: Interrelationships between mass, number of particles and moles.

Figure 2.2 illustrates that mass, number of particles, and moles are all interrelated.

To convert between mass and number of particles, a conversion to moles is required first.

2. Converting Mass to Number of Particles

Worked Example 2.9

How many molecules are present in a 17.5 g sample of P₄ O₁₀? [P = 31, O = 16]

Solution:

Step 1: List the known quantities and plan the problem.

Known:

sample mass = 17.5 g P₄ O₁₀ molar mass of O₂= 284 g/mol Unknown:

number of molecules of P₄ O₁₀ Working:

• First, convert the mass of P₄ O₁₀ to moles.

• Second, convert moles of P₄ O₁₀ to the number of molecules.

• Step 2: Calculate N( P₄ O₁₀) = n( P₄ O₁₀) × L = m( P₄ O₁₀)_ M( P₄ O₁₀) L = 17.5 g_ 284 g mol⁻¹× 6.02 × 10²³molecules mol⁻¹N( P₄ O₁₀) = 3.7 × 10²²molecules

Worked Example 2.10

Calculate the number of sodium ions present in 2.5 g of Na₂ S O₄.

[ Na = 23, S = 32, O = 16, L = 6.02 × 10²³particles mol⁻¹]

Solution:

· Step 1: Identify the molar mass of Na₂ S O₄ Na: 2 × 23.0 = 46.0 g / mol S O₄ ²⁻: 32. +(4 × 16.0) = 96.0 g / mol Total molar mass: 46.0 + 96.0 = 142.0 g / mol

• Step 2: Calculate the number of moles of Na₂ S O₄:

n(Na₂ S O₄) = m(Na₂ S O₄)_ M(Na₂ S O₄) = 2.5 g_ 142.0 ᵍ_ ₘₒₗ = 0.0176 mol

• Step 3: Identify the number of sodium ions (Na+) in the formula (Na₂ S O₄): 2

• Step 4: Calculate the total number of moles of sodium ions:

Na₂ S O₄ ⎯→2 Na+ + S O₄ ²⁻0.0176 mol Na₂ S O₄_______________ 1 mol Na₂ S O₄ × 2 mol 2 Na+ = 0.0352 mol Na+

• Step 5: Convert moles of sodium ions to number of particles using Avogadro’s number (L):

Number of particles = Moles × Avogadro′s number (L) = 0.0352 mol × 6.02 × 10²³ions mol⁻¹= 2.12 × 10²²ions Therefore, there are 2.12 × 10²²sodium ions (Na+) in 2.5 g of Na₂ S O₄.

3. Converting Mass to Moles

Worked Example 2.11

Calculate the number of moles present in 2.5 g of Na₂ S O₄.

Solution:

• Step 1: List the known quantities and plan the problem.

Known:

mass of Na₂ S O₄ produced = 2.81 g Unknown:

amount of Na₂ S O₄ produced in moles One conversion factor will allow us to convert from mass to moles.

• Step 2: Calculate.

First, it is necessary to calculate the molar mass of Na₂ S O₄. The molar mass is 142 g/mol.

n(Na₂ S O₄)= m(Na₂ S O₄)_ M(Na₂ S O₄) = 2.5 g_ 142 ᵍ_ ₘₒₗ = 0.0202 g

4. Converting Moles to Mass

Worked Example 2.12

Calculate the mass of 0.25 mol of Na₂ S O₄.

Solution:

• Step 1: List the known quantities and plan the problem.

Known:

0.25 mol mol Na₂ S O₄ molar mass of Na₂ S O₄ = 142.00 g/mol Unknown:

0.25 mol of Na₂SO₄= ? g The molar mass of Na₂ S O₄ will allow us to convert from moles of Na₂ S O₄ to grams.

• Step 2: Calculate.

n(Na₂ S O₄)= m(Na₂ S O₄)_ M(Na₂ S O₄) Make m(Na₂ S O₄) the subject:

m(Na₂ S O₄) = n(Na₂ S O₄) × M(Na₂ S O₄) = 0.25 mol × 142.00 ᵍ_ ₘₒₗ = 35.5 g

5. Converting moles to Number of Particles

Worked Example 2.13

Calculate the number of oxide ions contained in 0.5 mol of Al₂ O₃.

[ L = 6.02 × 10²³particles mol⁻¹]

Solution:

• Step 1: The number of oxide ions (O²⁻) in the formula = 3

• Step 2: Calculate the total number of moles of oxide ions using the balanced equation below:

Al₂ O₃ ⎯⎯⟶3 O²⁻+ 2 Al³⁺ 1 mol of Al₂ O₃ contains 3 mols O²⁻∴ 0.5 mol of Al₂ O₃ = 0.5 mol of Al₂ O₃_____________ mol of Al₂ O₃ × 3 mols O²⁻= 1.5 mol O²⁻• Step 3: Convert moles of oxide ions to number of particles using Avogadro’s number (L):

Number of particles = moles × Avogadro′s number (L) N(O²⁻) = n(O²⁻) × L = 1.5 mol × 6.02 × 10²³ions mol⁻¹= 9.03 × 10²³ions Therefore, there are 9.03 × 10²³oxide ions in 0.5 mol of Al₂ O₃.

6. Conversions between Moles and Gas Volume

a. Converting Gas Volume to Moles

Worked Example 2.14

Calculate the number of moles contained in 250 cm³of carbon dioxide gas at s.t.p.

[Vₘ = 22.4 dm³mol⁻¹]

Solution:

• Step 1: List the known quantities and plan the problem.

• Step 2: Convert the volume from cm³to litres volume of CO₂ = 250 cm³= 250/1000 = 0.250 dm³Unknown:

moles of CO₂ Use the molar volume to convert from dm³to moles:

1 mol = 22.4 dm³• Step 3: Calculate.

n(CO₂) = V( CO₂)_ Vₘ = 0.250 dm³___________ 22.4 dm³mol⁻¹= 0.0112 mol

Worked Example 2.15

Calculate the volume occupied by 0.25 moles of carbon dioxide gas at s.t.p.

[Vₘ = 22.4 dm³mol⁻¹]

Solution:

Multiply the number of moles by the molar volume to get the volume:

Volume = Moles × Molar volume = 0.25 mol × 22.4 dm³mol⁻¹l = 5.6 dm³Therefore, 0.25 moles of carbon dioxide (CO₂) occupy a volume of 5.6 dm³at stp.

b. Converting volume to number of particles

Worked Example 2.16

Calculate the number of carbon dioxide (CO₂) molecules in 500 cm³at stp, [ L = 6.02 × 10²³particles mol⁻¹; Vₘ = 22.4 dm³mol⁻¹]

Solution:

• Step 1: Convert the volume from cm³to dm³: 500 cm³= 0.5 dm³• Step 2: Calculate the number of moles:

moles = Volume_ Molar volume = 0.5 dm³___________ 22.4 dm³mol⁻¹= 0.0223 mol

• Step 3: Convert moles to number of particles (molecules):

Number of particles = moles x Avogadro′s number (L) = 0.0223 mol × 6.02 × 10²³molecules mol⁻¹= 1.34 × 10²²molecules Therefore, there are 1.34 × 10²²CO₂ molecules present in 500 cm³at stp.

Activity 2.9: Trial Questions

A chemical compound has a molar mass of 180 g/mol. Calculate:

1. The mass of 2 moles of the compound.

2. The number of molecules in this amount.

3. The volume occupied by these molecules at s.t.p.

A solution is a uniform mixture of a solute and a solvent. The quantity of solute per unit volume of solution is termed as concentration. To be able to compare the concentration of solutions, we use standard units.

Types of Concentration

1. Quantity of substance concentration (Molarity) It is defined as the quantity of solute (number of moles of solute) dissolved in one cubic decimetre of the solution.

Mathematically, it is expressed as:

Concentration in mol dm⁻³(C)= amount of substance in moles (n)______________________ Volume of solution in dm³(V) The number of moles of solute n can be made the subject as n = C ×V NB:

a. This equation can be used to calculate the number of moles of solute required for a given volume of a specified concentration.

b. The volume of solution required can be calculated given the number of moles and a specified concentration.

c. If the volume of solution is given in cm³, it must be converted to dm³by dividing it by 1000.

Worked Example 2.17

Calculate the number of moles in 250 cm³of 0.500 mol dm⁻³sulphuric acid

solution.

Solution:

Use the problem-solving approach

• Analyse the question Known:

Volume, V = 250 cm³Concentration, C = 0.500 mol dm⁻³Formula to use: n = C × V Unknown:

Number of moles, n =?

• Solve:

Convert the volume to dm³by dividing it by 1000.

V = 250/1000 = 0.250 dm³By definition, n = C × V Substituting the values, n = 0.500 mol dm⁻³× 0.250 dm³n = 0.125 mol

2. Mass Concentration (Concentration in g dm⁻³) It is defined as the mass of solute dissolved in one cubic decimetre of the solution.

It is denoted by ρ (not to be confused with density).

Mathematically, it is defined as:

Concentration in g dm⁻³(ρ)= mass of solute in grams (m)____________________ Volume of solution in dm³(V) The mass of the solute m can be made the subject as follows:

m = ρ ×V NB: This equation can be used to calculate:

a. The mass of solute in a given volume of a solution of a specified concentration.

b. The volume of solution needed when the mass of solute and specified concentration is given.

Mass of solute required to prepare a given volume of a standard solution Recall that, C = n_ V……. (1) butn = m_ M…….. (2) Substituting (2) into (1) gives C = m/M ×V Hence m = C ×M ×V This equation is used to calculate the mass of solute required to prepare a given volume of a standard solution.

Worked Example 2.18

Calculate the mass of NaOH required to prepare 250cm³of 0.50mol dm⁻³sodium hydroxide solution.

[Na = 23, H = 1, O = 16]

Solution:

Use the problem-solving approach:

• Analyse the question Known Volume of solution, V = 250 cm³Concentration, C = 0.50 mol dm⁻³Relative atomic masses: Na = 23, H = 1, O = 16 Formula to use: m = C × M × V Unknown:

Mass, m =?

• Solve: Apply the formula Calculate the M (NaOH) = 23 + 16 + 1 = 40 g mol⁻¹Convert the volume to dm³by dividing it by 1000.

V = 250/1000 = 0.250 dm³Substituting the values into the formula, m = 0.50 × 40 × 0.25 = 5g Therefore, the mass of NaOH required is 5g.

Relationship between molar concentration and mass concentration Consider the equation:

C = m_ M × V Also recall that, ρ = m/V . Combining the two equations yields, C = ρ__ M Where, c = Concentration in mol dm⁻³; ρ = mass concentration in g dm⁻³ M = Molar mass in gmol⁻¹Activity 2.10: Trial Questions Calculate the required quantity for the following questions:

1. A solution of hydrochloric acid (HCl) has a concentration of 0.5 mol/dm³.

Calculate the number of moles of HCl present in 250 cm³ (0.25 dm³) of this solution.

2. A solution of sodium hydroxide (NaOH) has a concentration of 0.1 mol/dm³. Determine the volume of this solution needed to contain 0.4 moles of NaOH.

3. Calculate the concentration of a solution that contains 2 moles of sulphuric acid (H₂SO₄) in a volume of 500 cm³ (0.5 dm³).

A standard solution is a solution whose concentration is accurately known.

Primary Standard

A primary standard is a substance that is usually available in pure form or a state of known purity, which is used in preparing a standard solution. Examples are sodium carbonate and potassium iodate Properties of a primary standard It should be available in pure form or easily purified.

It must be stable, that is, it must not lose weight or take up water during weighing.

It must have a reasonably high relative formula mass.

It must react speedily without side reactions with the substance being standardised.

It should have high solubility How to prepare a standard solution from a solid solute

1. Determine the mass of the solute required to make the appropriate concentration and volume of desired solution.

2. Weigh accurately the solute in a beaker.

3. Add distilled water and its contents to the beaker and swirl to dissolve the solid.

NB: The beaker must have a lower volume than the standard volumetric flask being used.

4. Transfer the solution to the required standard volumetric flask through a funnel.

5. Rinse the stirrer and the beaker used into the flask, then add more distilled water until the meniscus lies on the calibration mark.

6. Invert the stoppered flask a few times to mix.

7. Label the solution.

Activity 2.11: Practise the example of preparing a standard solution from a solid solute Task Example: Prepare 250 cm³of 2.0 mol dm⁻³NaOH solution.

Materials needed Beakers, volumetric flask, burette, electronic balance, stirring rods, funnels, and wash bottles, safety goggles and lab coats, stock solution, distilled water, and calculator.

Steps:

1. Perform calculations to determine the required quantities:

• Calculate the amount of the solute needed to prepare the solution.

Use the formula:

c(NaOH) = n(NaOH)_ V(NaOH)

• Convert volume from cm³to dm³250 cm³= 250_ 1000 = 0.250 dm³n(NaOH) = c(NaOH) × V(NaOH) = 2.0 mol dm⁻³× 0.250 dm³= 0.50 mol

• Convert moles to grams Find the molar mass of NaOH M(NaOH) = 23 + 16 + 1 = 40 g mol⁻¹m(NaOH) = n(NaOH) × M(NaOH) = 0.50 mol × 40.0 g mol⁻¹= 20.0 g

2. Accurately weigh the calculated mass of sodium hydroxide using the electronic balance.

3. Transfer the weighed solute to a beaker and add a small volume of distilled water.

4. Stir with a stirring rod until the solute is completely dissolved.

5. Transfer the small solution in the beaker into the right size volumetric flask (e.g. 250 cm³) with the aid of a funnel.

6. Rinse the beaker with distilled water and add the rinsing to the volumetric flask to ensure all solute is transferred.

7. Cover the volumetric flask and invert it several times to ensure thorough mixing of the solution.

8. Label the volumetric flask with the concentration, the solute, the date.

9. Place the prepared standard solution in a designated place for later use.

Preparation of Standard Solution from Concentrated Solution

1. Use the dilution formula (C₁V₁= C₂V₂) to calculate the volume of the concentrated solution required.

2. Pour some distilled water into the required standard volumetric flask.

3. Measure the stock or concentrated solution and transfer it into the distilled water in the volumetric flask.

4. Swirl the flask and its content and top the solution to the calibration mark with distilled water.

5. Label the solution.

Determination of the concentration of a stock solution The commercial stock solution usually contains chemical assay (that is the label on their container, specifying the purity, density, molecular mass, and other relevant information).

1. Calculate the mass of the substance in 1 dm³.

2. Calculate the mass of the pure substance in 1dm³by multiplying by the percentage purity.

3. Divide this mass by the molar mass to get the concentration.

Mathematically use the formula:

Concentration (C)= Density (ρ) × 1000 ×percentage purity (%)_______________________________ Molar mass (M) × 100

Activity 2.12: Preparing standard solutions of various concentrations Materials needed Beakers, volumetric flask, burette, electronic balance, stirring rods, funnels, and wash bottles, safety goggles and lab coats, stock solution, distilled water, and calculator.

Steps:

1. Observe safety precautions, such as wearing goggles and lab coats, handling chemicals carefully, and using apparatus properly.

2. Identify and label the set of apparatus required.

3. Prepare solutions of specific concentration such as 0.10 mol dm⁻³, 1.0 mol dm⁻³, 0.50 mol dm⁻³of sodium hydroxide, sodium chloride and hydrochloric acid.

Activity 2.13 Preparing a standard solution from a concentrated solution Materials needed Beakers, volumetric flask, burette, electronic balance, stirring rods, funnels, and wash bottles, safety goggles and lab coats, stock solution, distilled water, and calculator.

Task Example: Prepare a 250 cm³ solution of HCl with a concentration of 2.0 mol/dm³ using a stock solution of known density and percentage purity.

Steps:

Prepare a 250 cm³solution of HCl of concentration 2.0 mol dm⁻³from a stock HCl solution of specifications:

Density = 1.19 g cm⁻³Percentage purity = 37% Molar mass of HCl = 36.5 g mol⁻¹1. Calculate the concentration of the stock solution:

• Calculate the mass of HCl in 1 cm³of the stock solution:

Mass of solution = density × volume = 1.19 g cm⁻³× 1 cm⁻³= 1.19 g Given that the solution is 37% HCl by mass:

Percentage purity of HCl = 37% Mass of HCl = 1.19 g × 0.37 = 0.4403 g Now concentration of the stock solution:

n(HCl) = m(HCl)_ M(HCl) = 0.4403 g_ 36.5 g mol⁻¹= 0.01207 mol Since this is the amount in 1 cm³, the concentration of the stock

solution is:

1 cm³of the stock solution contains 0.01207 mol HCl ∴ 1000 cm³= 0.01207 mol_ 1 cm³× 1000 cm³= 12.07 mol HCl Concentration of the stock solution is 12.07 mol dm⁻³HCl

2. Calculate the volume of stock solution needed.

We want to prepare 250 cm³ (0.250 dm³) of a 2.0 mol/dm³ HCl solution.

