PDF CHEMISTRY F4

MODULE • Chemistry FORM 4 INTRODUCTION TO CHEMISTRY MATTER AND THE ATOMIC STRUCTURE ACIDS, BASES AND SALTS THE PERIODIC TABLE OF ELEMENTS CHEMICAL BOND RATE OF REACTION THE MOLE CONCEPT, CHEMICAL FORMULA AND EQUATION 1.1 Development in Chemistry and Its Importance in Daily Life 1.2 Scientific Investigation in Chemistry 1.3 Usage, Management and Handling of Apparatus and Materials 2.1 Basic Concepts of Matter 2.2 The Development of Atomic Model 2.3 Atomic Structure 2.4 Isotopes and Its Uses ENRICHMENT EXERCISE 6.1 The Role of Water in Showing Acidic and Alkaline Properties 6.2 The Strength of Acid and Alkali 6.3 Chemical Properties of Acid and Alkali 6.4 Concentration of Aqueous Solution 6.5 Standard Solution 6.6 PH Value 6.7 Neutralisation 6.8 Salt, Crystals and Uses of Salt in Daily Life 6.9 Preparation of Salts 6.10 Effect of Heat on Salt 6.11 Qualitative Analysis ENRICHMENT EXERCISE 4.1 The Development of the Periodic Table of Elements 4.2 The Arrangement in the Modern Periodic Table of Elements 4.3 Elements in Group 18 4.4 Elements in Group 1 4.5 Elements in Group 17 4.6 Elements in Period 3 4.7 Transition Elements ENRICHMENT EXERCISE 5.1 Basic of Compound Formation 5.2 Ionic Bonds 5.3 Covalent Bonds 5.4 Hydrogen Bonds 5.5 Dative Bonds 5.6 Metallic Bonds 5.7 Ionic and Covalent Compound ENRICHMENT EXERCISE 7.1 Determine Rate of Reaction 7.2 Factors Affecting the Rate of Reactions 7.3 Application of Factors that Affect the Rate of Reaction in Daily Life 7.4 The Collision Theory ENRICHMENT EXERCISE 3.1 Relative Atomic Mass and Relative Molecular Mass 3.2 Mole Concept 3.3 Chemical Formula 3.4 Chemical Equations ENRICHMENT EXERCISE CONTENTS UNIT 1 UNIT 2 UNIT 6 UNIT 4 UNIT 5 UNIT 7 UNIT 3 1 5 98 53 77 161 27 MANUFACTURED SUBSTANCES IN INDUSTRY THEME: INDUSTRIAL CHEMISTRY THEME: THE IMPORTANCE OF CHEMISTRY THEME: FUNDAMENTALS OF CHEMISTRY THEME: INTERACTION BETWEEN MATTERS 8.1 Alloy and its Importance 8.2 Composition of Glass and Its Uses 8.3 Composition of Ceramics and Its Uses 8.4 Composite Materials and Its Importance ENRICHMENT EXERCISE 196 UNIT 8 PDF Versi BM 00 Chem Contents Eng F4-csy1p.indd 1 20/12/2022 3:45 PM

MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. Performance Standard Performance Level (PL) Descriptor Mastered Not mastered U n i t 2 1 Recall knowledge and basic skills on matter and the atomic structure. 2 Understand and explain matter and the atomic structure. 3 Apply knowledge on matter and the atomic structure to explain the natural occurrences or phenomena and be able to carry out simple tasks. 4 Analyse knowledge on matter and atomic structure in the context of problem solving the natural occurrences or phenomena. 5 Evaluate knowledge on matter and atomic structure in the context of problem solving and decision-making to carry out a task. 6 Invent by applying the knowledge on matter and atomic structure in the context of problem solving and decision- making or when carrying out an activity/ task in new situations creatively and innovatively; giving due considerations to the social/ economic/ cultural values of the community. Guideline to Scan AR for 3D Model 1 Download the free QR reader application from the Play Store. 2 Download the free 'Nilam AR Module 2' application by scanning the QR code below. (Applicable for android only) 3 Find a page that have the following icon. To obtain a complete list of Performance Standard, you may scan this QR code. 4 Scan the icon on that page with your smartphone and enjoy the 3D model that appear. AR 00 Chem Contents Eng F4-csy1p.indd 2 20/12/2022 3:45 PM

1 UNIT 1 MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 1 UNIT INTRODUCTION TO CHEMISTRY DEVELOMENT IN CHEMISTRY FIELD AND ITS IMPORTANCE IN DAILY LIFE 1.1 CS 1.1 What is Chemistry? • Chemistry is a field ofscience thatstudy the structures, properties, compositions and interactions between matters. 1 Complete the following concept map. 2 State the career related to chemistry field. • Biomedical engineer • Biotechnology researcher • Doctor • Pharmacologist • Pharmacist • Nanotechnology engineer • Molecular engineer • Cosmetic scientist • Cosmetic consultant • Engineer • Geochemist Biotechnology Pharmaceutical Nanotechnology Cosmetic Green technology Food • Food colouring • Preservative • Stabiliser Industry • Paint • Glass • Polymer • Ceramic Medicine • Antibiotic • Antiseptic • Analgesic Agriculture • Herbicide • Pesticide • Fertiliser • Hormone Chemicals in daily life 01 U1 Chemistry F4(p01-4)csy2p.indd 1 20/12/2022 3:39 PM

© Nilam Publication Sdn. Bhd. 2 UNIT 1 MODULE • Chemistry FORM 4 1.2 SCIENTIFIC INVESTIGATION IN CHEMISTRY CS 1.2 PERFORMANCE LEVEL (PL) Mastered Not mastered PL1 Recall knowledge and basic skills in chemistry, scientific investigation, usage, management and handling of apparatus and materials. What is scientific investigation? • A scientific method used in solving problem in science. 1 Complete the following steps in scientific method. Making observations Collecting data Interpreting data Making inference Planning an experiment Making conclusion Identifying problem Controlling variables Preparing a report Making hypothesis Identifying variables USAGE, MANAGEMENT AND HANDLING OF APPARATUS AND MATERIALS 1.3 CS 1.3 PERFORMANCE LEVEL (PL) Mastered Not mastered PL2 Understand and explain the importance of chemistry, scientific investigation, usage, management and handling of apparatus and materials. 1 Compete the following label of personal protective equipment and its function. PL1 PL2 Prevents dust or splashes of chemicals from getting into the eyes Protecting respiratory organs from chemicals in the form of powder or fume Protection for the body and clothing against chemical spills such as acid, alkali and organic solvents Handling chemicals and protect hands from injuries and chemicals Protects the foot from injury caused by chemical spills, exposed to sharp objects or toxic substances Goggles Face mask Lab coat Glove Safety shoes 01 U1 Chemistry F4(p01-4)csy2p.indd 2 20/12/2022 3:39 PM

3 UNIT 1 MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 2 Name and state the function of the following safety equipment in the laboratory. Specially designed equipment to carry out experiments that release toxic vapour , cause combustion or produce pungent smell. Used to wash and clean the eyes when accidents occur on the parts of eyes. Used to wash, clean and extinguish fire on body when accidents occur on the parts of . body . Used for extinguishing fire in laboratory. Removing chemical substances, oil, dirt and microorganisms from the hands. (a) Fume chamber (c) Eyewash (b) Shower (d) Fire extinguisher (e) Hand wash 3 Complete the following storage for chemicals. Substances Storage methods (a) Reactive substances: Lithium, sodium, potassium • Stored in paraffin oil to prevent reaction with moisture in the air (b) Hydrocarbon and organic solvents: Highly volatile liquids, benzene, tetrachloromethane • Store far from sunlight and heat source (c) Substances that decompose easily: Concentrated nitric acid, hydrogen peroxide solution • Stored in dark bottles as it decomposed easily with the presence of sunlight (d) Substances with pH < 5 and pH > 9: Corrosive chemicals, strong acids, strong bases • Stored in special storage cabinets that are kept locked (e) Heavy metals and toxic substances: Chromium, cadmium, arsenic • Stored in special labelled containers and kept in locked room which is heat free PERFORMANCE LEVEL (PL) Mastered Not mastered PL3 Apply knowledge on chemistry, scientific investigations, usage, management and handling of apparatus and materials to explain the natural occurrences or phenomena and be able to carry out simple tasks. PL2 PL3 01 U1 Chemistry F4(p01-4)csy2p.indd 3 20/12/2022 3:39 PM