Using the dilution formula:

C₁ V₁ = C₂ V₂ C₁ = 12.07 mol dm⁻³V₁ = volume of stock solution needed C₂ = 2.0 mol dm⁻³V₂ = 250 cm³ V₁ = C₂ V₂_ C₁ = 2.0 mol dm⁻³× 250 cm³_________________ 12.07 mol dm⁻³= 41.4 cm³

3. Measure 41.4 cm³ of the stock solution using a burette.

4. Transfer the stock solution to a 250 cm³ volumetric flask.

5. Add distilled water to the flask up to the 250 cm³ mark.

6. Invert the flask to mix the solution thoroughly.

Review Question 2.1

The atomic mass unit of Carbon-12 is 1.6603 × 10⁻²⁴g . If the average mass of an atom X is 6.63310 × 10⁻²³g . Determine its relative atomic mass.

Review Questions 2.2

1. Calculate the relative molecular masses (Mᵣ) of the following substances:

a. Formic Acid (CH₂O₂) b. Acetic Acid (C₂H₄O₂)

c. Ethanol (C₂H₆O) d. Acetone (C₃H₆O)

e. Citric Acid (C₆H₈O₇) f. Sucrose (C₁₂H₂₂O₁₁) For reference, the atomic masses are approximately: H = 1; O = 16; C = 12

2. Determine the relative molecular mass of the following:

a. N H₃ b. C H₄

c. C₁₆ H₁₆ F₃ NO d. S²⁻e. SO₄ ²⁻f. Na₂ CO₃ . 10 H₂ O Aᵣ: H = 1.0; C = 12.0; N = 14.0; O = 16.0; F = 19.0; Na = 23.0; S = 32.0

Review Questions 2.3

Find the answers to the following questions:

1. How many moles of sodium chloride (NaCl) are present in 1.505 × 10²⁴formula units of NaCl?

2. Determine the number of moles in 2.409 × 10²³atoms of gold (Au).

3. How many moles are in 8.436 × 10²⁴molecules of glucose (C₆H₁₂O₆)?

4. Calculate the number of moles in 6.022 × 10²¹molecules of nitrogen gas (N₂).

5. What is a mole and why is it important in chemistry?

6. Explain how the mole relates to Avogadro’s number.

7. The number of molecules of ammonia gas is 12.04 × 10²³. Calculate the number of moles of ammonia gas. [L = 6.02 × 10²³]

8. Calculate the number of oxygen molecules in 0.5 mol of oxygen gas.

[ L = 6.02 × 10²³]

9. Calculate the number of atoms in 16 g of copper, Cu. [Cu = 63.5, L = 6.02 × 10²³]

Review Questions 2.4

1. Determine the number of moles in 88 g of carbon dioxide (CO₂).

2. How many moles are in 48 g of methane (CH₄)?

3. Calculate the number of moles in 180 g of glucose (C₆H₁₂O₆).

4. Determine the number of moles in 58.5 g of sodium chloride (NaCl).

5. Calculate the mass of 3 moles of water (H₂O).

6. Determine the mass of 2 moles of carbon dioxide (CO₂).

7. Find the mass of 4 moles of methane (CH₄).

8. Calculate the mass of 0.5 moles of glucose (C₆H₁₂O₆).

9. Determine the mass of 1.5 moles of sodium chloride (NaCl).

Review Questions 2.5

Calculate the required values for the following questions

1. Determine the volume occupied by 0.5 moles of carbon dioxide (CO₂) at STP, in dm³.

2. How many moles of hydrogen gas (H₂) are there in a 67.2 dm³ container at STP?

3. Calculate the volume occupied by 2 moles of nitrogen gas (N₂) at STP, in dm³.

Review Questions 2.6

1. A sample of carbon dioxide (CO₂) has a molar mass of 44 g/mol. Calculate:

a. The mass of 1.5 moles of carbon dioxide.

b. The number of molecules in this amount.

c. The volume occupied by these molecules at STP.

2. A sample of propane (C₃H₈) has a molar mass of 44 g/mol. Calculate:

a. The mass of 3 moles of propane.

b. The number of molecules in this amount.

c. The volume occupied by these molecules at STP.

3. The molar mass of CO₂is 44 g/mol. How many moles of CO₂are present in 124 g sample of CO₂.

4. What is the mass of 5.0 × 10²³molecules of NO₂?

[ N = 14.0; O = 16.0L = 6.02 × 10²³]

5. a. How many molecules are there in 4.00 mol of glucose, C₅ H₁₂ O₆?

b. How many atoms of carbon?

c. How many atoms of hydrogen?

[ L = 6.02 × 10²³]

Review Questions 2.7

1. A solution of potassium permanganate (KMnO₄) has a concentration of 0.02 mol/dm³. Find the volume of this solution required to contain 0.1 moles of KMnO₄.

2. Determine the volume of a 0.5 mol/dm³ solution of glucose (C₆H₁₂O₆) needed to obtain 0.15 moles of glucose.

3. Calculate the mass of sodium chloride (NaCl) dissolved in 500 cm³ (0.5 dm³) of a solution with a concentration of 0.4 g/dm³.

4. Find the concentration of a solution if 30 g of potassium nitrate (KNO₃) is dissolved in 150 cm³ (0.15 dm³) of water.

5. Determine the volume of a solution with a mass of 25 g and a concentration of 0.1 g/dm³.

6. Calculate the mass of copper sulphate (CuSO₄) in 250 cm³ (0.25 dm³) of

solution with a concentration of 0.8 g/dm³.

7. Find the concentration of a solution if 50 g of sucrose (C₁₂H₂₂O₁₁) is dissolved in 500 cm³ (0.5 dm³) of water.

8. A solution of hydrochloric acid (HCl) has a concentration of 0.4 mol/dm³.

Calculate the mass of HCl in 300 cm³ (0.3 dm³) of this solution.

9. Calculate the volume of a 0.2 mol/dm³ solution of sulphuric acid (H₂SO₄) needed to obtain 50 g of H₂SO₄.

10. Find the concentration of a solution if 150 g of sodium hydroxide (NaOH) is dissolved in enough water to make 500 cm³ (0.5 dm³) of solution.

11. Calculate the mass of a sample of copper sulphate (CuSO₄) in 200 cm³ (0.2 dm³) of solution with a concentration of 1.5 g/dm³.

12. A solution of ethanol (C₂H₆O) has a concentration of 0.8 mol/dm³. Find the volume of this solution required to obtain 100 g of ethanol.

13. List five (5) apparatus used in preparing a standard solution from solid solutes.

14. Work in small groups to prepare 250 cm³of 2.0 mol dm⁻³solution of sodium hydroxide in the laboratory. [H = 1.0, O = 16.0, Na = 23.0 ]

15. What are some potential sources of error when preparing standard solutions, and how can they be minimised?

In this section, you will master essential skills, including IUPAC nomenclature for inorganic compounds, writing compound formulas based on chemical laws, and balancing equations. You will also engage in stoichiometric calculations to deepen your understanding of chemical reactions.

At the end of this section, you should be able to:

• Use IUPAC nomenclature to name inorganic compounds, write the formulae of compounds based on the laws of chemical combination and write balanced chemical equations.

• Perform calculations involving stoichiometric relationship.

Key Ideas:

• Oxidation number is the charge of an ion in a compound.

• Empirical formula is the simplest ratio of atoms of each element in a molecule.

• Molecular formula is the actual number of atoms of each element in a molecule.

• Structural formula is a formula showing the arrangement of atoms in a molecule.

• Prefix indicates the number of atoms of an element in a compound (e.g., mono-, di-, tri-, tetra).

• Suffix indicates the type of compound or ion (e.g., -ide, -ate, ite).

• Reactant is the substance present at the start of a chemical reaction, which takes part in the reaction itself.

• Product is the substance formed as a result of a chemical reaction.

• Stoichiometry is the relationship between reactants and products in a chemical reaction.

• Coefficient is the number placed in front of a chemical formula in an equation to balance the equation.

• Mole ratio is the ratio of the number of moles between two or more substances.

• Yield refers to the amount (mass) of product obtained from a chemical reaction.

• Actual yield is the amount of (mass) product obtained from a reaction in practice.

• Theoretical yield: the maximum amount (mass) of product that could be produced from a given amount of reactant, based on stoichiometric calculations.

• Percent yield is the ratio of the actual yield to the theoretical yield, expressed as a percentage.

• Limiting reagents are the reactants in a chemical reaction that are consumed completely, and therefore limit the mass of the product formed.

• Excess reagents are the reactants that are left over when the reaction has stopped.

The nomenclature of inorganic compounds is based on the oxidation number system.

The oxidation number of an atom is the number of electrons gained or lost by an atom when forming a compound.

Rules for Assigning Oxidation Number

1. The oxidation number of elements in their elemental state is zero.

2. The oxidation state of oxygen in most compounds is -2 except in peroxide (-1) and superoxide (-½).

3. The oxidation state of hydrogen is (+1). When it is bonded to a non-metal and (-1) when bound to a metal.

4. The oxidation state of an ion is equal to the charge on the ion.

5. For a neutral molecule or polyatomic ion, the sum of the oxidation numbers of all the atoms must be equal to the total charge on it.

Rules for Naming Binary Ionic Compounds

1. Name the cation (positive ion) first followed by the name of the anion (negative ion).

2. The name of the cation is the name of the metal.

3. For metals with atomic numbers above 20, indicate the oxidation state in Roman numerals and brackets.

4. The anions are named by replacing the suffix with ‘ide’

Example:

NaCl – Sodium chloride FeCl₃– Iron (III) chloride Rules for Naming Simple Acids

1. Use the prefix ‘hydro’, then the root name of the central atom.

2. Add the suffix ‘ic’ to the root name.

3. Add the word ‘acid’.

Example:

HCl – Hydrochloric acid HI – Hydroiodic acid Rules for Naming Oxoacids

1. Use the prefixes ‘oxo’, ‘dioxo’, ‘trioxo’ and ‘tetraoxo’ to indicate the number of oxygen atoms present.

2. Add the root name of the central atom.

3. Add the suffix ‘ate’ followed by the oxidation state of the central atom in Roman numerals and brackets.

4. Add the word ‘acid’.

Example:

H₂CO₃– trioxocarbonate (IV) acid H₂SO₄– tetraoxosulphate (VI) acid Rules for Naming Acid Salts

1. Name the metal cation first followed by the name of the oxosalt.

1. If the cation has a relative atomic number above 20, its oxidation state should be indicated in Roman numerals and brackets.

2. Add the word ‘hydrogen’.

3. Name the oxoanion as usual without the word ‘ion’.

Example:

NaHSO₄– Sodium hydrogen tetraoxosulphate (VI) Rules for Naming Simple Non-Ionic Compounds

1. Name the electropositive element first.

2. Add the root name of the anion.

3. Add the suffix ‘ide’.

Example:

HCl – hydrogen chloride SiC – Silicon carbide Rules for Naming Molecular compounds where a pair of elements form different compounds with different number of oxygen atoms

1. Name the electropositive element first.

2. Indicate its oxidation state in Roman numerals and brackets.

3. Name the electronegative element as usual.

Example:

CO₂– Carbon (IV) oxide N₂O₃– Nitrogen (III) oxide Rules for Naming Hydrated Salt

1. Name the anhydrous part first.

2. Use the prefixes mono, di, tri, tetra, penta, hexa, hepta, octa, etc.to indicate the moles of water of crystallisation.

3. Add the suffix ‘hydrate’ to it.

Example:

CuSO₄.5H₂O – Copper (II) tetraoxosulphate (VI) pentahydrate

Activity 3.1: IUPAC naming conventions for various groups of inorganic compounds.

Materials needed:

Access to the internet or library resources, textbooks on inorganic chemistry, presentation tools (e.g., PowerPoint, poster boards), markers, paper, and other stationery Steps:

1. This Activity is best done in a classroom environment.

2. Your teacher will help you form groups of 5.

3. Your group will be assigned a specific type of inorganic compound to research into such as:

• Simple binary compounds (e.g., NaCl, CO₂)

• Oxides and hydroxides (e.g., H₂O, NaOH)

• Acids (e.g., H₂SO₄, HCl)

• Bases (e.g., KOH, NH₄OH)

• Salts (e.g., NaCl, K₂SO₄)

4. Research into the IUPAC rules for naming specific compounds.

5. Compile your findings into:

• The basic principles of IUPAC nomenclature for your compound type.

• Examples of compounds with their systematic names.

6. Prepare your presentation. This should include:

• An introduction to your inorganic compound assigned to your group.

• A detailed explanation of the IUPAC naming rules of that compound.

• Several examples with explanations.

• A list of exceptions to any IUPAC rules for certain chemicals.

• Visual aids such as diagrams, flowcharts, or models.

A sample outline for your presentation has been given at the end of this

activity to guide you.

7. Deliver your presentation to your peers in class.

8. Be sure to partake in any Questions and Answer sessions where other groups can ask questions or seek clarifications.

Sample Group Presentation Outline

1. Naming Simple Binary Compounds

IUPAC Naming Rules:

Rule 1: Naming the cation (positive ion) first.

Rule 2: Naming the anion (negative ion) second, modifying its name to end in “-ide”.

Example: AlCl₃is named aluminium chloride.

CrCl₃is chromium (II) chloride (Cr is in +3 oxidation state) SO₂is sulphur (IV) oxide (S is in +4 oxidation state) Aluminium chloride Al Cl Cl Cl Chromium (III) chloride CrCl Cl Cl

2. Naming Oxides and Hydroxides

IUPAC Naming Rules:

Rule 1: Naming the metal first.

Rule 2: Naming the oxide or hydroxide part.

Example: H₂O is named dihydrogen monoxide, NaOH is named sodium hydroxide SO₂is sulphur (IV) oxide (S is in +4 oxidation state)* Sulphur (IV) oxide S O O *It might be worth noting in the presentation that this is commonly called sulphur dioxide and could be considered an exception to the IUPAC rules as it is so common.

3. Naming Acids

IUPAC Naming Rules:

Binary Acids: Prefix “hydro-” + root of nonmetal + suffix “-ic” + “acid.”

Oxyacids: Root of polyatomic ion + suffix (“-ate” ions,” ions) + “acid.”

Example: HBr is hydrobromic acid, H₂PO₃is trioxophosphate (VI) acid.

A chemical formula is an expression which shows the chemical composition of a compound in terms of the symbols of the atoms involved.

Types of Chemical Formulae

There are three (3) types of chemical formulae:

1. Empirical formula

2. Molecular formula

3. Structural formula Empirical formula The empirical formula of a compound is the simplest whole number ratio of atoms present in a compound.

Determination of the empirical formula

1. State the mass or percentage mass of each element.

2. Divide each mass by its relative atomic mass.

3. Determine the simplest whole number ratio of each element by dividing the results by the least ratio number or scale up by X factor to get the simplest whole number ratio.

4. The simplest integer ratio is then used to write the empirical formula as a right-hand subscript.

Molecular formula Molecular formula is the formula that shows the actual number of atoms of each element in the simplest unit of a substance.

NB: Some compounds have their empirical formula being the same as the molecular formula.

Molecular formula is derived from the empirical formula using the relationship:

Molecular mass = (Empirical mass)ₙ The integer n obtained is used to multiply each element of the empirical formula to get the molecular formula.

Therefore Molecular formula = (Empirical formula)ₙ

Worked Example 3.1

In an experiment to determine the empirical formula of lead sulphide, the following results were obtained:

Mass of lead = 207 g Mass of sulphur = 32 g Calculate the empirical formula of lead sulphide [Pb = 207, S = 32].

Solution

Pb S Mass of element: 207 g 32 g Mole of element, ( ᵐ_ Ar) ²⁰⁷____ 207 32_ 32 Simplest ratio 1 1 Empirical formula: PbS

Activity 3.2: Trial Question

A compound is found to contain 1.92 g of carbon and 0.48 g of hydrogen.

Calculate the empirical formula for the compound.

Structural formula Structural formula is the formula that shows how the atoms in the molecule or compound are bonded to each other. For example, Ethane, with the molecular formula C₂H₆,has a structure formula:

Percentage Composition of Elements in a

Compound The method of calculating the percentage by mass composition of a compound in terms of its constituent element is as follows:

1. Calculate the molecular mass of the compound.

2. Calculate the mass of the specified elements in the compound, considering the number of atoms of each element in the formula:

Percentage of element in a compound = Relative atomic mass × Number of atoms of elements × 100__________________________________________ Relative molecular mass

Worked Example 3.2

Calculate the percentage by mass of nitrogen and hydrogen in NH₃.

Solution:

By definition, Percentage of element in a compound = Relative atomic mass × Number of atoms of elements × 100___________________________________________ Relative molecular mass % of nitrogen = mass of nitrogen_________________ Molar mass of ammonia × 100 % of nitrogen = 14/17 × 100 = 82.35 % % of hydrogen = mass of hydrogen_________________ Molar mass of ammonia × 100 % of hydrogen = 3/17 × 100 = 17.65 %

Activity 3.3: Trial Question

Calculate the percentage by mass of carbon in methane (CH₄).

Activity 3.4: Determining the empirical formula of a named compound This Activity will guide you to determine the empirical formula of a copper chloride compound using the conservation of mass, the law of definite proportions, and the law of multiple proportions.

Materials needed Periodic table, calculators, handouts with problem details, whiteboard and markers Steps:

1. Research and discover the laws of Conservation of Mass, Definite Proportions and Multiple Proportions. You may find the following useful in your search:

a. Conservation of Mass: In a chemical reaction, the total mass of reactants equals the total mass of products.

b. Law of Definite Proportions: A given chemical compound always contains its component elements in a fixed ratio by mass.

c. Law of Multiple Proportions: When two elements form more than one compound, the ratios of the masses of the second element that combine with a fixed mass of the first element are ratios of small whole numbers.

2. Determine the empirical formula of the named compound, e.g. a copper chloride compound with a mass composition of 47.4% copper and 52.6% chlorine.

Given:

Percentage composition: 47.4% Cu, 52.6% Cl Atomic masses: Cu = 64, Cl = 35.5

3. Calculate the Empirical Formula as follows:

Convert Percentage to Mass:

Assume you have 100 grams of the compound.

Mass of Cu = 47.4 grams Mass of Cl = 52.6 grams Convert Mass to Moles:

Moles of Cu = Mass of Cu______________ Atomic mass of Cu = 47.4 g/64 g / mol = 0.7406 mol Moles of Cl = 52.6 g/35.5 g / mol = 1.4817 mol

4. Determine the simplest ratio:

Divide the moles of each element by the smallest number of moles calculated:

Moles of Cu = 0.7406 mol/0.7406 mol = 1 Moles of Cl = 1.4817 mol/0.7406 mol = 2

5. Write the Empirical Formula:

The simplest whole-number ratio is 1:2.

Therefore, the empirical formula is CuCl₂.

Worked Example 3.3

The molar mass of the compound with the empirical formula CH₂O is 180 g/mol.

Determine the molecular formula.

Solution

1. Calculate the Empirical formula molar Mass:

Empirical formula mass of CH₂ O = (1 × 12)+ (2 × 1)+ (1 × 16) = 12 + 2 + 16 = 30

2. Determine the Ratio of Molar Masses to Empirical formula molar mass:

Ratio = Molar mass of compound_____________________ Empirical formula molar mass = 180g / mol/30 g / mol = 6

3. Calculate the Molecular Formula:

Multiply the subscripts in the empirical formula by the ratio calculated:

Molecular formula = C₁H₂O₁× 6 = C₆H₁₂O₆ The molecular formula is C₆H₁₂O₆.

Determining the percentage by mass of an element in a compound.

Worked Example 3.4

Determine the percentage by mass of carbon and hydrogen in C₂H₆. [C = 12 and H = 1.]

Solution

Calculate the Molar Mass of C₂H₆

Find the number of atoms of each element in the compound:

Carbon atoms: 2 Hydrogen atoms: 6 Calculate the total mass of each element in the compound:

The total mass of carbon: 2 × 12 = 24 g/mol Total mass of hydrogen: 6 × 1 = 6 g/mol Calculate the molar mass of C₂H₆ Molar mass of C₂H₆: 24 + 6 =30 g/mol Calculate the percentage by mass of each element %C = Total mass of C_______________ Molar mass of C₂ H₆ × 100 = 24/30 × 100 = 80% %H = Total mass of H_______________ Molar mass of C₂ H₆ × 100 = 6_ 30 × 100 = 20%

1. The Law of Conservation of Mass

The Law of Conservation of Mass states that mass is not created or destroyed in a chemical reaction, i.e. the total mass of the products made is equal to the total mass of the reactants.

2. The Law of Definite Proportion

It says that the proportion by amounts of each element in a pure compound is always the same, no matter how the compound is prepared.

3. The Law of Multiple Proportion

When two elements combine to form more than one compound, the different masses of one element that combine with a fixed mass of the other element are in a ratio of small whole numbers.

It is defined as an expression which uses chemical symbols in the formula to represent the elements and compounds that occur in a chemical reaction.

Types of Chemical Equations

Note: An equation may fall into multiple categories simultaneously.

1. Combustion Combustion is a chemical reaction in which a substance reacts with excess or limited oxygen to give oxides of the components of the substance.

Example: CH₄+ 2O₂→ CO₂+ H₂O

Application in everyday life Fuel combustion generates heat for homes, electricity for the power grid and to generate movement in engines.

2. Synthesis Synthesis a reaction in which two or more simple substances combine to form a more complex compound. That is, the reactants combine to form a single product.

It can be represented as: A + B → AB

Example: Na + Cl → NaCl Application in everyday life Synthesis is widely used in chemistry for the formation of salts, organic compounds, biomolecules, medicines, pesticides and polymers.

3. Displacement reaction A displacement reaction is one in which one atom or ion in a reactant is replaced by another atom or ion of another element.

Example: Mg + CuSO₄→ MgSO₄+ Cu Application in everyday life

a. Displacement reactions are essential in various chemical processes.

b. They have practical applications in metallurgy, electrochemistry, and extraction of metals such as gold from their ores.

4. Decomposition Decomposition is a chemical reaction in which a compound breaks down into two or more simpler substances under certain conditions. That is, a simple reactant undergoes a chemical change to produce multiple products.

It is illustrated as: AB → A + B Where AB is the initial reactant and A and B are the products.

The conditions under which the reaction occurs could be heat, light and the use of a catalyst.

Example: CaCO3 → CaO + CO₂ Application in everyday

a. It is important in natural and industrial processes.

b. They are essential in the fields of chemistry, biology and environmental science.

5. Ionic equation An ionic equation is a chemical equation involving at least one ionic species as a reactant or product, that is, species of dissolved ionic compounds in terms of their free ions; ions that exist in a chemical equation, but are not involved in the overall equation, are spectator ions.

Worked Example 3.5

Write the net ionic equation of the reaction, 2 KI(aq) + Pb(NO₃)₂(aq) → PbI₂(s) + 2 KNO₃(aq)

Solution:

Write the ionic equation 2 K(aq) + + 2I(aq) − + Pb(aq) 2+ + 2N O₃(aq) → P bI₂(s) + 2 K(aq) + + 2N O₃(aq) − Cancel the spectator ions to yield the net ionic equation 2 K_((aq)) + + 2I_((aq)) − + Pb_((aq)) ²⁺+ 2NO₃_((aq)) → P bI₂(s) + 2 K(aq) + + 2NO₃(aq) − Write the net ionic equation, 2I(aq) − + Pb(aq) 2+ → P bI₂(s) Applying and verifying the Laws of Chemical Combination Verifying the Law of conservation of matter The Law of conservation of matter states that mass is neither created nor destroyed in a chemical reaction. The total mass of reactants is equal to the total mass of products.

Worked Example 3.6

Consider the reaction between nitrogen and hydrogen to form ammonia:

N₂ + 3 H₂ 2NH₃

Solution:

Calculate the mass of reactants:

Mass of N₂ = (2 × 14) = 28 g Mass of 3 H₂ = 3(2 × 1) = 6 g Total mass of reactants = 28 + 6 = 34 g Calculate the mass of products:

Total mass of 2 NH₃ = 2 [(1 × 14) + (3 × 1)] = 34 g Verify the Law:

Total mass of reactants = Total mass of product 34 g = 34 g Verifying the Law of Definite Proportions The Law of definite proportions states that a chemical compound always contains the same proportion of elements by mass.

Example: methane (CH₄) always contains C and H in a mass ratio of 3:1. Here is how to verify this:

Calculate the mass ratio of elements in CH₄:

Molar mass of CH₄ = (1 × 12) + (4 × 1) = 16 Mass of C in CH₄ = 1 × 12 = 12 g Mass of H in CH₄ = 4 × 1 = 4 g Determine the mass ratio:

Mass ratio of C to H = 12/4 = 3/1 Verifying the Law of Multiple Proportions The Law of multiple proportions states that when two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element are in ratios of small whole numbers.

Example: copper forms two oxides: copper (I) oxide (Cu₂ O ) and copper (II) oxide ( CuO .

Compare the two Compounds:

Molar mass of Cu₂ O = (2 × 64) + 16 = 144 g mol⁻¹Molar mass of CuO = (1 × 64) + 16 = 80 g mol⁻¹Fixed Mass of oxygen:

Mass of Oxygen (O) in both compounds = 16 g Variable mass of Cu Mass of Cu in Cu₂ O = 128 g Mass of Cu in CuO = 80 g Find out the ratio:

Ratio of the masses of copper that combine with a fixed mass of oxygen (16 g):

128_ 64 = 16/8 = 2/1 The masses of copper in a simple whole number ratio of 2:1.

Activity 3.5: Trial Questions

1. Discuss the rules to be followed in balancing chemical equations.

2. Write and balance the following types of reactions:

a. Combustion Reaction

Example: to balance the chemical equation in which Ethyne (C₂ H₂) combusts in oxygen (O₂) to form carbon dioxide (CO₂) and water (H₂ O ), First, write the unbalanced equation:

C₂H₂+ O₂→ CO₂+ H₂O Count the atoms of each element:

Reactants: C = 2, O = 2, H = 2 Products: C = 1, O = 3, H = 2 Balance the equation:

2C₂H₂+ 5O₂→ 4CO₂+ 2H₂O Check your work:

Reactants: C = 4, O = 10, H = 4 Products: C = 4, O = 10, H = 4

b. Synthesis Reaction

Example: Aluminium (Al) reacts with chlorine (Cl₂) to form aluminium chloride ( Al Cl₃) Write the unbalanced equation:

Al + Cl₂ → Al Cl₃ Count the atoms of each element:

Reactants: Al = 1, Cl = 2 Products: Al = 1, Cl = 3 Balance the equation:

2Al + 3 Cl₂ → 2Al Cl₃ Check your work:

Reactants: Al = 2, Cl = 6 Products: Al = 2, Cl = 6

c. Displacement (Replacement) Reaction

Example: Magnesium chloride (MgCl₂) reacts with potassium phosphate (K₃ PO₄) to form magnesium phosphate (Mg₃ (PO₄ )₂) and potassium chloride ( KCl ) Write the unbalanced equation:

MgCl₂ + K₃ PO₄ → Mg₃ (PO₄ )₂ + KCl Count the atoms of each element:

Reactants: Mg = 1, Cl = 2, K = 3, P = 1, O = 4 Products: Mg = 3, Cl = 1, K = 1, P = 2, O = 8 Balance the equation:

3 MgCl₂ + 2 K₃ PO₄ → Mg₃ (PO₄ )₂ + 6KCl Check your work:

Reactants: Mg = 3, Cl = 6, K = 6, P = 2, O = 8 Products: Mg = 3, Cl = 6, K = 6, P = 2, O = 8

d. Decomposition Reaction

Example: Decomposition of calcium carbonate (CaCO₃) to form calcium oxide (CaO) and oxygen (CO₂).

Write the unbalanced equation:

CaCO₃ → CaO + CO₂ Count the atoms of each element:

Reactants: Ca = 1, C = 1, O = 3 Products: Ca = 1, C = 1, O = 3 Balance the equation:

CaCO₃ → CaO + CO₂ Check your work:

Reactants: Ca = 1, C = 1, O = 3 Products: Ca = 1, C = 1, O = 3

e. Ionic Equation

Example: Sodium sulphate (Na₂ S O₄) reacts with barium chloride (Ba Cl₂) to form barium sulphate (BaS O₄) and sodium chloride ( NaCl ).

Write the balanced molecular equation:

Pb (N O₃)₂(aq) + Ba Cl₂(aq) → Ba (N O₃)₂(aq) + Pb Cl₂(s) Write the complete ionic equation:

Pb²⁺(aq) + 2 NO₃ −(aq) + Ba²⁺(aq) + 2 Cl−(aq) → Ba²⁺(aq) + 2 NO₃ −(aq) + Pb Cl₂(s) Write the net ionic equation:

Pb²⁺(aq) + 2 Cl−(aq) → Pb Cl₂(s)

Activity 3.6: Trial Question

Balance the following reactions:

1. H₂ + O₂→ H₂ O

2. C₃ H₈ + O₂→ CO₂ + H₂ O

Activity 3.7: Demonstrating that mass is conserved in a chemical reaction Scenario Experiment: Reaction between Na₂ CO₃and Ca Cl₂.

Prepare 250 cm³of 1 mol dm⁻³Na₂ CO₃ and 250 cm³of 1 mol dm⁻³Ca Cl₂ solutions.

Materials needed:

Sodium carbonate solution (Na₂ CO₃), calcium chloride solution (Ca Cl₂), 2 beakers, a stirring rod, pipettes, an electronic balance Steps:

Preparation

1. Measure 25 cm³of 1 mol dm⁻³Na₂ CO₃ solution into a 250 cm³beaker labelled A.

2. Measure 25 cm³of 1 mol dm⁻³Ca Cl₂ solution into another beaker labelled B.

3. Record the initial mass of each solution using a balance.

Reaction

4. Slowly add the Ca Cl₂ solution to the Na₂ CO₃ solution while stirring.

5. Observe the reaction and take note of any changes.

6. Record the final mass of the mixture using a balance.

Na₂ CO₃ + Ca Cl₂ → Ca CO₃ (precipitate) + 2NaCl

7. Record your observations and measurements in the table below:

Measurement Description Mass (g)

Mass of empty Beaker A Mass of empty Beaker B Mass of Beaker A with sodium carbonate Mass of Beaker B with calcium chloride Total mass of reactants (A + B) Total mass of products (precipitate + solution).

If total mass of reactants = total mass of products Then mass is conserved in the chemical reaction.

Activity 3.8: Trial Question

Verify the Law of Conservation of Mass with the reaction:

C H₄ + 2 O₂ → C O₂ + 2 H₂ O STOICHIOMETRY Welcome to the world of stoichiometry, where you will explore the relationships between reactants and products in chemical reactions. You will learn to calculate reactant amounts (and masses), predict product yields, balance equations, and understand the mole concept. Get ready to enhance your problem-solving skills, critical thinking, and understanding of chemistry. Let us dive in and uncover the secrets of chemical reactions!

Stoichiometry is the relationship between quantities of reactants and products in a chemical reaction.

Mole Ratio

The relative quantity of any two substances that take part in a chemical reaction is termed as mole ratio. The stoichiometric coefficients in the balanced equation are considered as the number of moles of each reactant or product that would either be consumed or produced under ideal conditions.

Worked Example 3.7

In the reaction, 2C + O₂ → 2CO 2 mol 1 mol 2 mol The mole ratio between carbon and oxygen is written as:

n(C)_ n(O₂) = 2_ 1 n = N_ L Using Stoichiometric Quantities to Calculate Numbers of Entities Chemical Reactions

1. Write the mole ratio using the stoichiometric coefficient of the known substance and that of the substance being calculated.

2. Calculate the number of moles of the substance being calculated.

3. Use the calculated moles and the relation N = n × L to calculate the number of entities.

Worked Example 3.8

Consider the equation: N₂ + 3H₂ → 2NH₃ Calculate the number of molecules of ammonia gas produced if 3.01 × 10²³molecules of Hydrogen react with Nitrogen gas. [L = 6.02 × 10²³]

Solution

N₂ + 3H₂ → 2NH₃ N( H₂) = 3.01 × 10²³, n = ?

Determine the number of moles of Hydrogen using the formula n = N_ L n(H₂) = 3.01 × 10²³_ 6.02 × 10²³= 0.5 mol Write mole ratio between ammonia and Hydrogen n(N H₃)_ n(H₂) = 2_ 3 Calculate the number of moles of ammonia produced:

n(N H₃) = 2_ 3 × n(H₂) n(N H₃) = 2_ 3 × 0.5 = 0.33 mol Solve for the number of molecules of product using the formula N = n × L N(N H₃) = 0.33 × 6.02 × 10²³= 1.99 × 10²³molecules Calculating the mass of a substance

1. Write the correct balanced equation.

2. Convert the quantity of the known substance into the number of moles using the correct mole formula n = m/M.

3. Write the mole ratio using the stoichiometric coefficient of the known substance and substance being sought.

4. Calculate the number of moles of the substance being sought.

5. Use the calculated moles and the relation m = n × M to calculate the mass of the substance.

Worked Example 3.9

Consider the reaction, N₂ + 3H₂ → 2NH₃ Calculate the mass of ammonia produced if 7 g of Nitrogen reacts with excess Hydrogen gas.

[Aᵣ: N = 14, H = 1]

Solution

Given the equation, N₂ + 3H₂ → 2NH₃ m(N₂) = 7g, m(N H₃) = ?

Determine moles of Nitrogen using its mass and molar mass and the formula n = m_ M n( N₂) = 7/2 × 14 = 0.25 mol Determine the mole ratio between Nitrogen and Ammonia n(N H₃)_ n(N₂) = 2_ 1 n(N H₃) = 2 × n(N₂) n(N H₃) = 2 × 0.25 = 0.5 mol Aᵣ(NH₃) = 14 + 3(1) = 17 g mol⁻¹m = 0.5 × 17 = 8.5 g of Ammonia Calculate the concentration of substance (analyte)

1. Write the correct balanced equation.

2. Observe the units closely and convert any volumes or concentrations as required; for example, converting cm³to dm³.

3. Convert the quantity of the known substance into mole using the correct mole formula c = n/V.

4. Write the mole ratio using the stoichiometric coefficient of the known substance and substance being sought.

5. Calculate the number of moles of the substance being sought.

c = n_ V

6. Use the calculated moles and the relation to calculate the concentration of the unknown substance.

Worked Example 3.10

1. Consider the reaction, 2NaOH + H₂ SO₄ → Na₂ SO₄ + 2 H₂ O Given that 20 cm³of H₂SO₄reacts completely with 25 cm³of 0.5 mol dm⁻³ NaOH, calculate the concentration of H₂SO₄.