© Nilam Publication Sdn. Bhd. 4 UNIT 1 MODULE • Chemistry FORM 4 4 Match the substances with the correct way of disposal of the chemicals. Wastes with low concentration can be poured directly into the sink while high concentrated wastes has to be diluted with water before adding up with sodium sulphite before poured into the sink. Solid wastes The wastes insert into a plastic bag and the solution left to evaporate at fume chamber. The wastes tied carefully and put in a container. This type of wastes should be discarded according to standard procedures. Organic solvents and hydrocarbons The wastes stored in a closed containers and kept away from Sun and heat. Substances with pH < 5 and pH > 9 Disposed into the special containers. Heavy metals and toxic substances This type of wastes should be kept in closed and labelled containers during disposal. Volatile substances This wastes cannot be disposed directly into the sink because it may cause an environmental pollution. This wastes should be kept in a special containers made up of glass or plastic. Hydrogen peroxide PL3 Emergency Management Procedure in the Laboratory: Chemical spills 1 Quickly inform the accidents to teacher or laboratory assistant. 2 Prohibit others from entering the accident site. 3 Stop the spills from spreading to other area by using sand to make a border. 4 Clean the chemical spills. 5 Disposed the chemical spills by following the correct procedure. Mercury spills 1 Quickly inform the accidents to teacher or laboratory assistant. 2 Prohibit others from entering the accident site. 3 Sprinkle the sulphur powder to cover up the spills. 4 Contact the HAZMAT team of Fire and Rescue Department of Malaysia. 01 U1 Chemistry F4(p01-4)csy2p.indd 4 20/12/2022 3:39 PM

MODULE • Chemistry FORM 4 5 UNIT 2 © Nilam Publication Sdn. Bhd. Concept Map 2 UNIT MATTER Electron arrangement Nucleon number Proton number Molecule Electron Ion Neutron Atom Proton Liquid Gas Solid Atom Isotope A Z X Uses Meaning Calculate RAM Diffusion Kinetic Theory of Matter Particle Theory of Matter Three types of particle Three states of matter Meaning Standard representation of element History of Atomic Model Note: Differentiate between 1 State of matter 2 Type of particles 3 Sub atomic particles Three sub atomic particles MATTER AND THE ATOMIC STRUCTURE [Studied in Form 1] 02 U2 Chemistry F4(p05-26)csy2p.indd 5 20/12/2022 3:48 PM

MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 6 UNIT 2 1 Particle theory of matter Form 1, Unit 5 What is matter? LS 2.1.1 • Matter is any substance that has mass and occupies space. What is particle theory of matter? • Matter is made up of tiny and discrete particles. Three types of tiny particles are atoms , ions and molecules . Define element. • A substance made from only one type of atom. Define compound. • A substance made from two or more different elements which are bonded together. Describe Matter LS 2.1.1 2.1 BASIC CONCEPTS OF MATTER CS 2.1 PERFORMANCE LEVEL (PL) Mastered Not mastered PL1 Recall knowledge and basic skills on matter and the atomic structure. • Elements can be identified as metal or non-metal by referring to the Periodic Table of Elements at the end of the page. • Formation of molecule and ion will be further explained in Chapter 5, page 77. Define and give examples of substance that made up of: Ion • Positively or negatively charged particles, which are formed from metal atom and non-metal atom respectively. • The force of attraction between the two oppositely charged ions forms an ionic bond. Examples: Sodium chloride, NaCl Sodium chloride, NaCl Molecule • A neutral particle consists of similar or different non-metal atoms which are covalently bonded. Examples: (a) Molecule of an element: • Oxygen, O2 • Hydrogen, H2 (b) Molecule of a compound: • Carbon dioxide, CO2 • Water, H2O Oksigen, O2 Air, H2O Atom • The smallest neutral particle of an element. Examples: (a) Pure metal: • Copper, Cu • Lead, Pb Copper, Cu Carbon, C (b) Non-metal: • Carbon, C • Silicon, Si (c) Inert gases: • Neon, Ne • Helium, He Helium, He 02 U2 Chemistry F4(p05-26)csy2p.indd 6 20/12/2022 3:48 PM

MODULE • Chemistry FORM 4 7 UNIT 2 © Nilam Publication Sdn. Bhd. 3 Kinetic theory of matter State kinetic theory of matter. LS 2.1.2 • The tiny and discrete particles that made up matter are constantly moving. Remark: Particles represent atoms, ions or molecules. What is the significant of this theory? • This theory uses the particulate model to explain the properties of matter. • For example, diffusion and change of physical state of matter with the energy change (melting, boiling, freezing and condensation). 1 Determine the types of particle in the following substances. Substances Types of particle Substances Types of particle Substances Types of particle Hydrogen gas (H2) Molecule Sulphur dioxide (SO2) Molecule Tetrachloromethane (CCl4) Molecule Copper(II) sulphate (CuSO4) Ion Iron (Fe) Atom Zinc chloride (ZnCl2) Ion Argon (Ar) Atom Carbon (C) Atom Hydrogen peroxide (H2O2) Molecule Matter Element Compound Types of particles Types of particles Exercise 2 Complete the following blank space with the correct answers. Atoms Molecules Molecules Ions Examples: Cu, Ag, C, Ne, Ar Examples: O2, Br2, N2, Cl2 Examples: CO2, H2O, NO2 Examples: NaC1, KNO3, AgCl 02 U2 Chemistry F4(p05-26)csy2p.indd 7 20/12/2022 3:48 PM

MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 8 UNIT 2 State Kinetic Theory of Matter. • Matter that made up of tiny and discrete particles which are always in constantly moving . How does the matter exist? • Matter exists in three different states which are solid , liquid and gas . Complete the following table to compare particles in solid, liquid and gas. LS 2.1.2 State of matter Solid Liquid Gas Draw the particles arrangement. Each particle (atom / ion / molecule) is represented by Particles arrangement Closely packed in orderly manner. Closely packed but not in orderly manner. Very widely separated from each other. Particles movement Vibrate and rotate at their fixed position. Particles can vibrate , rotate and move throughout the liquid. Particles can vibrate , rotate and move freely. Attractive forces between the particles Very strong Strong Weak Energy content of the particles Very low Moderate Very high Define melting point. • The constant temperature at which a solid changes to become a liquid is called the melting point at a specific pressure. Remark: For the same pure substance, melting point = freezing point. For example, pure ice melts and freezes at 0 °C (the melting point of ice). The constant temperature at which a liquid changes to a solid is called freezing point . Define boiling point. • The constant temperature at which a liquid completely changes to become a gas at a specific pressure. When does a substance change its physical state? • A substance changes its physical state when the temperature reaches the melting point and boiling point. LS 2.1.3 LS 2.1.3 Explain the Change in Physical State of Matter LS 2.1.2 PERFORMANCE LEVEL (PL) Mastered Not mastered PL2 Understand and explain matter and the atomic structure. 02 U2 Chemistry F4(p05-26)csy2p.indd 8 20/12/2022 3:48 PM

MODULE • Chemistry FORM 4 9 UNIT 2 © Nilam Publication Sdn. Bhd. Simulation What causes a substance to change its physical state? • Matter undergoes changes of state when heat energy is absorbed or released : (i) When heat energy is absorbed by the matter (it is heated), the kinetic energy of the particles increases and they vibrate faster. (ii) When matter releases heat energy (it is cooled), the kinetic energy of the particles decreases and they vibrate less vigorously. How does the energy cause a substance to change its physical state? • Heat energy is absorbed to overcome the attractive forces between particles or released when the attractive forces between particles is formed. State the change in heat energy during physical state change. • During melting, boiling and sublimation, heat energy is absorbed from the surrounding. • During condensation, freezing and deposition heat energy is released to the surrounding. Melting Solid Liquid Gas Deposition Sublimation Boiling Freezing Condensation Heat is released to the surrounding Heat is absorbed from the surrounding Key: What happen to the temperature when a substance undergoes changes in physical states? Explain. • During the melting process, the temperature remains unchanged because heat energy absorbed by the particles is used to overcome the forces between particles so that the solid changes to a liquid . • During the freezing process, the temperature remains unchanged because the heat released to the surrounding is balanced by the heat released when the liquid particles attract one another to become a solid . Why does different substance have different melting and boiling points? • When the force of attraction between particles is stronger, more heat needed to overcome the force. • The melting and boiling points are higher . • When the force of attraction between particles is weaker, less heat needed to overcome the force. • The melting and boiling points are lower . 02 U2 Chemistry F4(p05-26)csy2p.indd 9 20/12/2022 3:48 PM

MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 10 UNIT 2 Determine the Melting Point and freezing point of Naphtalene LS 2.1.2 Describe experiment to determine the melting and freezing points of naphthalene. Materials: Naphthalene powder, water Apparatus: Boiling tube, conical flask, beaker, retort stand with clamp, thermometer 0 – 100 °C, stopwatch, Bunsen burner, wire gauze, tripod stand Procedures: I Heating of naphthalene (a) Filled 3 cm height of a boiling tube with naphthalene powder and a thermometer is placed into it. (b) Immersed the boiling tube in a water bath as shown in the diagram and make sure the water level in the water bath is higher than naphtalene powder in the boiling tube. (c) Heat the water and stir the naphthalene slowly with thermometer. (d) When the temperature of naphthalene reaches 60 °C, start the stopwatch. Record the temperature of naphthalene at 30 seconds intervals until the temperature of naphthalene reaches 90 °C. II Cooling of naphthalene (a) Remove the boiling tube and its content from the water bath and put into a conical flask as shown in the diagram. (b) Stir the content in the boiling tube constantly with thermometer throughout cooling process to avoid supercooling (the temperature of cooling liquid drops below freezing point, without the appearance of a solid). (c) Record the temperature of naphthalene every 30 seconds interval until the temperature drops to 60 °C. (d) Plot a graph of temperature against time for the heating and cooling process respectively. Results: I Heating of naphthalene Time / s 0 30 60 90 120 150 180 210 240 Temperature / °C II Cooling of naphthalene Time / s 0 30 60 90 120 150 180 210 240 Temperature / °C Naphthalene Thermometer / Termometer Boiling tube / Tabung didih Water / Air Naphthalene / Naftalena Heat Haba Thermometer Boiling tube Water Naphthalene Heat From the graph, the melting point of naphthalene is X °C Temperature / °C Time / s X From the graph, the freezing point of naphthalene is X °C Temperature / °C Time / s X 02 U2 Chemistry F4(p05-26)csy2p.indd 10 20/12/2022 3:48 PM