Solution

2NaOH + H₂ SO₄ → Na₂ SO₄ + 2 H₂ O c(NaOH) = 0.5 mol dm⁻³V(NaOH) = 25 c m³= 0.025 d m³c(H₂ SO₄) = ?

V(H₂ SO₄) = 20 c m³= 0.020 dm³Determine the number of moles of Sodium hydroxide solution using the formula, n = c × V n = 0.025 × 0.5 = 0.0125 mol Write the mole ratio:

n(H₂ SO₄)_ n(NaOH) = 1_ 2 n(H₂ SO₄) = 1_ 2 × n(NaOH) n(H₂ SO₄) = 1_ 2 × 0.0125 = 0.00625 mol Determine the concentration of sulphuric acid using the formula, C = ⁿ_ _(V):

C = 0.00625_ 0.020 = 0.313 mol dm⁻³Calculating for the volume of substance Procedure

1. Write the correct balanced equation.

2. Convert the quantity of the known substance into a number of moles using the correct mole formula n = m_ M n = N_ L n = V_ Vₘ

3. Write the mole ratio using the stoichiometric coefficients of the known substance and substance being calculated.

4. Calculate the number of moles of the substance being sought.

5. Use the calculated number of moles and the relationship V = n × Vₘ to calculate the volume of the gas.

Worked Example 3.11

10.5 g of methane reacts with excess Oxygen to produce carbon dioxide and water.

Calculate the volume of carbon dioxide gas produced at s.t.p. [Aᵣ: C = 12, H = 1, O = 16, Vₘ = 22.4 dm³/ mol ].

Solution

Use the problem-solving strategy:

1. Write the correct balanced equation.

CH₄ + 2 O₂ → CO₂ + 2 H₂ O

2. Convert the quantity of the known substance into a number of moles using the correct mole formula m(CH₄) = 10.5g, m(CH₄) = 12 + (1 × 4) = 16 g mol⁻¹n = m_ M = 10.5_ 16 = 0.66 mol

3. Write the mole ratio using the stoichiometric coefficients of the known substance and substance being calculated.

n(CO₂)_ n(CH₄) = 1_ 1 n(CO₂) = n(CH₄)

4. Calculate the number of moles of the substance being sought for.

n(CO₂) = n(CH₄) = 0.66 mol ∴ n(CO₂) = 0.66 mol

5. Use the calculated number of moles and the relationship V = n × Vₘ to calculate the volume of the gas.

n(CO₂) = V/Vₘ V = n(CO₂) × Vₘ = 0.66 × 22.4 = 14.78 dm³

Activity 3.9: Perform simple chemical reactions, write and balance equations, and perform stoichiometric calculations.

Materials needed:

• Chemicals: Sodium hydroxide (NaOH), Sodium bicarbonate (NaHC O₃), hydrochloric acid (HCl), Magnesium ribbon.

• Equipment: test tubes, test tube rack, beakers, measuring cylinders, safety goggles, gloves, lab coat.

Safety instructions:

• Follow the teacher’s instructions and ask questions if unsure.

• Wear safety goggles, gloves, and a lab coat always.

• Handle all chemicals with care.

Steps:

Reaction between Sodium bicarbonate (NaHCO₃) and hydrochloric acid ( 0.100 mol dm⁻³HCl ):

1. Measure 50 cm³of HCl using a measuring cylinder and pour it into a beaker.

2. Add one teaspoon of Sodium bicarbonate to the beaker with HCl.

3. Observe the reaction and record your observations.

4. Write the balanced chemical equation for the reaction.

5. Calculate the number of moles of NaHCO₃ used if 4.0 g of NaHCO₃ were reacted.

6. Using the balanced equation, determine the mole ratio of reactants to products.

7. Calculate the:

a. number of moles of CO₂ produced.

b. mass CO₂ produced.

c. volume of CO₂ produced.

d. concentration of CO₂ in g dm⁻³e. concentration of CO₂ in mol dm⁻³.

f. concentration of CO₂ in ppm (remember that ppm = gdm⁻³x 1000).

Steps:

Reaction between hydrochloric acid and Sodium hydroxide:

1. Measure 25 cm³of 0.100 mol dm⁻³HCl using a measuring cylinder and pour it into a beaker A.

2. Measure 25 cm³of NaOH using a measuring cylinder into another beaker B.

3. Add the solution in beaker B to the solution in beaker A.

4. Observe the reaction and record your observations.

5. Write the balanced chemical equation for the reaction.

6. Calculate the:

a. number of moles of HCl used if 25 cm³of 0.10 mol dm⁻³HCl

solution was reacted.

b. number of moles of NaCl produced.

c. mass of NaCl produced.

d. concentration of NaCl in g dm⁻³e. concentration of NaCl in mol dm⁻³.

f. concentration of NaCl in ppm.

Steps:

Reaction of Magnesium with hydrochloric acid:

1. Place 5 g of Magnesium ribbon into a test tube.

2. Add 20 cm³of hydrochloric acid to the test tube.

3. Observe the reaction and note any changes.

4. Write the balanced chemical equation for the reaction.

5. Calculate the:

a. number of moles of Magnesium used.

b. number of moles of H₂ produced.

c. mass H₂ produced.

d. volume of H₂ produced.

e. concentration of H₂ in g dm⁻³.

In a reacting system involving two reactants with initial quantities given or having information to determine their initial quantities, the reactant that is completely used up is called the limiting reagent. The reagent that is not completely used up is called the excess reagent.

The maximum quantity of the products formed is determined by the limiting reagent.

Procedure for Determining the Limiting and Excess

Reagents

1. Calculate the initial quantity of each reactant in moles.

2. If the stoichiometric ratio of the reactants is 1:1, the reagent with a lower value is the limiting reagent, and the other is the excess reagent.

3. If the stoichiometric ratio of the reactant is not 1:1, then write a mole ratio between the two reactants and solve for one of them.

4. Compare the calculated number of moles with the initial quantity of moles.

If the calculated number of moles is greater than the initial amount, then it is the limiting reagent, and if it is less, it is the excess reactant.

Worked Example 3.12

Consider the reaction, N₂ + 3H₂ → 2NH₃ If 12.0 g of Nitrogen and 8.0 g of Hydrogen react in the formation of ammonia,

a. Determine the limiting reagent.

b. Calculate the mass of ammonia (NH₃) produced. [Aᵣ: N = 14, H = 1]

Solution:

N₂ + 3H₂ → 2NH₃ m(N₂) = 12g, m(H₂) = 8g

a. Determine the initial moles of both reactants n( N₂) = m_ M = 12_ 28 = 0.43 mol n(H₂) = m_ M = 8_ 2 = 4 mol

b. Determine the limiting reagent n(N₂)_ n(H₂) = 1_ 3 n(N₂) = 1_ 3 × n(H₂) n(N₂) = 1_ 3 × 4 = 1.33 mol 1.33 moles of N₂are required to react with 4 moles of H₂,but we only have 0.43 moles, so N₂is the limiting reactant.

c. Write the mole ratio between the limiting reagent and the product (NH₃) n(NH₃)_ n(N₂) = 2_ 1 n(NH₃) = 2_ 1 × n(N₂) n(NH₃) = 2 × 0.43 = 0.86 mol Mass of NH₃, m = n × M = 0.86 × 17 = 14.62 g

Activity 3.10: Calculations for limiting reagents and excess reactants Materials needed:

• Worksheet, Periodic tables, calculator Given the reaction: 2 H₂ + O₂ → 2 H₂ O If 3 mol of H₂ react with 2 mol of O₂, determine the:

a. Limiting reagent.

b. Excess reactant.

c. Mass of water formed.

Steps to solve:

1. Identify the limiting and excess reactants:

Use the balanced equation to find the mole ratio.

n(O₂)_ n(H₂) = 1_ 2 n(O₂) = 1_ 2 × n(H₂) = 1_ 2 × 3 = 1.5 mol Compare the mole ratio with the actual moles available Since 2 mol of O₂ are available, but only 1.5 mol is needed to react with the available H₂, O₂ is in excess.

∴ The limiting reagent is H₂ The excess reactant is O₂ ≔ 2 − 1.5 = 0.50 mol Mass of H₂ O produced:

Use the moles of the limiting reagent to find the moles of H₂ O formed:

n(H₂ O)_ n(H₂) = 2_ 2 = 1 n(H₂ O) = n(H₂) = 3 mol Convert moles of H₂ O to gram M(H₂ O) = (2 × 1) + 16 = 18 g mol⁻¹m(H₂O) = n(H₂O) × M(H₂O) = 3 mol × 18 g mol⁻¹= 54.0 g Consider the reaction: 4Fe + 3O₂ → 2 Fe₂ O₃.

If 10 moles of Iron and 8 moles of Oxygen are available, determine the

(a) limiting reagent.

(b) excess reactant.

(c) mass of water formed.

Steps to solve:

1. Identify limiting and excess reactants:

Use the balanced equation to find the mole ratio.

n(O₂)_ n(Fe) = 3_ 4 n(O₂) = 3_ 4 × n(Fe) = 3_ 4 × 10 = 7.5 mol Compare the mole ratio with the actual moles available.

Since 8 mol of O₂ are available, but only 7.5 mol are needed to react with the available Fe , . O₂ is in excess.

∴ Limiting reagent is Fe Excess reactant is O₂ ≔ 8 − 7.5 = 0.50 mol Mass of Fe₂ O₃ produced:

Use the moles of the limiting reagent to find moles of Fe₂ O₃formed.

n(Fe₂ O₃)_ n(Fe) = 2_ 4 = 1_ 2 n(Fe₂ O₃) = 1_ 2 n(Fe) = 1_ 2 × 10 = 5 mol Convert moles of Fe₂ O₃ to gram M(Fe₂ O₃) = (2 × 56) + (3 × 16) = 160 g mol⁻¹m(Fe₂ O₃) = n(Fe₂ O₃) × M(Fe₂ O₃) = 5 mol × 160 g mol⁻¹= 800.0 g Calculate the volume of Chlorine required to react completely with 50 cm³of 1.0 mol dm³Sodium Bromide (NaBr) solution.

Steps to solve:

Writing and balancing equations Unbalanced: Cl₂+ NaBr → NaCl + Br₂ Balanced: Cl₂ + 2 NaBr → 2NaCl + Br₂ Find the number of mol of NaBr V(solution) = 50 cm³= 0.050 dm³, c(NaBr) = 0.10 mol dm⁻³n(NaBr) = c(NaBr) × V(solution) = 0.10 mol dm⁻³× 0.050 dm³= 0.050 mol Use the mole ratio to determine moles of Chlorine (Cl₂) needed:

n( Cl₂)_ n(NaBr) = 1_ 2 n(O₂) = 1_ 2 × n(NaBr) = 1_ 2 × 0.050 = 0.025 mol Calculate the volume of Chlorine Gas (Cl₂):

Assuming the reaction occurs at standard temperature and pressure, where 1 mole of any gas occupies 22.4 dm³.

V(Cl₂) = n(Cl₂) × Vₘ = 0.025 mol × 22.4 dm³mol⁻¹= 0.56 dm³

In a chemical reaction, the calculated amount of a product is usually not obtained due to the following:

1. The reaction may be reversible.

2. Some reactants may undergo side reactions.

3. Some products cannot be separated or recovered from the mixture.

The Actual Yield is the amount of a product obtained from a chemical reaction in practice.

The percentage yield of a reaction is the percentage of the product obtained compared to their theoretical maximum yield calculated from the balanced equation.

Percentage yield = Actual yield_____________ Theoretical yield × 100

Worked Example 3.13

Magnesium metal reacts with hydrochloric acid to produce Magnesium Chloride and Hydrogen gas.

a. If 12 g of Magnesium reacts with excess HCl, calculate the maximum theoretical mass of Magnesium Chloride formed.

b. If 42.0 g of purified anhydrous Magnesium Chloride was obtained, calculate the percentage yield. [Aᵣ: Mg = 24, Cl = 35.5]

Solution:

Use the problem-solving approach:

a. Analyse the question:

Mg + 2HCl → MgCl₂+ H₂ m(Mg) = 12 g m(MgCl₂) = ?

b. Determine the number of moles of Mg:

n(Mg) = m_ M = 12_ 24 = 0.50 mol

c. Write the mole ratio between MgCl₂and Mg:

n(MgCl₂)_ n(Mg) = 1_ 1 n(MgCl₂) = n(Mg) = 0.5 mol

d. Determine the theoretical mass of MgCl₂ formed:

m = n × M.

But, M (MgCl₂) = 24 +(2 × 35.5) = 95 g mol⁻¹m = 0.5 × 95 = 47.5 g

e. Determine the percentage yield:

Percentage yield = Actual amount obtained___________________ maximum theoritical yield × 100 = 42_ 47.5 × 100 = 88.42 %

Activity 3.11: Determining Actual Yield, Theoretical Yield, and

Percentage Yield

Step 1: Write the Balanced Chemical Equation

• Identify the reactants and products.

• Ensure the equation is balanced to reflect the conservation of mass.

Example: 2H₂+ O₂→ 2H₂O

Step 2: Convert Reactant Quantities to Moles

• Use molar masses to convert grams of reactants to moles.

Given: 10g H₂ Molar Mass H₂: 2 g/mol Moles of H₂ = 10g_ 2 g / mol = 5 mol

Step 3: Use Stoichiometry to Determine Theoretical Yield

• Use the balanced chemical equation to find the mole ratio between reactants and products.

• Calculate the moles of product expected from the given moles of reactants.

• Convert moles of product to grams using its molar mass.

Mole Ratio: H₂:H₂ O = 2 : 2 = 1 : 1 n( H₂ O) = n( H₂) = 5 mol H₂O Molar Mass H₂O = 18 g / mol Mass of H₂O = 5 moles × 18 g / mol = 90g

Step 4: Measure the Actual Yield

• Perform the experiment and measure the actual amount of product obtained.

Actual Yield from Experiment = 85g H₂O

Step 5: Calculate Percentage Yield

Percentage yield = Actual yield____________ Theoretical yield × 100 Percentage yield = 85 g_ 90 g × 100 = 94.4%

Worked Example 3.14

In a laboratory experiment, students aimed to produce 25 grams of copper sulphate (CuSO₄) according to the reaction: Cu + H₂ S O₄ → CuS O₄ + H₂ If they actually obtained 20 grams of copper sulphate, calculate the percentage yield of the reaction.

Solution:

• Identify the theoretical yield and actual yield:

Actual yield: The amount of product actually obtained, which is 20 grams of CuS O₄.

Theoretical yield: The amount of product that was aimed for, which is 25 grams of CuS O₄.

• Use the formula for percentage yield:

Percentage yield = Actual yield____________ Theoretical yield × 100 = 20 g_ 25 g × 100 = 80%

Worked Example 3.15

In a chemical synthesis, 30 grams of Calcium carbonate (CaC O₃) reacts with excess hydrochloric acid (HCl ) to produce Calcium Chloride (Ca Cl₂), Carbon dioxide (C O₂), and water (H₂ O ). If 22 grams of Calcium Chloride are obtained in the reaction, calculate the percentage yield.

Solution:

Write the balanced chemical equation: CaC O₃ + 2HCl → Ca Cl₂ + C O₂ + H₂ O Convert the mass of CaC O₃ to moles:

M(CaC O₃) = 40 + 12 + (3 × 16) = 100 g mol⁻¹n(CaC O₃) = m(CaC O₃)_ M(CaC O₃) = 30 g_ 100 g mol⁻¹= 0.30 mol Use stoichiometry to find the moles of Ca Cl₂ produced:

The balanced equation shows a 1:1 mole ratio between CaC O₃ and Ca Cl₂.

n(Ca Cl₂) = n(CaC O₃) = 0.030 mol Convert the moles of Ca Cl₂ to grams:

M(Ca Cl₂) = 40 + (2 × 35.5) = 111 g mol⁻¹ Theoretical yield = 0.30 mol × 111 g mol⁻¹= 33.3 g Actual yield of Ca Cl₂ = 22 grams Percentage yield = Actual yield____________ Theoretical yield × 100 = 22 g_ 33.3 g × 100 = 66.07%

Worked Example 3.16

In the preparation of aspirin (C₉H₈O₄), a student reacted 20 grams of salicylic acid (C₇H₆O₃) with excess acetic anhydride (C₄H₆O₃). The theoretical yield of aspirin is 24 grams. If the student obtained 18 grams of aspirin, calculate the percentage yield.

Solution:

Balanced chemical equation:

C₇ H₆ O₃ + C₄ H₆ O₃ → C₉ H₈ O₄ + C₂ H₄ O₂ Identify the theoretical yield and actual yield:

Actual Yield: 18 grams of aspirin (C₉ H₈ O₄) Theoretical Yield: 24 grams of aspirin (C₉ H₈ O₄).

Use the formula for percentage yield:

Percentage yield = Actual yield____________ Theoretical yield × 100 = 18 g_ 24 g × 100 = 75%

Review Questions 3.1

Calculate the empirical formula for the following compounds:

1. A compound contains 3.24 g of sulphur and 3.24 g of oxygen.

2. A sample of a compound contains 4.2 g of nitrogen and 12 g of oxygen.

3. A compound consists of 6.7 g of phosphorus and 8.5 g of oxygen.

4. Write the chemical formulae of the following compounds

a. Potassium chloride

b. Iron (II) bromide

c. Copper (II) tetraoxosulphate (VI)

Review Questions 3.2

1. Calculate the percentage by mass of oxygen in water (H₂O).

2. Determine the percentage by mass of sodium in sodium chloride (NaCl).

3. Calculate the percentage by mass of calcium in calcium carbonate (CaCO₃).

4. Determine the percentage by mass of carbon in glucose (C₆H₁₂O₆).

Review Questions 3.3

1. A compound contains 54.5% carbon, 9.1% hydrogen, and 36.4% oxygen.

The molar mass of the compound is 88 g/mol. Determine the empirical and molecular formulas.

a. Convert Percentage to Mass (Assume 100 g of compound):

Mass of C = ________ g Mass of H = ________ g Mass of O = ________ g

b. Convert mass to moles:

Moles of C = ________ Moles of H = ________ Moles of O = ________

c. Determine the simplest Ratio:

Moles of C = ________ Moles of H = ________ Moles of O = ________

d. Write the Empirical Formula:

Empirical formula =

e. Calculate the Empirical Formula Molar Mass:

Empirical formula molar mass = ________ g/mol

f. Determine the Ratio of Molar Masses:

Ratio = ________

g. Calculate the Molecular Formula:

Molecular formula =

2. An organic compound of relative molecular mass 46 on analysis was found to contain 52.0% carbon, 13.3% hydrogen and the remaining being oxygen. Determine its

a. Empirical formula

b. Molecular formula [H=1.0, C=12.0, O=16.0]

3. Determine the percentage by mass of nitrogen and hydrogen in ammonia (NH₃).

[N = 14, H = 1]

Review Questions 3.4

Balance the following reactions:

1. Fe + O₂→Fe₂ O₃

2. N₂ + H₂→N H₃

3. C₂ H₆ + O₂→CO₂ + H₂ O

4. KClO₃→KCl + O₂

5. Mg + HCl→MgCl₂ + H₂

6. NaOH + H₂ SO₄→Na₂ SO₄ + H₂ O

7. C₆ H₁₂ O₆ + O₂→CO₂ + H₂ O

8. Al + HCl→AlCl₃ + H₂

9. Balance the following chemical equation and use it to answer the questions that follow:

C₂H₆+ O₂→ CO₂+ H₂O

(a) How many moles of water would be obtained from 4 moles of C₂H₆?

(b) How many moles of C₂H₆would be needed to produce 3 moles of water?

Review Questions 3.5

1. Write a balanced chemical equation for the reaction between Zn and HCl

2. Zinc metal reacts with Hydrogen Chloride according to the reaction Zn + 2HCl → Zn Cl₂ + H₂.

Calculate the mass of zinc required to produce 2 moles of hydrogen gas.

[Aᵣ: Zn = 65 ]

3. Calculate the mass of CO₂ produced when 2 moles of propane (C₃ H₈) reacts with excess Oxygen. [Aᵣ: C = 12, H = 1, O = 16 ]

4. Balance the chemical equation H₂ + O₂ → H₂ O and determine the mole ratio of Hydrogen to Oxygen in the reaction.

5. Balance the chemical equation Al(s) + Fe₂ O₃(aq) → Al₂ O₃(aq) + Fe(s) and determine the mole ratio of Fe₂ O₃ to Fe in the reaction.

6. Consider the reaction, HCl + NaOH → NaCl + H₂O. If 25cm³of 0.25 mol dm⁻³HCl reacts completely with excess Sodium hydroxide, calculate the mass of Sodium Chloride produced.

[Aᵣ: Na = 23, Cl = 35.5]

7. Consider the reaction, 2KOH + H₂SO₄→ K₂SO₄+ 2H₂O. Calculate the volume of KOH of concentration 0.10 mol dm⁻³required to completely neutralise 20 cm³of a 0.25 mol dm⁻³H₂SO₄solution.

Review Questions 3.6

1. In a reaction, 20 grams of Hydrogen gas (H₂) reacts with 10 grams of Oxygen gas (O₂) to produce water (H₂ O ). Determine which reactant is the limiting reagent and which is the excess reagent.

2. In the combustion of propane (C₃ H₈), 40 grams of propane reacts with 100 grams of Oxygen gas (O₂).

a. Determine which reactant is the limiting reagent and which the excess reagent is.

b. Calculate the mass of Carbon dioxide (CO₂) produced.

c. Calculate the mass of the excess reagent remaining after the reaction is complete.

[C₃ H₈ + 5 O₂ ⎯⟶3 CO₂ + 4 H₂ O ]

3. In the synthesis of ammonia (N H₃), 50 grams of Nitrogen gas (N₂) reacts with 20 grams of Hydrogen gas (H₂).

a. Determine which reactant is the limiting reagent and which the excess reagent is.

b. Calculate the mass of ammonia produced.

c. Calculate the volume of nitrogen gas consumed at stp (standard temperature and pressure), assuming the reaction is complete.

[N₂ + 3H₂ ⎯⟶2N H₃]

Review Questions 3.7

1. In a laboratory experiment, 25 grams of Sodium Chloride (NaCl) are reacted with excess Silver nitrate (AgNO₃) to produce Silver Chloride (AgCl). If the theoretical yield of AgCl is 30 grams, and the actual yield obtained is 20 grams, calculate the percentage yield of the reaction, [ NaCl + AgN O₃ → AgCl + NaN O₃]

2. In the synthesis of aspirin, a student reacts 20 grams of salicylic acid (C₇H₆O₃) with 25 grams of acetic anhydride (C₄H₆O₃). The theoretical yield of aspirin is 24 grams. If the student obtains 18 grams of aspirin,

(a) State the actual yield of the reaction.

(b) Based on the actual yield obtained in (a), calculate the percentage yield of the reaction. [C₇ H₆ O₃ + C₄ H₆ O₃ → C₉ H₈ O₄ + C₂ H₄ O₂]

3. A chemical reaction between Hydrogen gas (H₂) and Nitrogen gas (N₂) produces ammonia (NH₃). If 15 grams of Hydrogen gas reacts with excess Nitrogen gas to produce 25 grams of ammonia,

(a) State the actual yield of the reaction.

(b) The theoretical yield of ammonia in the reaction described in part (a) is 30 grams. Calculate the percentage yield of the reaction based on the actual yield obtained.

[N₂ + 3H₂ ⎯⟶2N H₃]

4. In the production of a pain relief medication, 50 grams of the starting material reacts with excess reagent to produce 70 grams of the medication.

However, due to various inefficiencies, the actual yield obtained is 60 grams. Calculate the percentage yield of the reaction. Discuss the potential impacts of low yield on the cost and availability of the medication.

5. In the production of sulphuric acid (H₂SO₄), 100 grams of Sulphur dioxide (SO₂) reacts with excess oxygen to theoretically produce 120 grams of sulphuric acid. The actual yield obtained in the industrial process is 100 grams. Calculate the percentage yield of sulphuric acid. Consider the economic implications of yield on the industrial scale.

6. In a wastewater treatment plant, a reaction is designed to remove 80 grams of a contaminant. The theoretical yield of the process is 90 grams, but due to various factors, only 70 grams are actually removed. Calculate the percentage yield of contaminant removal. Discuss how efficiency in this process affects environmental health and compliance with regulations.

Hello learner! Welcome to the kinetic theory of matter. This fundamental concept in chemistry explains how different states of matter behave. It’s based on the idea that matter is made up of tiny particles—molecules, atoms, or ions—that are always in motion. In this section you will also focus on understanding and applying kinetic theory, which explains how particles behave in gases. You will also explore how gases are prepared in the laboratory and their practical uses in everyday life.

At the end of this section, you will be able to:

· Explain the kinetic theory of matter and apply it to distinguish between the properties of solids, liquids and gases.

· State and perform calculations involving various Gas Laws and analyse graphs based on the laws.

· State Graham’s Law of Diffusion/effusion and Dalton’s Law of partial pressure and apply them to perform calculations.

· Write the ideal gas equation and apply it in simple calculations using the different numerical values of R and units of Pressure and Volume.

· Explain why gases show deviation from ideal behaviour and suggest how the ideal gas equation could be modified to describe gas behaviour more accurately.

· Design and perform experiments to prepare and test for gases (hydrogen, ammonia and carbon dioxide gases).

Key Ideas

· Particles are small constituent units of matter, which can be atoms, molecules, or ions.

· Intermolecular forces are forces of attraction or repulsion between particles.

· Diffusion is the process of particles spreading out from an area of higher concentration to an area of lower concentration.

· Brownian motion is the random movement of particles suspended in a fluid (liquid or gas) resulting from collisions with fast-moving molecules of the fluid.

· Gas Law is a mathematical relationship between the pressure, volume, temperature, and quantity of a gas.

· Mole fraction is the ratio of the number of moles of a component to the total number of moles in a mixture.

· Standard temperature and pressure are defined as 0°C (273.15 K) and 1 atm pressure.

· Ideal gas equation is the equation describing the relationship between Pressure, Volume, Temperature, and the number of moles of a gas: PV = nRT · Universal Gas Constant (R) is a molar physical quantity.

· Displacement of water this method involves collecting a gas by displacing water in a container.

· Displacement of Air (Upward or Downward Delivery) gases are collected by displacing air, based on their density relative to air.

· Using gas syringe, a gas syringe is used to directly measure and collect a known volume of gas.

The kinetic theory of matter states that;

1. Matter is made of tiny particles which are in constant random motion.

2. Matter possesses kinetic energy due to the motion of the particles.

3. The difference between the different states of matter is due to the nature and extent of motion and the separation between the particles.

Solid State

1. Solids have a fixed shape and volume at a given temperature.

2. Particles of solids are closely packed in an orderly manner.

3. Solids have the greatest forces of attraction between their particles when compared to the energy possessed by the particles.

4. Particles of solids undergo vibration about their mean position.

5. Increasing the temperature of solids causes faster vibration of particles.

Using the kinetic model to explain the properties of solids

a. Solids tend to have the greatest density because the particles are usually closest together.

b. Solids have a fixed shape and volume because of the strong force of attraction between the

c. Particles hold the particles in their fixed positions.

d. Solids are difficult to compress because of the lack of empty space between the particles.

e. Solids expand on heating, due to the increased energy of the particles at higher temperatures. This allows the particles to exist at greater distances from each other because the intermolecular forces are constant.

The Liquid State

1. Liquids have fixed volumes that take the shape of the container at a given temperature.

2. Particles of liquids are close together and arranged randomly.

3. The particles of liquids move rapidly in all directions.

4. The forces of attraction between the particles are stronger than that of gases, but lower than that of solids when compared to the energy possessed by the particles.

5. Increasing the temperature of liquids makes their particles move faster due to a gain in kinetic energy.

Using the kinetic model to explain the properties of liquids

a. Liquids have greater densities than gases because the particles are closer due to the attractive forces.

b. Liquids have fixed volume but take the shape of their container because of the increased particle attraction.

c. Liquids are not easily compressed because there is so little empty space between the particles.

d. Liquids expand on heating, due to the increased energy of the particles at higher temperatures. This allows the particles to exist at greater distances from each other because the intermolecular forces are constant.

Activity 4.1: Representations to depict the arrangement and motion of particles in solids and liquids.

Solid (ice) Particles vibrating in place but not changing positions Liquid (water) Particles moving and sliding past each other.

Use the links below to watch the animation of particles of solid and liquid:

https://phet.colorado.edu/sims/html/states-of-matter-basics/latest/states-of- matter-basics_en.html or https://www.acs.org/content/acs/en/education/resources/k-8/inquiryinaction/ fifth-grade/particles-solid-hammer.html or https://www.youtube.com/watch?v=gPMVaAnij88

Activity 4.2: Investigating Properties of Solids and Liquids

Experiment 1: Observing the Shape and Volume of Solids and Liquids Materials needed:

· Transparent containers of different shapes and sizes · A solid object (e.g. A rubber ball) · Different liquids (e.g. Water, oil, syrup).

Procedure:

1. Place the solid object in different containers and observe if its volume and shape change.

2. Pour the liquids into different containers and observe how they take the shape of each container.

3. Measure the volume of the liquids before and after transferring to different containers to see if the volume changes.

Presentation Solid Liquid

Shape Volume

4. Explain the differences between the shape and volume of solids and liquids.

Extension task Consider the material toothpaste, is it a solid or a liquid? Prepare a balanced argument which outlines the reasoning for it being a solid and a liquid.

Experiment 2: Compare the density of various solids and liquids.

Materials needed:

· Different solid objects (e.g., metal cubes, wooden blocks, plastic objects), · Various liquids (e.g., water, oil, syrup) · Graduated cylinders · Electronic balance · Measuring cups Procedure:

1. Measure the mass of each solid object using the electronic balance.

2. Measure the volume of each solid by water displacement in a graduated cylinder.

3. Calculate the density of each solid using the formula:

Density = Mass______ Volume

4. Measure the volume of each liquid using measuring cups.

5. Measure the mass of each liquid by pouring it into the graduated cylinder and weighing it.

6. Calculate the density of each liquid using the same formula.

Presentation Solid Liquid

Solid 1 Solid 2 Solid 3 Liquid 1 Liquid 2 Liquid 3 Mass Volume Density

7. Compare the densities of solids and liquids.

Extension task If the density of water is 1000 kgm⁻³, which of these objects (both solid and liquid) will float or sink?

Experiment 3: Behaviour of particles in solids Materials needed: Marbles, beads, trays and spoons Procedure

1. Place marbles or beads tightly packed into a tray.

2. Gently shake the tray to show that the marbles/beads vibrate but do not change their positions significantly.

3. Use a spoon to apply gentle pressure and rearrange the marbles/beads, showing that while their positions can be changed by an external force, they still maintain a fixed arrangement overall.

4. This simulates how particles in a solid can be rearranged by external forces but generally stay in a fixed structure.

Experiment 4: Comparing the compressibility of various solids and liquids.

Materials needed: Flexible plastic bottle with cap, water and beads Procedure

1. Fill the bottle with water to the very top and put the cap on securely.

2. Squeeze the bottle.

3. Hold the bead between two fingers and squeeze.

4. Were you able to squeeze the bottle filled with water or the bead?

Extension task When a solid sponge is compressed its volume changes drastically. Why does this not constitute a solid being compressed?

Safety: Wear safety glasses or goggles and be sure to follow all safety instructions given by your teacher. Wash your hands after completing the

activity.

Experiment 5: Comparing the Viscosity of Liquids

Materials needed: Different liquids (e.g., water, oil, honey, syrup), stopwatch, inclined planes, measuring cups.

Procedure:

1. Pour equal amounts of each liquid down the inclined plane.

2. Measure how long it takes for each liquid to reach the bottom of the inclined plane.

3. Rank the liquids based on their flow times to determine their viscosity.

Type of liquid Time taken for liquid to flow Water Oil Syrup Honey

4. Discuss why some liquids might flow faster than others.

Extension task Consider why viscosity is an important property for an engineer to understand about a liquid.

· What applications could there be for liquids with very low viscosities?

· What applications could there be for liquids with very high viscosities?

· What issues might arise from handling such liquids (high or low viscosity) on an industrial scale? Think about any issues that might arise from pumping, pouring or mixing.

1. Gases have no fixed shape or volume but fill a container.

2. Forces of attraction between the particles of gases are negligible.

3. Particles are so small that the actual volume of individual particles is negligible compared to the volume of the container.

4. Particles are widely spaced and scattered and undergo random and rapid motion.

5. The average kinetic energy of the gas particles is directly proportional to the absolute temperature of the particles.

6. The collision of the gas particles with the surface of the container causes gas pressure.

Using the kinetic model to explain the properties of gases

a. Gases have very low densities because the particles will space out in the container.

b. Gases have no fixed volume and shape because of the negligible force of attraction.

c. Gases are easily compressed because of the space between the particles

d. Order of ease of compression: gas > liquid > solid

e. Gases exert pressure because of the collision of the particles with the walls of the container.

Activity 4.3: Gas Behavior Under Different Conditions

Materials needed: software - PhET Interactive Simulations (or similar) computer and internet access Use the link below to watch the simulation of gases:

https://phet.colorado.edu/sims/html/states-of-matter-basics/latest/states-of- matter-basics_en.html Procedure:

Temperature and pressure effects

1. Increase and decrease the temperature of the gas in the simulation.

2. Observe how changing temperature affects the speed and kinetic energy of gas particles.

3. Adjust the pressure settings in the simulation.

4. Observe how changes in pressure affect the volume and density of the gas.

5. Discuss the relationship between pressure and the frequency of particle collisions.

6. Discuss how temperature impacts the volume and pressure of the gas.

7. Record your observations and findings from each task.

Volume Effects

1. Change the volume of the gas chamber in the simulation.

2. Observe how altering the volume affects the pressure and density of the gas.

3. Record your observations and findings from each task.

Extension task Push the model to its limits. Describe and then explain what you see, how would this look in a real-world scenario?

Activity 4.4: Comparing space, shape and compressibility of gasses Material needed: Empty balloon Procedure:

1. Fill the entire space inside the balloon.

2. Discuss what you observed.

3. Squeeze the balloon between.

4. Discuss what you observed.

Extension task How does the compressibility of a gas change as it is being compressed? Use your knowledge of particles to answer this question.

Melting When solids are heated, particles gain kinetic energy and vibrate more strongly.