MODULE • Chemistry FORM 4 11 UNIT 2 © Nilam Publication Sdn. Bhd. Interpretation of heating curve LS 2.1.3 PERFORMANCE LEVEL (PL) Mastered Not mastered PL3 Apply knowledge on matter and the atomic structure to explain the natural occurrences or phenomena and be able to carry out simple tasks. What is heating curve? • It is a plot of the temperature against time to show how temperature change as a substance is heated up. Sketch the heating curve of a substance with the melting point P °C and the boiling point Q °C from solid to gas. Label the part on the graph where melting point and boiling point take place. Temperature / °C Time / s Q P Points AB BC CD DE EF States of matter Solid Solid and Liquid Liquid Liquid and Gas Gas Explanations 1 Heat energy is absorbed by the particles in the solid naphthalene. 2 The heat energy absorbed causes kinetic energy of the particles to increase and the particles vibrate faster . 3 The temperature increases . 1 Heat energy is absorbed by the particles in the solid naphthalene. 2 The heat energy absorbed is used to overcome forces of attraction between particles so that the solid turns to liquid. 3 The temperature remains constant . The constant temperature is melting point . 1 Heat energy is absorbed by the particles in the liquid naphthalene. 2 The heat energy absorbed causes the kinetic energy of the particles to increase and the particles move faster . 3 The temperature increases . 1 Heat energy is absorbed by the particles in the liquid naphthalene. 2 The heat absorbed is used to overcome the forces of attraction between particles so that the particles begin to move freely to form a gas . 3 The temperature remains constant . The constant temperature is boiling point . 1 Heat energy is absorbed by the particles in the gas . 2 The heat absorbed causes their kinetic energy of the particles to increase and the particles move faster . 3 The temperature increases . Study the heating curve of a substance. State and explain the changes in physical state of the substance at the following region AB, BC, CD, DE, EF. A B C D E F Temperature / °C Time / s 02 U2 Chemistry F4(p05-26)csy2p.indd 11 20/12/2022 3:48 PM

MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 12 UNIT 2 Interpretation of cooling curve LS 2.1.3 The diagram shows how the physical state of substance is related to its melting and boiling points. Solid Liquid Gas Melting point Boiling point Solid + liquid Liquid + gas Temperature increase Points PQ QR RS States of matter Liquid Liquid and Solid Solid Explanations 1 Heat is released to the surrounding by the particles in the liquid naphthalene. 2 The particles in the liquid lose their heat energy and move slower . 3 The temperature decreases . 1 Heat is released to the surrounding by the particles in liquid naphthalene. 2 The heat released is balanced by the heat energy released as the particles attract one another to form a solid . 3 The temperature remains constant . The constant temperature is freezing point . 1 The particles in the solid naphthalene releases heat. 2 The heat released causes the particles vibrate slower . 3 The temperature decreases . Study the cooling curve of a substance. State and explain the changes in physical state of the substance at the following region PQ, QR, RS. Temperature / °C Time / s P Q R S 02 U2 Chemistry F4(p05-26)csy2p.indd 12 20/12/2022 3:48 PM

MODULE • Chemistry FORM 4 13 UNIT 2 © Nilam Publication Sdn. Bhd. Exercise 1 The table below shows substances and their chemical formula. Substances Chemical formula Types of particle Silver Ag Atom Potassium oxide K2O Ion Ammonia NH3 Molecule Chlorine Cl2 Molecule (a) State the type of particles that made up each substance in the table. (b) Which of the substances are element? Explain your answer. Silver and chlorine. Silver and chlorine are made up of only one type of atom. (c) Which of the substances are compound? Explain your answer. Potassium oxide and ammonia. Potassium oxide and ammonia are made up of two different elements. 2 The table below shows the melting and boiling points of substances P, Q and R. Substances Melting point / °C Boiling point / °C P –36 6 Q –18 70 R 98 230 (a) (i) What is meant by ‘melting point’? The constant temperature at which a solid changes to a liquid at particular pressure. (ii) What is meant by ‘boiling point’? The constant temperature at which a liquid changes to a gas at particular pressure. (b) Draw the particles arrangement of substances P, Q and R at room conditions. Substance P Bahan P Substance Q Bahan Q Substance R Bahan R (c) (i) State the substance/substances that exist in the form of liquid at 0 °C. Q (ii) Explain your answer in 2(c)(i). The temperature 0 °C is above the melting point of Q and below the boiling point of Q. PL2 PL2 PL2 PL1 PL1 PL2 02 U2 Chemistry F4(p05-26)csy2p.indd 13 20/12/2022 3:48 PM

MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 14 UNIT 2 (d) (i) Substance Q is heated from room temperature to 100 °C. Sketch a graph of temperature against time for the heating of substance Q. Temperature / °C Time / s 70 (ii) What is the state of matter of substance Q at 70 °C? Liquid and gas (e) Compare the melting point of substances Q and R. Explain your answer. • The melting point of substance R is higher than substance Q. • The attraction force between particles in substance R is stronger than substance Q. • More heat is needed to overcome the forces between particles in substance R. (f) The diagram shows a set up of apparatus to separate a mixture of water and ethanol. During the process, the vapour produces at part X and pure ethanol is collected into the conical flask. Thermometer Water out Liebig condenser Water in Pure ethanol Porcelain chips Heat up Water + ethanol X (i) Which compound is being distilled first? Explain your answer. Ethanol. Boiling point of ethanol is lower than boiling point of water. (ii) Draw the arrangement of particles at part X. (iii) Explain how the state of matter at part X changes during the process. Condensation process occur // vapour is condensed. Gas changes into liquid. PL4 PL4 PL4 HOTS 02 U2 Chemistry F4(p05-26)csy2p.indd 14 20/12/2022 3:48 PM

MODULE • Chemistry FORM 4 15 UNIT 2 © Nilam Publication Sdn. Bhd. 3 The melting point of acetamide can be determined by heating solid acetamide until it melts as shown in the diagram on the right. The temperature of acetamide is recorded every three minutes when it is left to cool down at room temperature. (a) What is the purpose of using water bath in the experiment? To ensure even heating of acetamide. Acetamide is easily combustible. (b) State the name of another substance which its melting point can also be determined by using water bath as shown in the above diagram. Naphthalene (c) Sodium nitrate has a melting point of 310 °C. Can the melting point of sodium nitrate be determined by using the water bath as shown in the diagram? Explain your answer. No, because the boiling point of water is 100 °C which is lower than the melting point of sodium nitrate. (d) Why do we need to stir the acetamide in the boiling tube in above experiment? To make sure the heat is distributed evenly. (e) The graph of temperature against time for the cooling of liquid acetamide is shown below. Temperature / °C Time / s T1 T2 T3 Q R (i) What is the freezing point of acetamide? T2 °C (ii) The temperature between Q and R is constant. Explain. The heat lost to the surrounding is balanced by the heat released by the liquid when the liquid acetamide particles rearrange themselves to become solid. (f) Acetamide exists as molecules. State the name of another compound that is made up of molecules. Water / naphthalene (g) What is the melting point of acetamide? T2 °C (h) A housewife bought a mothball as insect repellent at her house. She put the mothballs pieces into a cupboard. After a month, she found the mothballs becoming smaller. Explain why the mothballs becoming smaller. • Mothballs are made of naphthalene. • Naphthalene is a substance that undergoes sublimation. PL2 PL2 PL4 PL3 PL3 Thermometer Water Acetamide Boiling tube Heat PL2 PL4 PL5 HOTS 02 U2 Chemistry F4(p05-26)csy2p.indd 15 20/12/2022 3:48 PM

MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 16 UNIT 2 Scientist Atomic Model Discovery Dalton (i) Matter is made up of particles called atom . (ii) Atoms cannot be created , destroyed or divided . (iii) Atoms from the same element are identical . Thomson Positively charged sphere Negatively charged electron (i) Discovered the electrons , the first subatomic particle. (ii) Atom is sphere of positive charge which is embedded with negatively charged particles called electrons . Rutherford Electron moves outside the nucleus Nucleus that contains proton (i) Discovered the nucleus as the centre of an atom and positively charged . (ii) Proton is a part of the nucleus. (iii) Electron moves outside the nucleus. (iv) Most of the mass of the atom found in the nucleus . Neils Bohr Shell Nucleus that contains proton Electron (i) Discovered the existence of electron shells . (ii) Electrons move in the shells around the nucleus . James Chadwick Shell Nucleus that contains proton and neutron Electron (i) Discovered the existence of neutron . (ii) Nucleus of an atom contains neutral particles called neutron and positively charged particles called proton . (iii) The mass of a neutron and proton is almost the same. LS 2.2.3 Series of Atomic Model based on Atomic Model by Dalton, Thomson, Rutherford, Bohr and Chadwick 2.2 THE DEVELOPMENT OF ATOMIC MODEL CS 2.2 02 U2 Chemistry F4(p05-26)csy2p.indd 16 20/12/2022 3:48 PM