Attractive forces weakening in comparison to particle energy, particles become free to move around.

Freezing When liquids are cooled, particles lose kinetic energy. Attractive forces strengthen in comparison to particle energy, causing particles to be restricted to a fixed position they still move but this is now a vibration.

Evaporation and Boiling

1. Evaporation occurs when the particles of a liquid escape to form a vapour.

2. On heating, surface particles gain more kinetic energy, move faster, and break away from intermolecular forces.

3. Boiling is rapid vaporisation anywhere in the bulk liquid at a fixed temperature (the boiling point of the liquid).

Condensation It is the change in the physical state of matter from the gaseous form into the liquid form. This occurs when the temperature of the gas is lowered to the point where its particles lose enough kinetic energy to form bonds and transition into a liquid. Condensation is the reverse process of vaporisation.

Melting Point Determination

1. Heat one end of the capillary tube in a Bunsen flame.

2. Fill about one-quarter of the capillary tube with the substance and tie it to a thermometer.

3. Insert the tube and thermometer into an oil bath.

4. Heat with constant stirring and record the temperature at which the first crystal melts and the

5. temperature at which the last crystal melts

Figure 4.1: Phase changes of matter Image source: https://www.britannica.com/science/phase-state-of-matter#/media/1/455270/155241

Activity 4.5: Demonstrating melting and boiling processes using ice cubes and a heat source Materials Needed: Ice cubes, Bunsen burner or hot plate, or electric kettle, thermometers, beakers, stopwatch, safety goggles and gloves.

Melting of ice

1. Place a few ice cubes in a beaker and record its initial temperature.

2. Allow the ice to melt at room temperature first, observing the process and recording the temperature change every minute.

3. Use a heat source to speed up melting after initial observations. (be careful if using a direct heat source such as a Bunsen as the glass may shatter)

4. Note the time it takes for the ice to melt completely with and without the heat source.

Melting Observation

The initial temperature of the ice cube Time taken for ice to melt at room temperature Time taken for ice to melt with a heat source The final temperature of melted water Boiling of water

1. Pour the melted water into a beaker and place it on the heat source.

2. Measure and record the temperature of the water every minute until it starts boiling.

3. Observe the boiling process and note the boiling point of water (100°C).

Boiling Observation

The initial temperature of the water Time taken to reach boiling point Boiling point observed Observations Extension task How could this experiment be improved to be more accurate (closer to the value of 100°C)?

Activity 4.6: Demonstrating change of state using simulations of the water cycle or industrial process like distillation.

1. The simulation allows users to explore how water molecules move and change state under different temperatures and pressures. Use the link below to observe the animation.

https://phet.colorado.edu/sims/html/states-of-matter-basics/latest/ states-of-matter-basics_en.html

2. The video provides an engaging animation of the water cycle, demonstrating processes like evaporation, condensation, and precipitation. Use the link below to observe the animation.

https://www.youtube.com/watch?v=ncORPosDrjI GAS LAWS Hello learner! In this lesson, you will learn about the gas laws, which help predict how gases behave under different conditions. You will understand how pressure, volume, and temperature affect gases and relate the amount of gas to these factors.

These laws have many practical applications in Engineering, Chemistry, Physics, and more.

Boyle’s Law

It states that the volume of a fixed mass of gas at constant temperature is inversely proportional to the pressure of that gas.

Mathematically Boyle’s law is expressed as, P₁ V₁ = P₂ V₂ Where P₁and P₂are initial and final pressures respectively And V₁and V₂are initial and final volumes respectively.

Graphically Boyle’s law is represented as follows;

Figure 4.2: Graphs Showing Different Representations of Boyle’s Law.

Example 4.1

10 m³volume of a gas at a pressure of 101,300 Pa was compressed to a volume of 6 m³at constant temperature, calculate the final pressure.

Answer: (Use problem solving strategy)

a. Analyse the question Known Initial volume, V₁= 10 m³Initial pressure, P₁= 101,300 Pa Final volume, V₂= 6 m³Unknown Final pressure, P₂=?

b. Apply the problem-solving strategy P₁V₁= P₂V₂ P₂ = P₁ V₁_ V₂ = 101,300 × 10/6 = 168,833 Pa Charles’ Law It states that the volume of a fixed mass of a gas at constant pressure is directly proportional to its absolute temperature.

Mathematically Charles’ law is represented as, V₁_ T₁ = V₂_ T₂ Where V₁and V₂are initial and final volumes respectively, and T₁and T₂are initial and final temperatures respectively.

NB: The temperature must always be converted to Kelvin.

Graphically Charle’s law is represented as follows;

Figure 4.3: Graphical Representation of Charles Law.

Example 4.2

10 m³volume of a gas in a cylinder is heated from 250 K to 300 K at constant pressure, calculate the final volume of the gas in the cylinder.

Answer (Use problem solving strategy)

a. Analyse the question Known Initial volume, V₁= 10 m³Initial temperature, T₁= 250 K Final temperature, T₂= 300 K Unknown Final volume, V₂=?

b. Apply the problem-solving strategy V₁_ T₁ = V₂_ T₂ V₂ = V₁ T₂_ T₁ = 10 × 300/250 = 12 m³Gay-Lussac’s Law It states that for a fixed mass of gas at constant volume, the pressure of that gas is directly proportional to its absolute temperature (K).

Mathematically Gay-Lussac’s law is represented as, P₁_ T₁ = P₂_ T₂

Example 4.3

A fuel and air mixture in a car engine cylinder of volume 1000 cm³increases from 20 ⁰C to 2000 ⁰C upon combustion. If the normal atmospheric pressure is 100 kPa, calculate the final pressure.

Answer: (Use problem-solving strategy)

a. Analyse the question Known Initial temperature, T₁= 20 + 273 = 293 K Final temperature, T₂= 2000 + 273 = 2,373 K Initial pressure, P₁= 100 kPa Unknown Final pressure, P₂=?

b. Apply the problem-solving strategy P₁_ T₁ = P₂_ T₂ P₂ = 100 × 2273/293 = 775.8 kPa Evaluate: Check the answer to see if it makes sense.

Combined gas Law It is the combination of the Boyle’s law, Charle’s law and the Gay-Lussac’s law.

Mathematically it is expressed as P₁ V₁_ T₁ = P₂ V₂_ T₂

Example 4.4

20 cm³of a gas at 1 atm and 25 ⁰C was compressed to 16 cm³at 40 ⁰C, calculate the final pressure of the gas.

Answer: (Use problem-solving strategy)

a. Analyse the question Known Initial pressure, P₁= 1 atm Initial volume, V₁= 20 cm³Final volume, V₂= 16 cm³Initial temperature, T₁= 25 + 273 = 298 K Final temperature, T₂= 40 + 273 = 313 K Unknown Final pressure, P₂=?

b. Apply the problem-solving strategy P₁ V₁_ T₁ = P₂ V₂_ T₂ P₂ = P₁ V₁ T₂_ T₁ V₂ = 1 × 20 × 313/298 × 16 = 1.31 atm It is worth noting in all of these mathematical laws that the units that are chosen for each quantity need to be consistent. If the volume is initially measured in m³, then it should be in m³for the final volume. If the volume is, however, in cm³ initially and in m³for the final volume, then one of them will need to be converted to the other.

Avogadro’s Law

It states that equal volumes of gases at the same temperature and pressure contain the same number of molecules or moles of gas.

Mathematically Avogadro’s law is expressed as, V_ n = K Where V = Volume occupied by the gas n = number of moles of the gas K = constant of proportionality

Example 4.5

Consider the reaction, 2C O(g) + O₂(g) → 2 CO₂(g) If the rate of production of CO in an industrial plant is at 20 dm³per minute, what is the rate of production of oxygen to produce CO₂?

Answer 2C O(g) + O₂(g) → 2 CO₂(g) V(CO)_ V(O₂) = 2/1 V(O₂) = V(CO)______ 2 = 20/2 = 10 dm³min⁻¹

Activity 4.7:Using simulations to visualise the relationships between variables in gas laws.

1. Boyle’s Law

Use the link below to observe the simulations.

https://phet.colorado.edu/sims/html/gas-properties/latest/gas-properties_ en.html Steps:

1. Open the PhET “Gas Properties” simulation and select the “Boyle’s Law” experiment setup.

2. Adjust the volume of the container and observe the changes in pressure while keeping the temperature constant.

3. Take note of pressure and volume readings at different volumes.

4. Plot a graph of Pressure (P) on the y-axis and Volume (V) on the x-axis.

Note: The plot should show that as volume decreases, pressure increases, demonstrating the inverse relationship between P and V.

2. Charles’s Law

Steps:

1. Open the PhET “Gas Properties” simulation and select the “Charles’s Law” experiment setup.

2. Adjust the temperature while keeping the pressure constant and observe the changes in volume.

3. Take note of volume and temperature readings at different temperatures.

4. Plot a graph of Volume (V) on the y-axis and Temperature (T) on the x-axis (in Kelvin).

Note: The plot should show that as temperature increases, volume increases, demonstrating the direct relationship between V and T.

3. Avogadro’s Law

Steps:

1. Open the PhET “Gas Properties” simulation and select the “Avogadro’s Law” experiment setup.

2. Add or remove gas particles (moles of gas) while keeping the temperature and pressure constant and observe the changes in volume.

3. Take note of volume and number of moles readings at different amounts of gas.

4. Plot a graph of Volume (V) on the y-axis and Number of Moles (n) on the x-axis.

Note: The plot should show that as the number of moles increases, volume increases, demonstrating the direct relationship between V and n.

4. Gay-Lussac’s Law

Steps:

1. Open the PhET “Gas Properties” simulation and select the “Gay-Lussac’s Law” experiment setup.

2. Adjust the temperature while keeping the volume constant and observe the changes in pressure.

3. Plot a graph of Pressure (P) on the y-axis and Temperature (T) on the x-axis (in Kelvin).

Note: The plot should show that as temperature increases, pressure increases, demonstrating the direct relationship between P and T. Take note of pressure and temperature readings at different temperatures.

DIFFUSION Hello learner! Get ready to explore Graham’s law of diffusion, which explains how gases spread, and Dalton’s law of partial pressures, which shows how gases behave in a mixture. Exciting insights into the world of gases await you!

What is Diffusion?

It is the random motion of molecules by which there is a net flow of matter from a region of high concentration to a region of low concentration. A familiar example is the perfume of a flower that quickly permeates the still air of a room.

Graham’s law of diffusion/effusion It states that the rate of diffusion, or effusion, for a gas is inversely proportional to the square root of its density at constant temperature and pressure.

Mathematically, R ∝ 1___ √_ D or R ∝ 1___ √_ M R = Rate of diffusion or effusion D = Density of the gas M = Molecular mass of the gas Diffusion involves the movement of molecules from an area of high concentration to an area of low concentration due to random motion of molecules. It occurs in gases, liquids and solids and does not require a boundary. On the other hand, effusion specifically refers to the escape of gaseous molecules through a tiny hole into a vacuum or region of lower pressure. It is a type of diffusion, but it is specifically about gaseous molecules moving through a tiny opening.

For 2 gases diffusing at the same time:

R₁__ R₂ = √__ D₂__ D₁ and R₁__ R₂ = √__ M₂___ M₁ Where R₁and R₂are rates of diffusion of gases 1 and 2.

Where D₁and D₂are densities of gases 1 and 2.

Where M₁and M₂are molecular masses of gases 1 and 2 Graham’s law of diffusion can also be expressed in terms of time.

For two gases, t₁__ t₂ = √__ M₂___ M₁ Where t₁ and t₂ are times for the diffusion of gases 1 and 2 Graham’s law of diffusion can also be expressed in terms of volume. For two gases, V₁_ V₂ = √M₂_ M₁ Where V₁ and V₂ are volumes for gases 1 and 2

Activity 4.8: Experiment to Demonstrate Graham’s law of Diffusion

Figure 4.4: A diagram Showing Graham’s Law of Diffusion.

Procedure

1. Soak a piece of cotton wool in concentrated NH₃solution and soak another piece of cotton wool in concentrated HCl.

2. Put at opposite ends of a dry glass tube.

3. Ensure that the glass tube is horizontally mounted to ensure that the diffusion of the gas is not under the influence of gravity.

4. After a few minutes, a white cloud of ammonium chloride appears. This shows the position at which the two gases react.

5. The white cloud of NH₄Cl forms at a position closer to the cotton wool soaked in HCl because particles of NH₃are lighter than that of HCl and so move faster and where they meet react to give that white fume.

Example 4.6

If 100 cm³of methane (CH₄) gas diffuses through a membrane in 40 s, what time will it take 150 cm³of ammonia (NH₃) gas to diffuse through the same membrane?

[C = 12, H = 1, N = 14] Answer: (Use problem-solving strategy)

a. Analyse the question Known [C = 12, H = 1, N = 14] Time, t (CH₄) = 40 s Volume, V(CH₄) = 100 cm³Volume, V(NH₃) = 150 cm³Unknown Time, t (NH₃) =?

b. Apply the problem-solving strategy Mᵣ(NH₃) = 14 + 3(1) = 17 Mᵣ(CH₄) = 12 + 4(1) = 16 R(NH₄) = V __ = 100/40 = 2.5 cm³s⁻¹R(NH₃)_ R(CH₄) = √____ M(CH₄)_____ M(NH₃) R(NH₃)_ 2.5 = √16_ 17 R(NH₃) = 0.9696 × 2.5 = 2.4 cm³s⁻¹t(NH₃) = 1 50_ 2.4 = 62.5 s

Activity 4.9: Discussing Graham’s Law of Diffusion/Effusion

Materials needed: Whiteboard, markers, calculators, sample problems for practice, gas diffusion simulation software or videos.

Group discussion

1. Discuss the meaning of diffusion and effusion.

2. Explain how Graham’s Law mathematically relates the rate of diffusion/ effusion to the molar mass of gases.

Mathematical Expression:

R₁_ R₂ = √D₂_ D₁ and R₁_ R₂ = √M₂_ M₁

3. Discuss real-life examples where diffusion or effusion occurs (e.g., the smell of perfume spreading in a room, gases escaping from a balloon).

4. Watch a simulation demonstrating gas diffusion/effusion https://phet.colorado.edu/sims/html/diffusion/latest/diffusion_all.html Extension task If gases diffuse from high concentrations to low concentrations why do the gases in the Earth’s atmosphere now diffuse into space?

Activity 4.10: Effect of Relative Molecular Mass on the Rate of Diffusion/ Effusion

1. Lighter gases or gases with lower molecular mass diffuse or effuse more rapidly. For example, Hydrogen (H₂) with a molar mass of 2 g/mol diffuses faster than Oxygen (O₂) with a molar mass of 32 g/mol.

2. Heavier gases or gases with higher molecular mass diffuse or effuse more slowly.

This relationship arises because lighter gas molecules move faster at a given temperature than heavier molecules due to their lower mass.

Illustration Consider two gases, Helium (He) with a molar mass of 4 g/mol, and Argon (Ar) with a molar mass of 40 g/mol. Using Graham’s Law:

Rate of diffusion of He________________ Rate of diffusion of Ar = √M_(Ar)_ M_(He) Rate of diffusion of He________________ Rate of diffusion of Ar = √40_ 4 = √10 = 3.16 This means helium diffuses approximately 3.16 times faster than argon.

Activity 4.11: Investigating the Rate of Diffusion of Ammonia and Hydrogen Chloride Gas Materials needed: Cotton wool, ammonia solution (NH₃), hydrochloric acid (HCl) solution, two droppers, long glass tube, rubber stoppers or corks, measuring tape or ruler, stopwatch, safety goggles and gloves Safety Precautions:

· Conduct the experiment in a well-ventilated area or under a fume hood.

· Wear safety goggles and gloves to protect against chemical splashes.

· Ammonia and hydrochloric acid are corrosive and can cause irritation.

Handle with care.

Setting up the apparatus

1. Use a long glass tube for the experiment

2. Insert a piece of cotton wool soaked in ammonia solution at one end of the tube.

3. Insert a piece of cotton wool soaked in hydrochloric acid solution at the other end of the tube.

4. Seal both ends of the tube with rubber stoppers or corks immediately after placing the cotton wool.

5. Observe the formation of a white ring inside the tube, which is the product of the reaction between NH₃and HCl, forming ammonium chloride (NH₄Cl).

6. Use a stopwatch to time how long it takes for the white ring to form after sealing the tube.

7. Measure the distance from each end of the tube to the white ring.

8. Record the distances and time.

Distances travelled by NH₃ Distances travelled by HCl time taken for the white ring to form Calculation Rate of diffusion of NH₃ = Distances travelled by NH₃____________________ time Rate of diffusion of HCl = Distances travelled by HCl___________________ time Extension task Use your periodic table to determine the heaviest and lightest gaseous elements, then determine their relative rates of diffusion.

Example 4.8

Hydrogen gas (H₂) and Nitrogen gas (N₂) are allowed to effuse through a small hole. Given that the molar mass of hydrogen is 2 g/mol and the molar mass of nitrogen is 28 g/mol, calculate the rate of effusion of Hydrogen compared to Nitrogen.

Answer M_(H)₂ = 2 g mol⁻¹, M_(N)₂ = 28 g mol⁻¹Write the formula for Graham’s Law:

Rate of diffusion of H₂_________________ Rate of diffusion of N₂ = √M_(N)₂_ M_(H)₂ Substitute the given values and calculate the ratio:

Rate of diffusion of H₂_________________ Rate of diffusion of N₂ = √28_ 2 = √14 = 3.74 Conclusion: Hydrogen effuses approximately 3.74 times faster than Nitrogen.

Example 4.9

Given:

100 cm³ of Nitrogen (N₂) effuses through a membrane in 50 seconds. How long will it take for 100 cm³ of Chlorine to effuse through the same membrane?

M_(N)₂ = 28 g mol⁻¹, M_(Cl)₂ = 71 g mol⁻¹Answer Rate of effusion of N₂_________________ Rate of effusion of Cl₂ = √M_(Cl)₂_ M_(N)₂ Rate of effusion of N₂_________________ Rate of effusion of Cl₂ = √71_ 28 = 1.59 Rate of diffusion of N₂ = Volume of N₂__________ time for of N₂ = 100 cm³_______ 50 s = 2 cm s⁻¹⟹ 2cm s⁻¹________________ Rate of effusion of Cl₂ = 1.59 ∴ Rate of effusion of Cl₂ = 2cm s⁻¹_______ 1.59 = 1.26 cm s⁻¹time for Cl₂ = V olume of Cl₂________________ Rate of diffusion of N₂ = 100 cm³________ 1.26 cm s⁻¹= 79.4 s

It states that, in a mixture of gases which do not react, the total pressure exerted is equal to the sum of the partial pressures of the individual gases at constant temperature.

For a mixture of gases 1 and 2, P_(T) = P₁ + P₂ where P_(T) = total pressure P₁ and P₂ are the partial presures of gases 1 and 2 P₁= X₁P_(T) P₂= X₂P_(T) where X₁ and X₂ are the mole fractions of gases 1 and 2 X₁ = n₁_ n₁+ n₂ X₂ = n₂_ n₁+ n₂ , where n₁ and n₂ are the number of moles of gases 1 and 2

Example 4.10

Consider the reaction, N₂ + 3H₂ → 2NH₃ If the total pressure is 150 atm, calculate the partial pressure of Nitrogen gas.

Answer: Use problem – solving approach Determine the mole ratio of the gaseous species involved Mole fraction of N₂, X_(N)₂ = n_(N)₂____________ n_(N)₂ + n_(H)₂ + n_(NH)₃ = 1/1 + 3 + 2 = 1/6 P_(N)₂ = X_(N)₂ × P_(T) = 1/6 × 150 = 25.05 atm

Example 4.11

Gases A, B and C have partial pressures 2 atm, 3 atm and 1 atm respectively.

Calculate the total pressure of the gases.

Answers Pₜₒₜₐₗ = P_(A) + P_(B) + P_(C) = 2 + 3 + 1 = 6 atm Finding partial pressure

Example 4.12

The total pressure of a mixture of gases A, B and C is 10 atm. The partial pressure of A is 4 atm and B is 5 atm. Calculate the partial pressure of gas C.

Answer Pₜₒₜₐₗ = P_(A) + P_(B) + P_(C) 10 = 4 + 5 + P_(C) P_(C) = 10 − 9 = 1 atm

Example 4.13

A gas mixture contains 20% Oxygen (O₂), 30% Nitrogen (N₂) and 50% Carbon dioxide (CO₂). If the total pressure is 750 mmHg, Calculate the partial pressures of Oxygen, Nitrogen and Carbon dioxide Answers Pᵢ = Xᵢ × Pₜₒₜₐₗ P_(O)₂ = X_(O)₂ × Pₜₒₜₐₗ = 0.20 × 750 mmHg = 150 mmHg P_(N)₂ = X_(N)₂ × Pₜₒₜₐₗ = 0.30 × 750 mmHg = 225 mmHg P_(CO)₂ = 0.50 × 750 mmHg = 375 mmHg

The ideal gas equation, also known as the ideal gas law, is a fundamental equation in chemistry that describes the behaviour of gases under certain conditions. It relates the Pressure (P), Volume (V), Temperature (T) and amount of gas in moles (n) of an ideal gas sample. It combines Boyle’s law, Charles’s Law and Avogadro’s law.

PV = nRT Where:

P = Pressure of gas Pascals;

V = Volume of gas meters cubed;

n = Moles of gas (moles);

T = Temperature of gas (Kelvin) R = Ideal gas constant = 8.314 J mol⁻¹K⁻¹(S . I. unit of R), Other units of R = 0.082057 L atm mol⁻¹K⁻¹, 62.364 L Torr mol⁻¹K⁻¹NB: Because of the different units of R, it is important to always match the units of pressure, volume, number of moles and temperature given with the units of R.

If the value of R is given as 0.082057 L atm mol ⁻¹K ⁻¹, the unit for pressure must be atm, the unit for volume must be litre and for temperature must be Kelvin.

If the value of R is given as 62.364 L Torr mol⁻¹K⁻¹, the unit for pressure must be Torr, for volume the unit must be litre, and for temperature must be Kelvin.

The ideal gas equation is essential for understanding and predicting the behaviour of gases in various chemical reactions and processes.

Example 4.14

What is the volume of 10 g of nitrogen gas at 25 ⁰C and 101 kPa? [N =14, R = 8.314 Jmol⁻¹K⁻¹] Answer (Use problem-solving strategy) Known N =14, R = 8.314 Jmol⁻¹K⁻¹; Temperature, T = 25 + 273 = 298 K; Mass, m = 10 g Pressure, P = 101 kPa = 101 000 Pa; R = 8.314 Jmol⁻¹K⁻¹Unknown Volume, V =?

M( N₂) = 2(14) = 28 g n = m/M = 10/28 = 0.357 mol PV = nRT V = nRT/P = 0.357 × 8.314 × 298/101000 = 0.0088 m³∴ volume = 0.0088 m³Activity 4.12: To share understanding of Boyle’s Law, Charles’ Law, and Avogadro’s Law.

In your group:

1. Discuss and explain (Boyle’s, Charles’, or Avogadro’s) law in your own words.

a. How does pressure relate to volume in Boyle’s Law?

b. What happens to the volume of gas when the temperature increases according to Charles’ Law?

c. How does adding more gas particles affect volume according to Avogadro’s Law?

2. Give a simple example to illustrate the law.

3. Think of a real-life situation where Boyle’s Law applies.

4. How do hot air balloons relate to Charles’ Law?

5. When might Avogadro’s Law be important in the real world?

6. Share your understanding and examples with the whole class.

7. Discuss how these laws (Boyle’s, Charles’, or Avogadro’s) relate to one another.

a. How might changing the temperature in Charles’ Law affect the pressure in Boyle’s Law?

b. If you increase the number of particles according to Avogadro’s Law, what might happen to the pressure or volume?

c. How do these laws together help us understand the behaviour of gases?

Activity 4.13: Deriving the Ideal Gas equation Boyle’s Law: Pressure increases as volume decreases (like squeezing a balloon).

PV = constant, when T and n are constant.

Charles’ Law: Volume increases as temperature rises (like a hot air balloon).

V ∝ T, so V = kT, when P and n are constant.

Avogadro’s Law: Volume increases with more gas particles (like inflating a balloon).

V ∝ n, so V = kn, when P and T are constant.

Combining these three relationships into one equation that relates pressure (P), volume (V), temperature (T), and the number of moles (n) of a gas:

V ∝ nT/P To remove the proportionality sign, introduce a constant, R, so:

PV = nRT

Example 4.15

A car tyre is filled with 0.050 moles of air at a temperature of 20°C. The tyre has a volume of 10.0 liters. What is the pressure in the tyre in kPa?

Answer Given Data:

n = 0.050 mol ; V = 10.0 L ; T = 20 ∘ C = 20 + 273.15 = 293.15 K R = 8.314 J ⋅ mol⁻¹⋅ K⁻¹= 8.314 kPa ⋅ L ⋅ mol⁻¹⋅ K⁻¹Use the Ideal Gas Law:

PV = nRT Solve for P P (pressure):

P = nRT/V Substitute the values:

P = 0.050 mol × 8.314 kPa ⋅ L ⋅ mol⁻¹⋅ K⁻¹× 293.15 K/10.0 L = 12.2 kPa The pressure in the tyre is 12.2 kPa.

Example 4.16

You need to inflate a car tyre to a pressure of 2.5 atm. The tyre has a volume of 30.0 litres, and the temperature is 300 K. How many moles of air are required to achieve the desired pressure?

Answer Use the Ideal Gas Law:

PV = nRT Given:

P = 2.5 atm ; V = 30.0 L ; T = 300 K ; R = 0.0821 L ⋅ atm / mol ⋅ K Solve for n:

n = PV___ RT = 2.5 atm × 30.0 L/0.0821 L ⋅ atm / mol ⋅ K × 300 K = 3.05 mol The implications of using the ideal gas equation to predict the behaviour of a gas at extremely high pressures or low temperatures The Ideal Gas Law is a simplified model that works well for predicting gas behavior under a wide range of conditions but fails under extremes of high pressure and low temperature. In these situations, molecular interactions and phase changes cause significant deviations from ideal behavior. To accurately describe gases under such conditions, more complex equations like the Van der Waals equation or empirical models that account for real gas behavior must be used. Understanding these limitations is crucial for applications where precision is necessary, such as in chemical engineering, material science, and thermodynamics.

A real gas behaves differently from what is expected in ideal conditions at:

1. High-pressure

2. Low temperature This is due to the following:

a. Real gases have an actual volume of molecules, which is significant at very high pressure. At extremely high pressures, the value of PV becomes greater than the ideal value. The deviation increases when the relative molecular mass (Mr) increases.

b. Intermolecular forces always exist in real gases. At lower temperatures, the kinetic energy of the molecules is at their lower value, intermolecular forces increase, which reduces the pressure P, making the PV in value less than the ideal value. Examples of gases with greater intermolecular forces are those which are more polar.

Figure 4.5: Deviations from ideal gas law for select gases.

Example 4.17

Arrange these gases in their order of deviation from ideal behaviour and explain the order: O₂, Ne, N H₃ Answer Ne < O₂ < N H₃ Increasing order of deviation Reasons: Neon, a noble gas, exhibits the least deviation from ideal behaviour.

This is because noble gases have very weak intermolecular forces (it is non- polar), and their molecules are far apart, resulting in minimal interactions between them. Oxygen, while also a non-polar molecule like neon, has slightly stronger intermolecular forces due to its higher molar mass. However, these forces are still relatively weak, compared to other gases, resulting in a moderate deviation from ideal behaviour. Ammonia, a polar molecule with hydrogen bonding, experiences the highest deviation from ideal behaviour. Hydrogen bonding leads to stronger intermolecular attraction between ammonia molecules, compared to neon and oxygen, causing greater deviation from the ideal gas law.

Assumptions of the Ideal Gas Law

Ideal Gas Law: PV = nRT

1. Particles have zero volume: Gas particles are considered as point particles with no volume, meaning they occupy no space.

2. No intermolecular forces: The law assumes that there are no attractive or repulsive forces between gas particles, so they move independently of one another.

3. The gas consists of a large number of molecules: These molecules are in constant random motion.

4. Collisions between molecules (and with the walls) are perfectly elastic:

No kinetic energy is lost during collisions, meaning the total kinetic energy of the system remains constant.

5. The time taken for a collision is negligible: The duration of collisions between molecules is so short that it can be ignored compared to the time between collisions.

6. The motion of molecules follows Newton’s laws: Specifically, Newtons second law applies to the motion of the molecules between collisions, meaning their motion is governed by classical mechanics.

Use the link below to watch a video on ideal gases:

https://www.youtube.com/watch?v=Hr5Baj3lXFA&t=106s Relating assumptions of Ideal Gas Law to Real-Gases The ideal gas law works well under many conditions, it doesn’t perfectly describe real gases, especially at very high pressures or very low temperatures, where gas particles do interact and have volume.

Activity 4.14: The Ideal Gas Law equation and its underlying assumptions.

Group Discussion: In small groups discuss the conditions under which real gases deviate from ideal behavior

1. High Pressure: Discuss why real gases deviate from ideal behaviour at high pressures.

a. What happens to the volume of gas particles under high pressure?

b. How does this affect the distance between particles and their interactions?

2. Low Temperature: Discuss why real gases deviate from ideal behaviour at low temperatures.

a. How does a decrease in temperature affect the kinetic energy of gas particles?

b. What role do intermolecular forces play when gas particles move more slowly?

3. Large Gas Molecules: Discuss how the size of gas molecules can cause deviations from ideal behaviour.

a. How does the physical size of gas molecules affect their volume?

b. Why might larger molecules not behave ideally compared to smaller molecules?

4. Strong Intermolecular Forces: Discuss how strong intermolecular forces lead to deviations from ideal behaviour.

a. What are intermolecular forces, and how do they affect gas particles?

b. In what way do these forces influence the behaviour of gases under different conditions?

Activity 4.15: Demonstrating Deviations from Ideal Gas Behavior High-Pressure Demonstration with a Carbonated Drink Objective: To show how gases dissolve in liquids under high pressure, which deviates from the Ideal Gas Law.

Materials Needed: Carbonated drink (e.g., soda), safety goggles and gloves.

Carbonated drinks contain dissolved CO₂gas. At high pressure (inside the sealed bottle), CO₂stays dissolved in the liquid.

Procedure:

1. Shake a sealed bottle of soda and then carefully open it.

2. State what you observe.

Discussion: Explain your observation.

Use the link below to watch a video comparison of real gas behavior versus ideal gas law predictions:

https://www.khanacademy.org/science/ap-chemistry-beta/ x2eef969c74e0d802:intermolecular-forces-and-properties/ x2eef969c74e0d802:deviation-from-ideal-gas-law/v/real-gases-deviations- from-ideal-behavior

Example 4.18

Explain why real gases deviate from ideal behaviour?

Answer Real gases deviate from ideal behaviour because the assumptions made in the ideal gas law do not hold under certain conditions. These assumptions are:

1. Particles have zero volume

2. No intermolecular forces

3. The gas consists of a large number of molecules

4. Collisions between molecules (and with the walls) are perfectly elastic

5. The time taken for a collision is negligible

6. The motion of molecules follows Newton’s laws

The Van der Waals equation is an equation of state that extends the ideal gas law to include the non-zero size of gas molecules and the interactions between them.

For one mole of gas, the equation is (P + a/V ²)(V– b) = RT and for n moles of gas, the Van der Waals equation is (P + a n²___ V ²)(V– nb) = n RT Where;

a = a measure of the strength of the intermolecular forces.

b = The excluded molar volume

Activity 4.16: Understanding the Van der Waals Equation and Its Purpose in Describing the Behaviour of Real Gases Objective: To explore the Van der Waals equation, understand how it accounts for the behaviour of real gases, and learn why it differs from the Ideal Gas Law.

Steps:

1. Write the Ideal Gas Law

2. Discuss the assumptions of the Ideal Gas Law.

3. State the assumptions might not hold true for real gases.

4. Write the Van der Waals Equation.

5. Explain the significance of each term in the equation.

6. Discuss how the Van der Waals equation modifies the Ideal Gas Law to describe real gases more accurately.

7. Think about how the constants a and b would change under different conditions.

Extended task

1. State differences between the ideal gas law and the Van der Waals equation.

2. Use the Van der Waals equation to calculate the pressure of 1 mole of carbon dioxide gas in a 2-litre container at 273 K. Given: a = 3.592, b = 0.0427 and R = 8.314 J mol⁻¹K⁻¹3. Compare this value to that for the ideal gas equation, express the difference as a percentage.

Equation for preparation:

Mg(s) + 2HCl(aq) → Mg Cl₂(aq) + H₂(g)

Figure 4.6: Preparation of Hydrogen gas Test for hydrogen gas: Put a burning/lighted splint in the gas; a “pop” sound indicates the presence of hydrogen gas.

How to dry hydrogen gas: Pass the gas produced through anhydrous CaCl₂.

Physical properties of hydrogen gas

a. It is the lightest gas and has the lowest density.

b. It is colourless and odourless.

c. It is insoluble in water.

Chemical properties of hydrogen gas

a. It is neutral to litmus.

b. It is unreactive under normal conditions.

c. It does not support combustion.

d. It burns in oxygen (air) with the pop sound to produce water.

2H₂(g) + O₂(g) → 2H₂O (l)

e. It reacts with halogens e.g. H₂(g)+ F₂(g)→2HF (g) Uses of hydrogen gas in everyday life

a. It is used to produce ammonia gas in the Haber process.

b. It is used to manufacture margarine by hydrogenation of unsaturated fats.

c. It is used in oxy-hydrogen flame for cutting and welding of metals.

d. It is used in fuel cells.

Activity 4.17: Exploring the uses of hydrogen gas in everyday life Objective: Investigate and present the various uses of hydrogen gas in everyday life.

Materials: Internet access for research (if available), handouts or printed resources on hydrogen gas.

Steps In small groups find the use of hydrogen gas in areas such as:

1. Fuel and energy

2. Industrial processes

3. Food industry

4. Chemical industry

Activity 4.18: Preparation and Collection of Hydrogen Gas

Materials Needed: Zinc granules or magnesium ribbon, dilute hydrochloric acid (HCl) or sulfuric acid (H₂SO₄), conical flask (250 mL), thistle funnel or dropping funnel, rubber stopper with a single hole, delivery tube, water-filled trough or basin, gas jar or test tubes, wooden splints, matches or lighter, safety goggles, lab coat or apron, gloves Safety Precautions:

· Ensure you wear safety goggles and a lab coat.

· Handle acids with care.

· Conduct the experiment in a well-ventilated area or under a fume hood.

Procedure

1. Weigh out a defined mass of zinc granules or a piece of magnesium ribbon and place into the conical flask.

2. Insert the thistle funnel into the flask through the rubber stopper, ensuring the end of the funnel is submerged in the acid when added.

3. Connect the delivery tube to the rubber stopper and place the other end under the water in the trough or basin.

4. Position a gas jar or test tube upside down over the end of the delivery tube in the trough to collect the gas.

5. Slowly add a defined volume dilute hydrochloric acid (of known concentration) through the thistle funnel. The acid will react with the zinc, producing hydrogen gas.

6. Observe the formation of bubbles as the gas is generated.

7. The hydrogen gas will travel through the delivery tube and displace water in the gas jar or test tube, filling it with hydrogen gas.

8. Once enough gas is collected, carefully remove the gas jar or test tube from the water, keeping it upside down.

9. Light a wooden splint and bring it near the mouth of the jar/test tube. The presence of hydrogen gas will be confirmed by a characteristic “pop” sound when it ignites.

Chemical reaction Zn(s) + 2HCl(aq) → Zn Cl₂(aq) + H₂(g) Mg(s) + 2HCl(aq) → Mg Cl₂(aq) + H₂(g) Zn(s) + H₂ SO₄(aq) → Zn SO₄(aq) + H₂(g) Extended task: Discussions on Activity 4.18

1. What did you observe when the acid was added to the zinc?

2. How did you confirm the presence of hydrogen gas?

3. Why does hydrogen gas produce a “pop” sound when ignited?

4. State the method of collecting hydrogen gas and explain why it is collected by this method.

5. Calculate the number of moles of each reactant and determine which was in excess.

6. Calculate the number of moles of gas produced using the ideal gas equation.

7. Calculate the number of moles of gas produced using a mass and mole balance calculation

8. Compare solution from question 6 with that in question 7

9. State two properties of hydrogen that enable it to be used to fill balloons.

10. State three practical uses of hydrogen in everyday life.

11. You are provided with the thistle funnel, delivery tube, split cork, conical flask, gas jar, beehive stand, water trough, Magnesium ribbon and dilute HCl. Design an experiment to prepare and test for hydrogen gas.

Equation for preparation: CaC O₃(s) + 2HCl (aq) → Ca Cl₂ (aq)+ H₂ O(l) + C O₂(g):

Figure 4.7: Preparation of Carbon dioxide Test for carbon dioxide gas: Pass the gas through lime water (saturated solution of calcium hydroxide), which turns milky.

How to dry carbon dioxide gas: Pass the gas produced through concentrated sulphuric acid. As an alternative (as concentrated sulphuric acid is hazardous to handle) it can be passed over anhydrous CaCl₂.

Physical properties of carbon dioxide gas

1. It is colourless and odourless.

2. It is denser than air.

3. It is soluble in water.

4. It condenses into a white solid called dry ice.

Chemical properties of carbon dioxide gas

1. It turns moist blue litmus red.

2. It turns lime water (Ca(OH)₂solution) milky.

C O₂ (g)+ Ca (OH)₂(aq) → CaC O₃ (s)+ H₂ O(l)

3. Excess passage of CO₂causes milkiness to disappear.

CaC O₃ (s)+ H₂ O(l) + C O₂(g) → Ca(HC O₃)₂(aq)

4. It reacts with water in the presence of sunlight and chlorophyll to produce glucose and oxygen (photosynthesis).

6C O₂ (g)+ 6 H₂ O (l) → C₆ H₁₂ O₆ (s)+ 6 O₂(g)

5. It does not support combustion.

Uses of Carbon dioxide in everyday life

1. It is used in photosynthesis to produce glucose and oxygen.

2. It is used in fire extinguishers to extinguish fires.

3. It is dissolved into fizzy drinks.

4. It is used to manufacture refrigerants.

Activity 4.19: Preparation and Collection of Carbon Dioxide Gas Materials Needed: marble chips (CaCO₃), dilute hydrochloric acid (HCl), Conical flask (250 mL), thistle funnel or dropping funnel, rubber stopper with a single hole, delivery tube, water-filled trough or basin, gas jar or test tubes, Wooden splints, matches or lighter, safety goggles, lab coat.

Procedure:

1. Weight out a defined mass of marble chips (calcium carbonate) and place them into the conical flask.

2. Insert the thistle funnel into the flask through the rubber stopper, ensuring the end of the funnel is submerged in the acid when added.

3. Connect the delivery tube to the rubber stopper and position the other end under the water in the trough or basin.

4. Place a gas jar or test tube over the end of the delivery tube to collect the gas.

5. Slowly add a defined volume of dilute hydrochloric acid (of a known concentration) through the thistle funnel. The acid will react with the marble chips, producing carbon dioxide gas.

6. Observe the formation of bubbles as the gas is generated.

7. The carbon dioxide gas will travel through the delivery tube and displace water in the gas jar or test tube, filling it with carbon dioxide gas.

8. Test for Carbon Dioxide:

Limewater Test: Pour a small amount of limewater into another test tube.

Use the gas collected to bubble through the limewater. The presence of carbon dioxide will be confirmed if the limewater turns milky or cloudy.

9. Reaction: CaC O₃ (s)+ 2HCl (aq) → Ca Cl₂ (aq)+ H₂ O (l)+ C O₂(g) Discussion:

10. What did you observe when the acid was added to the marble chips?

11. How did you confirm the presence of carbon dioxide gas?

12. Why does carbon dioxide gas make limewater turn cloudy?

13. Why does carbon dioxide gas extinguish a flame?

14. Calculate the number of moles of each reactant and determine which was in excess.

15. Calculate the number of moles of gas produced using the ideal gas equation.

16. Calculate the number of moles of gas produced using a mass and mole balance calculation

17. Compare solution from question 6 with that in question 7

18. What are some uses of carbon dioxide gas in daily life or industry?

Equation for preparation: Ca (OH)₂(aq)+ 2N H₄ Cl(aq) → Ca Cl₂(aq) + 2 H₂ O (l)+ 2N H₃(g)

Figure 4.8: Preparation of Carbon dioxide Test for gas:

a. It turns moist red litmus blue.

b. It forms white fumes with concentrated HCl vapour to form NH₄Cl.

How to dry the gas: Pass the gas produced over solid calcium oxide (CaO).

Physical properties

a. It is a colourless gas with a pungent, choking smell.

b. It is less dense than air.

c. It is soluble in water.

Chemical properties

a. It turns moist red litmus blue.

b. It forms dense white fumes with HCl; NH₃+ HCl → NH₄Cl

c. It burns in oxygen with a pale-yellow flame.

Uses of Ammonia gas in everyday life

a. It is used to manufacture fertilisers.

b. It is used to manufacture explosives.

c. It is used to manufacture nylon.

d. It is used to manufacture plastics.

e. It is used to manufacture pigment.

Activity 4.20: Preparation and Collection of Ammonia Gas

Materials Needed: Ammonium chloride (NH₄Cl), Calcium hydroxide (Ca(OH)₂) or sodium hydroxide (NaOH) pellets, test tube or small conical flask, Bunsen burner, rubber stopper with a single hole, delivery tube, water- filled trough or basin, gas jar, red litmus paper, concentrated hydrochloric acid (HCl) glass rod or dropper, safety goggles, Lab coat or apron, gloves.

Safety Precautions:

· Ensure you wear safety goggles and a lab coat.

· Handle chemicals, especially concentrated acids and bases, with care.

· Conduct the experiment in a well-ventilated area or under a fume hood, as ammonia gas is pungent and irritating.

Procedure

1. Mix ammonium chloride and calcium hydroxide (or sodium hydroxide) in a test tube or small conical flask.

2. Insert the delivery tube into the test tube or flask through the rubber stopper, ensuring a tight fit.

3. Collect the gas by upward displacement of air (since ammonia is less dense than air).

4. Discuss the Reaction:

Ca (OH)₂(aq)+ 2N H₃Cl(aq) → Ca Cl₂(aq) + 2 H₂O (l)+ 2N H₃(g)

5. Heat the Mixture:

a. Gently heat the test tube containing the mixture using a Bunsen burner. The heat will cause the ammonium chloride and calcium hydroxide to react, releasing ammonia gas.

b. Observe the release of gas, which can be recognized by its pungent smell.

6. The ammonia gas will travel through the delivery tube. If using water displacement, it will collect in the gas jar or test tube, displacing the water. If collecting by upward displacement of air, it will fill the gas jar.

7. Observe the smell of the gas, the reaction with litmus paper, and the formation of ammonium chloride fumes.

8. Test for Ammonia:

a. Litmus Paper Test: Hold a piece of moist red litmus paper near the mouth of the gas jar or test tube. The paper will turn blue, indicating the presence of ammonia gas (a basic gas).

b. Fume Formation: Dip a glass rod or dropper into concentrated hydrochloric acid and hold it near the mouth of the gas jar. White fumes of ammonium chloride will form, confirming the presence of ammonia.

Discussion:

1. What did you observe when the ammonium chloride and calcium hydroxide were heated?

2. How did you confirm the presence of ammonia gas?

3. Why does ammonia gas turn red litmus paper blue?

4. Why is ammonia collected by upward displacement of air?

5. What are some uses of ammonia gas in daily life or industry?

6. Why is it particularly important to conduct this experiment under a fume hood?

Review Questions 4.1

1. Draw a simple diagram illustrating the arrangement and motion of particles in a solid and a liquid.

2. Compare and contrast the behaviour of particles in solids and liquids according to the kinetic theory of matter.

3. In a laboratory experiment, students investigate the effect of temperature on the viscosity of a liquid. Describe how the kinetic theory of matter can be applied to explain the observed changes in viscosity as temperature increases.

4. What real-world phenomena can be explained by the Kinetic Theory of Matter, and how does understanding this theory help in practical applications?

Review Questions 4.2

1. Describe how the kinetic theory of matter explains the behaviour of gases.

Provide one example to illustrate how gas particles move and interact with each other.

2. A sealed container of gas is heated, causing an increase in temperature.

Explain how the kinetic theory of matter can be used to predict the changes in gas pressure and volume inside the container.

3. A gas cylinder is compressed to half its original volume while maintaining a constant temperature. Using the principles of the kinetic theory of matter, explain how the gas pressure changes because of compression.

Review Questions 4.3

1. Explain the difference between condensation and sublimation, providing examples of substances that undergo each of these changes of state.

2. Describe how a solid can change directly into a gas without passing through the liquid state, using examples from everyday life.

3. Analyse the impact of changes in atmospheric pressure on the boiling point of a liquid. Explain how altitude affects the boiling point of water and its implications for cooking and food preparation.

Review Questions 4.4

1. State the definition of the following gas laws:

a. Boyles’ law

b. Charles’ law

c. Gay-Lussac’s law

d. Combined gas law

2. Express any two (2) of the gas laws mathematically taking into consideration the conditions associated with each law.

a. Identify the variables that are related to each law.

b. Illustrate the relationships among the various variables of the law.

3.

a. Explain how Boyles’ law can be applied in real life situation.

b. Derive the combined gas law from Boyle’s, Charles’ and Gay– Lussac’s laws.

4.

a. 20 m3 of gas at a pressure of 100,000 Pa was compressed to a pressure of 400,000 Pa at constant temperature. Calculate the final volume of the gas.

b. 20 dm3 of gas in a cylinder is heated from 150 K to a certain temperature. If its volume expands to 35 dm3, Calculate the final temperature of the gas.

c. 35 cm3 of a gas at 1.5 atm at 40 0C was compressed to 20 cm3at 50 0C, calculate the final pressure of the gas.

d. A balloon was filled to a volume of 2.0 dm3 with 0.082 moles of helium gas. Suppose 0.015 moles of helium is added to the balloon with constant temperature and pressure, what will be the new volume of the balloon?

e. The temperature of a gas contained in a cylinder undergoes combustion in constant volume, increasing from 30 0C to 1500 0C.

If the normal atmospheric pressure is 101 kPa, calculate the peak pressure reached after combustion.

5. A 250 cm3 sample of a gas has a pressure of 20 Pa. What will be its volume if the pressure is raised to 45 Pa at the same temperature?

6. A balloon can hold 800 cm3 of air before bursting. If the balloon contains 500 cm3 of air at 10 0C, determine whether the balloon would burst when it is taken into a room of temperature of 20 0C, assuming that the pressure of the gas in the balloon is kept constant.

Review Questions 4.5

1. State Graham’s law of diffusion or effusion.

2. Given that 100 cm3 of ethane (C2H6) diffuses through a membrane in 40 s, what time will it take 80 cm3 of propane (C3H8) to diffuse through the same membrane at the same temperature and pressure? [C = 12, H = 1].

3. Given a cylindrical glass tube, stoppers, two retort stands, cotton wool, concentrated NH3, concentrated HCl, and tweezers, design an experiment to determine the rate of diffusion of a gas.

4. A gas mixture contains methane (CH₄) and Sulfur dioxide (SO₂). The molar mass of methane is 16 g/mol, and the molar mass of Sulfur dioxide is 64 g/mol. Determine the ratio of their effusion rates.

5. Two gases, A and B, have molar masses of 4 g/mol and 36 g/mol respectively. If gas A effuses through a membrane in 30 seconds, how long will it take gas B to effuse the same amount?

6. An unknown gas X effuses at half the rate of Oxygen. What is the molar mass of gas X?

Review Questions 4.6

1. State Dalton’s law of partial pressures.

2. A mixture of gases contains 4.76 mole of Ne, 0.74 mole of Ar and 2.5 mole of Xe. Calculate the partial pressure of the gases if the total pressure is 2 atm, at a fixed temperature.

3. A mixture of 40.0 g of Oxygen and 40.0 g of Helium has a total pressure of 0.900 atm. What is the partial pressure of each gas? [H = 2, O = 16]

4. Three gases (8 g of CH4, 18 g of C2H6, and unknown amount of C3H8) were added to the same 10.0 dm3 container. At 23°C , the total pressure in the container was measured to be 4.43 atm. Calculate the partial pressure of each gas in the container. [C = 12, H = 1]

5. A gaseous mixture of O2 and N2 contains 32.8% Nitrogen by mass. What is the partial pressure of Oxygen in the mixture if the total pressure is 785.0 mmHg? [ N = 14, O = 16]

6. A gas mixture contains 40% Helium (He), 30% and 60% Argon (Ar). If the total pressure is 400 mmHg. Calculate the partial pressures of He, and Ar

7. A gas mixture of gases A, B and C contain 2 mol, 3 mol and 5 mol respectively. The total pressure of the gases is 10 atm. Calculate the:

a. mole fraction of gases A, B and C.

b. partial pressures of gases A, B and C.

Review Questions 4.7

1. A helium balloon with a volume of 2.5 litres is filled to a pressure of 1.2 atm and a temperature of 290K. How many grams of helium are in the balloon? (Molar mass of helium = 4.00 g/mol)

2. You have a balloon that can hold 2.50 litres of air. If you want to inflate the balloon at room temperature (25°C) and a pressure of 1.00 atm, how many moles of air are needed?

3. A room has a volume of 100 L. Due to a gas leak, the pressure inside the room rises to 1.1 atm. The temperature in the room is 295 K. If the room was initially filled with air at 1.0 atm and the same temperature, how many moles of gas leaked into the room?

4. Under what conditions are deviations from ideal gas behaviour most likely to occur?

5. A real gas behaves more like an ideal gas at high temperatures and low pressures. Explain why this is the case.

6. Evaluate the significance of intermolecular forces in causing deviations from ideal behaviour in real gases. Discuss how these forces affect gas behaviour under different conditions.

7. Why do real gases deviate from the ideal gas law at high pressures?

8. Explain how the volume of gas molecules contributes to the limitations of the ideal gas law at high pressures.

9. How does the Van der Waals equation address the limitations of the ideal gas law?

10. What is the Van der Waals equation, and how does it differ from the ideal gas law?

11. How do you calculate the pressure of 1 mol of a real gas using the Van der Waals equation?

12. What is the significance of the constants a and b in the Van der Waals equation?

13. How do you determine the volume (V) of a gas using the Van der Waals equation?

14. How do the Van der Waals constants a and b vary between different gases, and why is this important?

15. What safety precautions should be taken when preparing hydrogen gas?

16. Why is hydrogen considered a potential clean energy source?

17. What safety precautions should be taken when preparing carbon dioxide gas?

18. How can you test the presence of carbon dioxide gas?

19. How does carbon dioxide contribute to the greenhouse effect?

20. What role does carbon dioxide play in the carbon cycle?

21. What safety precautions should be taken when preparing ammonia gas?

22. How can you confirm the presence of ammonia gas?

23. State the properties of CO2 that allow it to be used in the manufacture of:

a. fizzy drinks

b. gas refrigerants.

24. In the laboratory preparation of ammonia gas, the round bottom flask containing the reagents is slanted downwards whilst being heated.

Explain why these actions are necessary in the preparation of the gas.

25. What role do intermolecular forces play in causing deviations from ideal gas behaviour?