MODULE • Chemistry FORM 4 17 UNIT 2 © Nilam Publication Sdn. Bhd. What are the sub atomic particles in an atom based on the development of atomic model? • The sub atomic particles are proton , neutron and electron . What are the characteristics of the subatomic particles? Subatomic particles Symbol Charge Relative mass Position Electron e – (negative) 1 1 840 ≈ 0 In the shells Proton p + (positive) 1 In the nucleus Neutron n neutral 1 In the nucleus The Sub Atomic Particles LS 2.2.2 Form 1, Unit 6: Periodic Table Describe atomic structure based on history of the atomic structure. Atom Shell Electron Nucleus that contains protons and neutrons (i) An atom has a central nucleus and electrons that move in the shells around the nucleus. (ii) The nucleus contains protons and neutrons. (iii) The relative mass of a neutron and a proton which are in the nucleus is 1. (iv) The mass of an atom is obtained mainly from the number of proton and neutron . Explain how the electrons are filled in specific shells. Every shell can be filled only with a certain number of electrons. For the elements with atomic numbers 1 – 20: • First shell can be filled with a maximum of 2 electrons. • Second shell can be filled with a maximum of 8 electrons. • Third shell can be filled with a maximum of 8 electrons. First shell is filled with 2 electrons (duplet) Second shell is filled with 8 electrons (octet) Third shell is filled with 8 electrons (octet) 2.3 ATOMIC STRUCTURE CS 2.3 The Diagram of Atomic Structure and the Electron Arrangement LS 2.3.4 PERFORMANCE LEVEL (PL) Mastered Not mastered PL4 Analyse knowledge on matter and atomic structure in the context of problem solving the natural occurrences or phenomena. 02 U2 Chemistry F4(p05-26)csy2p.indd 17 20/12/2022 3:48 PM

MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 18 UNIT 2 What are valence electrons? • Valence electrons are the electrons in the outermost shell of an atom. Atom or element? Examples: Na Na Na Na Na Na Na Na Na Na Na Na Na Na Sodium element Sodium element Sodium element Sodium atom • Atom is the smallest neutral particle of an element. • An element is made from only one type of atom. Define proton number. LS 2.3.1 • Proton number of an element is the number of proton in the nucleus of its atom . Remark: Every element has its own proton number (Refer to Periodic Table of Elements). Define nucleon number. LS 2.3.1 • Nucleon number of an element is the total number of protons and neutrons in the nucleus of its atom . Remark: – Nucleon number is also known as mass number. (Refer to Periodic Table of Elements) Why atoms are neutral? • Each proton has charge of +1 . • Each electron has an electrical charge of –1 . • The neutron has no charge (it is neutral ). • An atom has the same number of protons and electrons, so the overall charge of atom is zero . • Atom is neutral . Remark: If an atom loses or gains electrons it is called an ion – formation of ion will be studied in Chapter 5. How to calculate the number of protons, neutrons and electrons in an atom? LS 2.3.2 In an atom: • Number of protons = Proton number • Number of electrons = Number of proton • Number of neutrons = Nucleon number – Proton number Examples (i) • Proton number of potassium, K is 19. • Potasium atom has 19 protons in the nucleus and 19 electrons in the shells. (ii) • Proton number of oxygen, O is 8. • Oxygen atom has 8 protons in the nucleus and 8 electrons in the shells. 02 U2 Chemistry F4(p05-26)csy2p.indd 18 20/12/2022 3:48 PM

MODULE • Chemistry FORM 4 19 UNIT 2 © Nilam Publication Sdn. Bhd. What is the symbol of an element? • The symbol of an element represents an element. • If the symbol has only one letter, it must be a capital letter. • If it has two letters, the first is always a capital letter, while the second is always a small letter. Examples: Elements Symbols Elements Symbols Elements Symbols Oxygen O Nitrogen N Calcium Ca Magnesium Mg Sodium Na Copper Cu Hydrogen H Potassium K Chlorine Cl • For example KCl. • There are two elements chemically bonded in KCl because there are two capital letters represent potassium and chlorine. How to write the standard representation of an element? LS 2.3.3 • The standard representation of an atom of an element can be written as: Nucleon number Proton number Symbol of an element A X Z Example: 27 13 Al (i) What information can be obtained from the standard representation of the element? (ii) Construct the structure and the electron arrangement of Aluminium atom. (i) Information from the standard representation of element: • The element is Aluminium. • The nucleon number of Aluminium is 27 . • The proton number of Aluminium is 13 . • Aluminium atom has 13 protons , 14 neutrons and 13 electrons. (ii) Atomic structure and electron arrangement of Aluminium atom The structure of Aluminium atom The electron arrangement of Aluminium atom 13 p 14 n AI 02 U2 Chemistry F4(p05-26)csy2p.indd 19 20/12/2022 3:48 PM

MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 20 UNIT 2 2 Draw the structure of sodium atom and electron arrangement of sodium atom. 23 11 Na The structure of sodium atom The electron arrangement of sodium atom 11 p 12 n Na 1 Complete the following table: Elements Standard representation for an atom Number of proton Number of electron Number of neutron Proton number Nucleon number Electron arrangement of atom Number of valence electron Hydrogen 1 1 H 1 1 0 1 1 1 1 Helium 4 2 He 2 2 2 2 4 2 2 Boron 11 5 B 5 5 6 5 11 2.3 3 Carbon 12 6 C 6 6 6 6 12 2.4 4 Nitrogen 14 7 N 7 7 7 7 14 2.5 5 Neon 20 10 Ne 10 10 10 10 20 2.8 8 Sodium 23 11 Na 11 11 12 11 23 2.8.1 1 Magnesium 24 12 Mg 12 12 12 12 24 2.8.2 2 Calcium 40 20 Ca 20 20 20 20 40 2.8.8.2 2 3 The diagram below shows the symbol of atoms P, R and S. 35 17 P 12 6 R 37 17 S (a) What is meant by nucleon number? Nucleon number of an element is the total number of protons and neutrons in the nucleus of its atom. (b) What is the nucleon number of P? 35 (c) State the number of neutron in atom P. 18 (d) State number of proton in atom P. 17 PL2 PL3 PL2 Exercise 02 U2 Chemistry F4(p05-26)csy2p.indd 20 20/12/2022 3:48 PM

MODULE • Chemistry FORM 4 21 UNIT 2 © Nilam Publication Sdn. Bhd. Define isotope. LS 2.4.1 • Isotopes are atoms of the same element with same number of protons but different number of neutrons. Or • Isotopes are atoms of the same element with same proton number but different nucleon number. What causes atoms of the same element to be isotopes? • Number of neutron Examples: 17 p 18 n 17 p 20 n Chlorine-35 Chlorine-37 In any isotopes; • Number of neutron is different. Hence, nucleon number is different . • Number of proton is the same. Hence, proton number is the same. • Number of electron is the same . Atoms of isotopes have same electron arrangement . Compare the chemical properties and physical properties of isotope. • Isotopes have same chemical properties. • Different physical properties (such as mass, density, melting and boiling points) due to different number of neutron or nucleon number. What is natural abundance of isotopes? • Natural abundance is the percentage of isotopes present in a natural sample of the element. Examples: Elements Name of isotope Natural abundance Isotope Bromine Bromine-79 50% 79 35 Br Bromine-81 50% 81 35 Br Chlorine Chlorine-35 75% 35 17 Cl Chlorine-37 25% 37 17 Cl How to calculate the relative atomic mass of the isotope? Relative atomic mass of isotope = ∑(% isotope × Relative mass of isotope) 100 Relative atomic mass of Bromine = (50% × 79) + (50% × 81) 100 = 80 Relative atomic mass of Chlorine = (75% × 35) + (25% × 37) 100 = 35.5 Remark: Relative mass of isotope is equal to nucleon number. 2.4 ISOTOPES AND ITS USES CS 2.4 LS 2.4.2 PERFORMANCE LEVEL (PL) Mastered Not mastered PL5 Evaluate knowledge on matter and atomic structure in the context of problem solving and decision-making to carry out a task. 02 U2 Chemistry F4(p05-26)csy2p.indd 21 20/12/2022 3:48 PM

MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 22 UNIT 2 Example 1 Neon is naturally found in three isotopes: 90.9% 20 10Ne, 0.3% 21 10Ne and 8.8% 22 10Ne. Calculate the relative atomic mass of Neon. Relative atomic mass of Neon = (20 × 90.9%) + (21 × 0.3%) + (21 × 8.8%) 100 = 20.18 2 Relative atomic mass of Copper is 63.62. Isotopes of Copper: 69% 63 29Cu and 31% q 29Cu. Calculate the nucleon number of isotope q 29Cu. 63.62 = (69% × 63) + (31% × q) 100 q = 65 Nucleon number of isotope q 29Cu is 65. Give examples of the usage of isotopes. (i) Medical field • To detect brain cancer . • To detect thrombosis (blockage in blood vessel). • Iodine-131 is used to measure the rate of iodine absorption by thyroid gland. • Cobalt-60 is used to destroy cancer cells. • Cobalt-60 is used to kill microorganism in the sterilising process. (ii) In the industrial field • To detect wearing out in machines. • To detect any blockage in water, gas or oil pipes. • Sodium-24 detect leakage of pipes underground. • To detect defects/cracks in the body of an aeroplane. (iii) In the agriculture field • Phosphorus-32 is used to detect the rate of absorption of phosphate fertilizer in plants. • To sterile insect pests for plants. (iv) In the archeology field • Carbon-14 can be used to estimate the age of artifacts. What are the disadvantages of the usage of isotopes? (i) Some isotopes can stay in the human body for a long time and cause harmful effects by their radiation. The tissues will be damaged and cause burns, nausea, diseases such as leukemia and cancer. (ii) The disposal of isotope residues not according to the procedure will result in radioactive radiation to humans. (iii) The production of isotopes requires a nuclear reactor where the cost is very high . LS 2.4.3 02 U2 Chemistry F4(p05-26)csy2p.indd 22 20/12/2022 3:48 PM

MODULE • Chemistry FORM 4 23 UNIT 2 © Nilam Publication Sdn. Bhd. 1 Draw the atomic structure and electron arrangement for the atom of each element: Standard Representation of an Element The structure of an atom Electron arrangement of an atom 1 1 H 1 p 0 n H 23 11 Na 11 p 12 n Na 16 8 O 8 p 8 n O 2 Choose the correct statement for the symbol of element X. 23 11 X Statements Tick ( 3 / 7 ) Statements Tick ( 3 / 7 ) Element X has 11 proton number. 7 Nucleon number of atom X is 23. 3 The proton number of element X is 11. 3 Number of nucleon of element X is 23. 7 The proton number of atom X is 11. 3 Atom X has 23 nucleon number. 7 The number of proton of element X is 11. 7 Neutron number of atom X is 12. 7 The number of proton of atom X is 11. 3 Number of neutron of atom X is 12. 3 Nucleon number of element X is 23. 3 Number of neutron of element X is 12. 7 PL2 PL2 Exercise 02 U2 Chemistry F4(p05-26)csy2p.indd 23 20/12/2022 3:48 PM

MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 24 UNIT 2 1 The diagram shows the symbol of atom X, Y and Z. 35 17 X 37 17 Z 12 6 Y (a) (i) What is meant by isotope? Isotopes are atoms of the same element with same number of protons but different number of neutrons (ii) State a pair of isotope in the diagram shown. X and Z (iii) Give reason for your answer in 1(a)(ii). Atoms X and Z have same proton number but different nucleon number//number of neutrons (b) An isotope of Y has 8 neutrons. Write the symbol for the isotope Y. 14 6 Y (c) The diagram shows an object discovered by an archeologist. An isotope is used by this archeologist in his research. (i) Name the isotope used. Carbon-14 (ii) State one advantage and one disadvantage of using the isotope. Advantage: Estimate the age of artifact Disadvantage: The isotope is expensive 2 The graph below shows the natural abundance of Germanium, Ge. Natural abundance (%) Nucleon number 20.6% 27.4% 36.7% 7.6% 40 20 30 70 75 80 10 7.7% 0 (a) State the name of heaviest isotope of Germanium. Germanium-76 (b) Use the natural abundance of each isotope to calculate the relative atomic mass of Germanium. Relative atomic mass = (70 × 20.6%) + (72 × 27.4%) + (73 × 7.7%) + (74 × 36.7%) + (76 × 7.6%) 100 = 72.7 Subjective Questions PL1 PL2 PL3 PL4 PL2 PL4 ENRICHMENT EXERCISE Quiz 02 U2 Chemistry F4(p05-26)csy2p.indd 24 20/12/2022 3:48 PM

MODULE • Chemistry FORM 4 25 UNIT 2 © Nilam Publication Sdn. Bhd. 3 The table below shows the number of proton and neutron of atoms of elements P, Q and R. Elements P Q R Number of protons 1 1 6 Number of neutrons 0 1 6 (a) Which of the atoms in the above table are isotope? Explain your answer. P and Q. Atoms P and Q have same number of protons but different number of neutrons // nucleon number. (b) (i) Write the standard representation of element Q. 2 1 Q (ii) State three informations that can be deduced from your answer in 3(b)(i). • The proton number of element Q is 1. • Nucleon number of element Q is 2. • Number of neutron = 2 – 1 = 1 • Nucleus of atom Q contains 1 proton and 1 neutron (c) (i) Draw atomic structure for atom of element R. 6 p 6 n (ii) Describe the atomic structure in 3(c)(i). • The atom consists of 2 parts: The centre part called nucleus and the outer part called electron shell. • The nucleus consists of 6 protons which are positively charged and 6 neutrons which are neutral. • The electrons are in two shells, the first shell consists of two electrons and the second shell consists of four electrons. Electrons move around nucleus in the shells. (d) Element P reacts with oxygen and produces liquid Z at room temperature. The graph below shows the sketch of the graph when liquid Z at room temperature, 27 °C is cooled to –5 °C. Temperature / °C 0 Time / s 27 −5 t1 t 2 (i) What is the state of matter of liquid Z from t1 to t2? Explain why is the temperature remain unchanged from t1 to t2. Liquid and solid. Heat lost to the surrounding is balanced by the heat released when the liquid particles rearrange themselves to become solid. (ii) Describe the change in the particles movement when Z is cooled from room temperature to –5 °C. The particles move slower PL4 PL2 PL3 PL4 02 U2 Chemistry F4(p05-26)csy2p.indd 25 20/12/2022 3:48 PM

MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 26 UNIT 2 1 The diagram below shows the graph of temperature against time when a liquid Y is cooled. Temperature / °C P Q T3 T2 T1 R S Time / s Which of the following statements are true about the curve? I At Q, liquid Y begins to freeze. II At PQ, particles in Y absorb heat from the surroundings. III Liquid Y freezes completely at S. IV The freezing point of Y is T2°C. A I and III only C II and III only B I and IV only D II and IV only 2 The diagram below shows the standard representation of beryllium atom. 9 4 Be What is the number of valence electrons of beryllium atom? A 2 C 4 B 3 D 7 3 The diagram below shows the graph of temperature against time when solid Z is heated. Temperature / °C Time / s 80 0 1 2 3 4 5 6 7 8 9 Which of the following is true during the fourth minute? A All the molecules are in random motion. B All the molecules are vibrating at fixed positions. C All the molecules are closely packed and in random motion. D Some of the molecules are vibrating at fixed positions but some are in random motion. 4 The table below shows the melting point and boiling point of substances S, T, U, V and W. Substance Melting point/°C Boiling point/°C S –182 –162 T –23 77 U –97 65 V 41 182 W 132 290 Which substance exists as liquid at room temperature? A S only C T and U only B S and T only D V and W only 5 Petrol and paraffin are accidentally mixed at an oil refinery. What is the best method to separate and collect them? A Condensation B Evaporation C Filtration D Fractional distillation 6 The table below shows the proton number and the number of neutrons for atoms of elements W, X, Y and Z. Elements Proton number Number of neutrons W 7 7 X 8 8 Y 8 9 Z 9 10 Which of the following pair of elements is isotope? A W and X C X and Y B W and Y D Y and Z 7 An inflated balloon will shrink if it is placed in refrigerator. Which of the following is the effect of lower temperature to the gas particles in the balloon? A Move faster and become closer together B Move faster and become further apart C Move slower and become closer together D Move slower and become further apart Objective Questions SPM PRACTICE Quiz Objective Questions PL2 PL2 PL3 PL3 PL4 PL3 PL4 02 U2 Chemistry F4(p05-26)csy2p.indd 26 20/12/2022 3:48 PM

MODULE • Chemistry FORM 4 27 UNIT 3 © Nilam Publication Sdn. Bhd. 3 UNIT Remark: – Molar mass is mass of one mol of a substance. The unit is g mol–1. – Molar volume of gas is volume occupied by one mol of any gas, 24 dm3 mol–1 at room conditions or 22.4 dm3 mol–1 at standard temperature and pressure (s.t.p.) – Avogadro constant is number of particles in one mol of any substance. The value is 6.02 × 1023 mol–1. • Reactants • Products • Physical state Interpret qualitatively Numerically equal ÷ Avogadro constant × Avogadro constant ÷ Molar Mass (g mol–1) × Molar Mass (g mol–1) NUMBER OF MOLE Empirical formula Molecular formula Formula of ionic compound ÷ Molar Volume (dm3 mol–1) × Molar Volume (dm3 mol–1) RELATIVE MASS • Relative Atomic Mass • Relative Molecular Mass • Relative Formula Mass Numerical problem involving chemical equation (Interpret quantitatively) THE MOLE CONCEPT, CHEMICAL FORMULA AND EQUATION Concept Map Number of particles Volume of gas (dm3 ) Chemical formula Calculate percentage by mass of element in a compound Chemical equation Mass (g) 03 U3 Chemistry F4(p27-52)csy2p.indd 27 21/12/2022 3:04 PM

MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 28 UNIT 3 What is relative mass? • The mass of an atom is determined by the mass of atom of an element to the mass of another element that is chosen as standard. What atom is chosen as a standard atom? Explain how Relative Atomic Mass is calculated. LS 3.1.1 • Carbon-12 (an isotope of carbon) is chosen as a standard atom because its mass can be determined very accurately using mass spectrometer. Carbon-12 isotope is given a mass of exactly 12.00. • Relative atomic mass based on the carbon-12 scale is the mass of one atom of the elements compared with 1 12 of the mass of an atom of carbon-12: • Relative atomic mass of an element (RAM) = The average mass of one atom of the element 1 12 × The mass of atom of carbon-12 What can be interpreted? C-12 Mass of an atom of carbon-12 = 12.00 1 12 of the mass of an atom carbon-12 = 1 Example: Mg C C Relative atomic mass of magnesium = 24 = Mass of magnesium atom 1 12 × Mass of an atom carbon-12 Mass of magnesium atom = 24 ( 1 12 × mass of an atom carbon-12) = 24 What is the unit for relative atomic mass? Give reason. • It has no unit. • The relative atomic mass of an element can also be considered as the number of times the mass of one atom of that element is heavier than 1 12 of carbon-12. Example: Relative atomic mass of helium is 4. – 3 atoms of helium have the same mass as one carbon-12 3.1 RELATIVE ATOMIC MASS AND RELATIVE MOLECULAR MASS CS 3.1 PERFORMANCE LEVEL (PL) Mastered Not mastered PL1 Recall knowledge and basic skills on mole concept, chemical formulae and equations. 03 U3 Chemistry F4(p27-52)csy2p.indd 28 21/12/2022 3:04 PM

MODULE • Chemistry FORM 4 29 UNIT 3 © Nilam Publication Sdn. Bhd. Define Relative Molecular Mass (RMM). • The average mass of one molecule of a substance when compared to 1 12 of the mass of carbon-12. • Relative Molecular Mass = The average mass of one molecule 1 12 × The mass of an atom of carbon-12 Remark: The Relative Molecular Mass is used for covalent molecules. How to calculate Relative Molecular Mass (RMM)? • RMM is obtained by adding up the RAM of all the atoms that are present in the formula. Exercise 1 Calculate Relative Molecular Mass (RMM) for the following molecular substances: LS 3.1.2 Molecular substances Molecular formula Relative molecular mass Oxygen O2 2 × 16 = 32 Water H2O 2 × 1 + 16 = 18 Carbon dioxide CO2 12 + 2 × 16 = 44 Ammonia NH3 14 + 3 × 1 = 17 [Relative atomic mass: O = 16, H = 1, C = 12, N = 14] 2 Calculate Relative Formula Mass (RFM) for the following ionic substances: Substance Chemical formula Relative formula mass Sodium chloride NaCl 23 + 35.5 = 58.5 Potassium oxide K2O 2 × 39 + 16 = 94 Copper(II) sulphate CuSO4 64 + 32 + 4 × 16 = 160 Ammonium carbonate (NH4)2CO3 2 [14 + 4 × 1] + 12 + 3 × 16 = 96 Aluminium nitrate Al(NO3)3 27 + 3 [14 + 3 × 16] = 213 Calcium hydroxide Ca(OH)2 40 + 2 [16 + 1] = 74 Lead(II) hydroxide Pb(OH)2 207 + 2 [16 + 1] = 241 Hydrated copper(II) sulphate CuSO4•5H2O 64 + 32 + 4 × 16 + 5 [2 × 1 + 16] = 250 [Relative atomic mass : Na = 23, Cl = 35.5, K = 39, O = 16, Cu = 64, S = 32, N = 14, H = 1, C = 12, Al = 27, Ca = 40, Pb = 207] 3 The formula of metal oxide of M is M2O3. Its relative formula mass is 152. What is the relative atomic mass of metal M? M = Relative atomic mass for M 2M + 3 × 16 = 152 M = 52 LS 3.1.1 PL2 PL2 PL3 03 U3 Chemistry F4(p27-52)csy2p.indd 29 21/12/2022 3:04 PM

MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 30 UNIT 3 4 Phosphorus forms a chloride with a formula PClx. Its relative molecular mass is 208.5. Calculate the value of x. [Relative atomic mass: P = 31, Cl = 35.5] 31 + x × 35.5 = 208.5 35.5x = 208.5 – 31 35.5x = 177.5 x = 5 5 Relative atomic mass of calcium is 40 based on the carbon-12 scale. (a) State the meaning of the statement above. Mass of one calcium atom is 40 times greater than 1 12 mass of one carbon-12 atom. (b) How many times is one calcium atom heavier than one oxygen atom? [Relative atomic mass: O = 16] Relative atomic mass of calcium Relative atomic mass of oxygen = 40 16 = 2.5 times (c) How many calcium atoms have the same mass as two atoms of bromine? [Relative atomic mass: Br = 80] Number of calcium atom × 40 = 2 × 80 Number of calcium atom = 2 × 80 40 = 4 PL3 PL3 What is mole? • A mole is a quantity of substance that contains as many particles as the number of atoms in exactly 12 g of carbon-12. • A mole of a substance is the quantity of substance which contains a constant number of particles (atoms, ions, molecules), which is 6.02 × 1023. What is Avogadro constant, NA? • A fixed quantity of 6.02 × 1023 of particles in one mole substance. • It can be as 6.02 × 1023 mol–1. Remark: The particles can be atom, ion or molecules Why is Avogardro constant, NA useful? • It is a way for counting the particles (atoms, ions, or molecules). • This is because the size of particles is too small, so it is not possible to count physically. Example: • The concept of mole is the same as the concept of a dozen in our everyday life. • Dozen is used to represent a quantity: 1 dozen of pencil 1 × 12 pencils 2 dozens of pencil 2 × 12 pencils 3 dozens of pencil 3 × 12 pencils 1 mol of atom 1 × 6.02 × 1023 atoms 2 mol of atom 2 × 6.02 × 1023 atoms 3 mol of atom 3 × 6.02 × 1023 atoms LS 3.2.2 LS 3.2.1 3.2 MOLE CONCEPT CS 3.2 Mole and the Number of Particles PERFORMANCE LEVEL (PL) Mastered Not mastered PL2 Understand and explain the mole concept, chemical formulae and equations. 03 U3 Chemistry F4(p27-52)csy2p.indd 30 21/12/2022 3:04 PM

MODULE • Chemistry FORM 4 31 UNIT 3 © Nilam Publication Sdn. Bhd. 1 Complete the following table: LS 3.2.7 Substances Formula Number of atom per molecule/ Number of positive and negative ion Number of particles in 1 mol of substance Chlorine Cl2 Cl : 2 6.02 × 1023 Cl2 molecules 2 × 6.02 × 1023 Cl atoms Water H2O H : 2 O : 1 6.02 × 1023 H2O molecules 2 × 6.02 × 1023 H atoms 1 × 6.02 × 1023 O atoms Ammonia NH3 N : 1 H : 3 6.02 × 1023 NH3 molecules 1 × 6.02 × 1023 N atoms 3 × 6.02 × 1023 H atoms Sulphur dioxide SO2 S : 1 O : 2 6.02 × 1023 SO2 molecules 1 × 6.02 × 1023 S atoms 2 × 6.02 × 1023 O atoms Magnesium chloride MgCl2 Cl– : 2 1 × 6.02 × 1023 Mg2+ ions 2 × 6.02 × 1023 Cl– ions Aluminium oxide Al2O3 Al3+ : 2 O2– : 3 2 × 6.02 × 1023 Al3+ ions 3 × 6.02 × 1023 O2– ions Why is mole concept useful? • The mole concept is useful for determining the number of particles in an ionic compound for or a covalent molecule. Example: Methane has a formula CH4. 1 methane molecule, CH4 is made up of 1 C atom and 4 H atoms which are covalently bonded. H C H H H C H H H H Is made up of 1 CH4 molecule 1 carbon atom and 4 hydrogen atoms • Hence, 6.02 × 1023 CH4 contains 1 × 6.02 × 1023 C atoms and 4 × 6.02 × 1023 H atoms. • Applying the mole concept: 1 mol of CH4 molecules consists of 1 mol C atoms and 4 mol of H atoms. What is the relationship between number of moles and number of particles (atoms/ions/molecules)? LS 3.2.2 Number of moles Number of particles × Avogadro Constant, NA ÷ Avogadro Constant, NA PL2 Exercise Mg2+ : 1 03 U3 Chemistry F4(p27-52)csy2p.indd 31 21/12/2022 3:04 PM

MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 32 UNIT 3 Define molar mass. • Molar mass is the mass of one mole of any substance. State how to obtain the molar mass for any substance. • Molar mass of any substance is numerically equal to its relative mass (Relative atomic mass/relative formula mass/relative molecular mass). What is the unit of molar mass? • Molar mass is the relative atomic mass (RAM), relative molecular mass (RMM) and relative formula mass (RFM) of a substance in g mol–1. 2 × (6.02 × 1023) = 1.20 × 1024 molecules of SO2 3 × 2 = 6 mol of atoms (e) 2 mol of SO2 [Sulphur dioxide] 2 mol of S atoms, number of S atoms = 2 × (6.02 × 1023) = 1.20 × 1024 4 mol of O atoms, number of O atoms = 4 × (6.02 × 1023) = 2.41 × 1024 6.02 × 1023 molecules of ammonia, NH3 4 mol atoms (b) 1 mol of NH3 [Ammonia gas] 1 mol of nitrogen atom, N 3 mol of hydrogen atoms, H 6.02 × 1023 molecules of chlorine, Cl2 2 × (6.02 × 1023) = 1.20 × 1024 atoms of chlorine, Cl (a) 1 mol of Cl2 [Chlorine gas] 1 4 × (6.02 × 1023) = 1.51 × 1023 molecules of ammonia, NH3 1 mol of atom (c) 1 4 mol of NH3 [Ammonia gas] 1 4 or 0.25 mol of N atoms, number of N atoms = 0.25 × (6.02 × 1023) = 1.51 × 1023 3 4 or 0.75 mol of H atoms, number of H atoms = 0.75 × (6.02 × 1023) = 4.52 × 1023 2 mol of Mg2+ ions, number of Mg2+ ions = 2 × (6.02 × 1023) = 1.20 × 1024 4 mol of Cl– ions, number of Cl– ions = 4 × (6.02 × 1023) = 2.41 × 1024 (d) 2 mol of MgCl2 [Magnesium chloride] 2 Complete the following: [Differentiate between “mole” dan “molecule”] LS 3.2.3 PL1 Number of Moles and Mass of Substance 03 U3 Chemistry F4(p27-52)csy2p.indd 32 21/12/2022 3:04 PM

MODULE • Chemistry FORM 4 33 UNIT 3 © Nilam Publication Sdn. Bhd. Calculate: LS 3.2.7 PL3 1 Mass of 3 mol of sodium hydroxide, NaOH Molar mass of NaOH = (23 + 16 + 1) g mol–1 = 40 g mol–1 Mass of 3 mol of sodium hydroxide, NaOH = 3 mol × 40 g mol–1 = 120 g Answer: 120 g 2 Number of moles in 20 g of sodium hydroxide, NaOH Number of moles of sodium hydroxide, NaOH = 20 g 40 g mol–1 = 0.5 mol Answer: 0.5 mol Examples Substance Carbon, C Water, H2O Aluminium, Al Sodium chloride, NaCl Relative mass 12 2(1) + 16 = 18 27 35.5 + 23 = 58.5 1 mol substance 12.00 g 12C OH H 18.00 g AI 27.00 g Cl – Na+ 58.50 g Mass for 1 mol 12.01 g 12 g 18.00 g 18 g 27.00 g 27 g 58.00 g 58.5 g Molar mass 12 g mol–1 18 g mol–1 27 g mol–1 58.5 g mol–1 What is the relationship between number of moles and any given mass of a substance? Number of moles Mass in gram × (RAM/RFM/RMM) g mol–1 ÷ (RAM/RFM/RMM) g mol–1 Examples (i) Calculate mass of 2 mol of water Relative molecular mass of H2O = 18 Molar mass of 1 mol of H2O = 18 g mol–1 Mass of 2 mol of H2O = Number of moles × Molar mass = 2 mol × 18 g mol–1 = 36 g (ii) Calculate number of moles of 45 g of water, H2O Number of moles of 45 g of H2O = Mass of H2O Molar mass = 45 g 18 g mol–1 = 2.5 mol LS 3.2.4 Exercise 03 U3 Chemistry F4(p27-52)csy2p.indd 33 21/12/2022 3:04 PM

MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 34 UNIT 3 Define molar volume of gas. • Volume occupied by 1 mol of a gas. Remark: The volume of gas is affected by temperature and pressure. State the molar volume of any gases at room conditions and at standard temperature and pressure (s.t.p). • Molar volume of any gases at room conditions is 24 dm3 mol–1. • Molar volume of any gases at standard temperature and pressure (s.t.p) is 22.4 dm3 mol–1. Example: The diagram shows the molar volume of three gases at room conditions. Oxygen gas, O2 Ammonia gas, NH3 Carbon dioxide gas, CO2 Volume of gas 24 dm3 24 dm3 24 dm3 Number of mol 1 mol 1 mol 1 mol Mass 32 g 17 g 44 g Number of particles 6.02 × 1023 O2 molecules 6.02 × 1023 NH3 molecules 6.02 × 1023 CO2 molecules Remark: 1 dm3 = 1 000 cm3 3 Mass of 2.5 mol of oxygen gas, O2 Molar mass of oxygen gas, O2 = (16 + 16) g mol–1 = 32 g mol–1 Mass of 2.5 mol of oxygen gas, O2 = 2.5 mol × 32 g mol–1 = 80 g Answer : 80 g 4 Mass of 0.5 mol of sodium chloride, NaCl Molar mass of NaCl = (23 + 35.5) g mol–1 = 58.5 g mol–1 Mass of 0.5 mol of sodium chloride, NaCl = 0.5 mol × 58.5 g mol–1 = 29.25 g Answer: 29.25 g 5 Number of moles in 37.8 g of zinc nitrate, Zn(NO3)2 Molar mass of zinc nitrate, Zn(NO3)2 = [65 + 2 (14 + 3 × 16)] g mol–1 = 189 g mol–1 Number of moles of zinc nitrate, Zn(NO3)2 = 37.8 g 189 g mol–1 = 0.2 mol Answer: 0.2 mol 6 Mass of 3.01 × 1023 copper atoms, Cu Number of moles of Cu = Number of copper atom Avogadro constant = 3.01 × 1023 6.02 × 1023 = 0.5 mol Mass of Cu = Number of moles × Molar mass = 0.5 mol × 64 g mol–1 = 32 g Answer: 32 g LS 3.2.5 Number of Moles and Volume of Gas 03 U3 Chemistry F4(p27-52)csy2p.indd 34 21/12/2022 3:04 PM

MODULE • Chemistry FORM 4 35 UNIT 3 © Nilam Publication Sdn. Bhd. Formula for conversion of unit: LS 3.2.6 Relationship between number of moles and any given volume of gas. × 24 dm3 mol–1 / 22.4 dm3 mol–1 ÷ 24 dm3 mol–1 / 22.4 dm3 mol–1 Number of moles of gas Volume of gas in dm3 Example (i) 2 mol of carbon dioxide gas occupies 44.8 dm3 at STP. (ii) 16 g of oxygen gas = 0.5 mol of oxygen gas. Therefore, 16 g of oxygen gas occupies a volume of 12 dm3 at room conditions. [Relative atomic mass: O =16] 1 A sample of chlorine gas weighs 14.2 g. Calculate [Relative atomic mass: Cl = 35.5] (a) Number of moles of chlorine atoms. Number of moles of chlorine atoms, Cl = 14.2 g 35.5 g mol–1 = 0.4 mol (b) Number of moles of chlorine molecules (Cl2). Number of moles of chlorine molecules, Cl2 = 14.2 g 71 g mol–1 = 0.2 mol (c) Volume of chlorine gas at room conditions. [Molar volume of gas = 24 dm3 mol–1 at room temperature and pressure] Volume of chlorine gas = 0.2 mol × 24 dm3 mol–1 = 4.8 dm3 LS 3.2.6 LS 3.2.7 PL3 Volume of gas in dm3 × 24 dm3 mol–1/ 22.4 dm3 mol–1 ÷ 24 dm3 mol–1/ 22.4 dm3 mol–1 Mass in gram (g) ÷ (RAM/RFM/RMM) g mol–1 × (RAM/RFM/RMM) g mol–1 Number of moles Number of particles Exercise ÷ 6.02 × 1023 mol–1 × 6.02 × 1023 mol–1 03 U3 Chemistry F4(p27-52)csy2p.indd 35 21/12/2022 3:04 PM

MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 36 UNIT 3 2 (a) Calculate the number of atoms in the following substances: [Relative atomic mass: N = 14; Zn = 65; Avogadro Constant, NA = 6.02 × 1023 mol–1] (i) 13 g of zinc Number of mol of zinc atom = 13 g 65 g mol–1 = 0.2 mol Number of zinc atom = 0.2 × 6.02 × 1023 = 1.204 × 1023 (ii) 5.6 g of nitrogen gas Number of moles of N atom = 6.5 g 14 g mol–1 = 0.4 mol Number of N atom = 0.4 × 6.02 × 1023 = 2.408 × 1023 (b) Calculate the number of molecules in the following substances: [Relative atomic mass: N = 14, H = 1, Cl = 35.5, Avogadro Constant, NA = 6.02 × 1023mol–1] (i) 8.5 g of ammonia gas, NH3 Molar mass of ammonia gas, NH3 = (14 + 3) g mol–1 = 17 g mol–1 Number of moles of ammonia gas, NH3 = 8.5 g 17 g mol–1 = 0.5 mol Number of molecules in ammonia gas, NH3 = 0.5 mol × 6.02 × 1023 = 3.01 × 1023 (ii) 14.2 g of chlorine gas, Cl2 Molar mass of chlorine gas, Cl2 = 35.5 × 2 g mol–1 = 71 g mol–1 Number of moles of chlorine gas, Cl2 = Mass of chlorine Molar mass = 14.2 g 17 g mol–1 = 0.2 mol Number of chlorine molecules = 0.2 mol × 6.02 × 1023 = 1.204 × 1023 3 A gas jar contains 240 cm3 of carbon dioxide gas. Calculate: [Relative atomic mass: C = 12, O = 16; Molar volume of gas = 24 dm3 mol–1 at room conditions] (a) Number of moles of carbon dioxide gas: Number of moles of CO2 = 240 cm3 24 000 cm3 mol–1 = 0.01 mol (b) Number of molecules of carbon dioxide gas: Number of molecules of CO2 = 0.01 × 6.02 × 1023 = 6.02 × 1021 PL3 PL3 03 U3 Chemistry F4(p27-52)csy2p.indd 36 21/12/2022 3:04 PM

MODULE • Chemistry FORM 4 37 UNIT 3 © Nilam Publication Sdn. Bhd. (c) Mass of carbon dioxide gas: Mass of CO2 = 0.01 mol × [12 + 2 × 16] g mol–1 = 0.44 g 4 What is the mass of chlorine molecules (Cl2) that contains twice as many molecules as found in 3.6 g of water? [Relative atomic mass: H = 1, O = 16, Cl = 35.5] Number of moles of chlorine molecule, Cl2 = 2 × Number of moles of water, H2O Number of moles of H2O = 3.6 g 18 g mol–1 = 0.2 mol Number of moles of chlorine molecule = 2 × 0.2 mol = 0.4 mol Mass of Cl2 = 0.4 mol × 71 g mol–1 = 28.4 g 5 Calculate the mass of carbon that has the same number of atoms as found in 4 g of magnesium. [Relative atomic mass: C = 12, Mg = 24] Number of moles of magnesium = 4 g 24 g mol–1 = 1 6 mol Number of moles of carbon = Number of moles of magnesium = 1 6 mol Mass of carbon = 1 6 mol × 12 g mol–1 = 2 g 6 Compare the number of molecules in 32 g of sulphur dioxide (SO2) with 7 g of nitrogen gas (N2). Explain your answer. [Relative atomic mass: S = 32, O = 16, N = 14] Number of moles of molecules in 32 g SO2 = 32 g 64 g mol–1 = 0.5 mol Number of moles of molecules in 7 g N2 = 7 g 28 g mol–1 = 0.25 mol Number of molecules in 32 g SO2 is two times more than 7 g N2. Number of mole in sulphur dioxide molecules is two times more than number of mole of nitrogen molecules. 7 Compare number of atoms in 1.28 g of oxygen to the number of atoms in 1.3 g of zinc. Explain your answer. [Relative atomic mass: O = 16, Zn = 65] Number of moles of O atoms in 1.28 g O2 = 1.28 g 16 g mol–1 = 0.08 mol Number of moles of Zn atoms in 1.3 g Zn = 1.30 g 65 g mol–1 = 0.02 mol Number of oxygen atoms in 1.28 g oxygen is 4 times more than number of zinc atoms in 1.3 g zinc. Number of mol of oxygen atom is 4 times more than zinc atom. PL3 PL3 PL4 PL4 03 U3 Chemistry F4(p27-52)csy2p.indd 37 21/12/2022 3:04 PM

MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 38 UNIT 3 Define chemical formula. • A representative of a substance using letters to show the type of atoms and subscripts numbers to show the number of atoms in that substance. Example: Substances Chemical formula Water H2O Ammonia NH3 Propane C3H8 What information can be obtained from the chemical formula? Example: Substance Chemical formula Informations Ammonia NH3 (i) Elements present in the substance Ammonia is made up of nitrogen and hydrogen (ii) Number of atoms of each element in the compound Ammonia molecule consists of 1 nitrogen atom and 3 hyrogen atoms (iii) Relative formula mass Relative molecular mass = 14 + (3 × 1) = 17 Define empirical formula. • A formula that shows the simplest whole number ratio of atoms of each element in a compound. Define molecular formula. • Molecular formula of a compound is a formula that shows the actual number of atoms of each element that are present in a molecule of the compound. What is the relationship between molecular formula and empirical formula? Molecular formula = (Empirical formula)n, where n is an integer. Determine the empirical formula and the value of n. Compounds Molecular formula Empirical formula Value of n Water H2O H2O 1 Carbon dioxide CO2 CO2 1 Sulphuric acid H2SO4 H2SO4 1 Ethene C2H4 CH2 2 Benzene C6H6 CH 6 Glucose C6H12O6 CH2O 6 Remark: The molecular formula and the empirical formula of a compound will be the same if the value of n = 1 but different if the value of n > 1. LS 3.3.1 LS 3.3.1 LS 3.3.1 3.3 CHEMICAL FORMULA CS 3.3 PERFORMANCE LEVEL (PL) Mastered Not mastered PL3 Apply knowledge on mole concept, chemical formulae and equations to explain the natural occurrences or phenomena and be able to carry out simple tasks. 03 U3 Chemistry F4(p27-52)csy2p.indd 38 21/12/2022 3:04 PM

MODULE • Chemistry FORM 4 39 UNIT 3 © Nilam Publication Sdn. Bhd. 2 Experiment to Determine Empirical Formula of Magnesium Oxide LS 3.3.2 In this experiment, magnesium reacts with oxygen in the air to form white fumes, magnesium oxide: Magnesium + Oxygen → Magnesium oxide Material: Magnesium ribbon, sand paper Apparatus: Crucible with lid, tongs, Bunsen burner, pipe-clay triangle, balance and tripod stand Set-up of apparatus: Magnesium ribbon Heat 1 Experiments to determine empirical formula of metal oxide: Empirical formula of magnesium oxide Empirical formula of copper(II) oxide Set-up of apparatus: Heat Magnesium Set-up of apparatus: Copper(II) oxide powder Hydrogen gas Heat Reaction occurs: Magnesium is burnt in a crucible to react with oxygen to form magnesium oxide. Reaction occurs: Hydrogen gas is passed through the heated copper(II) oxide. Hydrogen reduces copper(II) oxide to form copper and water. Balanced equation: 2Mg + O2 → 2MgO Balanced equation: CuO + H2 → Cu + H2O This method can also be used to determine the empirical formula of reactive metal oxide such as aluminium oxide and zinc oxide. This method can also be used to determine the empirical formula of less reactive metal oxide such as lead oxide and tin oxide. Example The empirical formula for chlorinated hydrocarbon is CHCl2. The relative formula mass of this compound is 168. Find the molecular formula of this compound. (CHCl2)n = 168 Molecular formula (12 + 1 + [2 × 35.5])n = 168 = (Empirical formula)n (84)n = 168 = (CHCl2)2 n = 2 = C2H2Cl4 03 U3 Chemistry F4(p27-52)csy2p.indd 39 21/12/2022 3:04 PM

MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 40 UNIT 3 Procedures: (a) A crucible and its lid are weighed and the mass recorded. (b) 10 cm of magnesium ribbon is cleaned with sand paper. (c) The magnesium ribbon is coiled loosely and placed in the crucible. (d) The crucible together with the lid and magnesium ribbon are weighed again. (e) The apparatus is set up as shown in the diagram. (f) The crucible is heated strongly without its lid. When the magnesium starts to burn, the crucible is covered with its lid. (g) The lid of the crucible is lifted from time to time using a pair of tongs. (h) When the magnesium ribbon stops burning, the lid is removed and the crucible is heated strongly for another 2 minutes. (i) The crucible, the lid and its content are allowed to cool down to room temperature. (j) The crucible, lid and its content are weighed again and the mass recorded. (k) The process of heating, cooling and weighing are repeated until a constant mass is obtained. Precaution steps: Steps taken Purposes • Magnesium ribbon is cleaned with sand paper . • To remove the oxide layer on the surface of the magnesium ribbon. • The crucible lid is lifted from time to time. • The crucible lid is then replaced quickly. • To allow oxygen from the air to react with magnesium . • To prevent fumes of magnesium oxide from escaping. • The process of heating , cooling and weighing is repeated until a constant mass is obtained. • To ensure magnesium reacts completely with oxygen to form magnesium oxide . Observation: Magnesium burns brightly to release white fumes and white solid is formed. Inference: Magnesium is a reactive metal. Magnesium reacts with oxygen in the air to form magnesium oxide . Result: Descriptions Mass (g) Mass of crucible + lid x Mass of crucible + lid + magnesium y Mass of crucible + lid + magnesium oxide z Calculation: Elements Mg O Mass (g) y – x z – y Number of mole of atoms y – x 24 z – y 16 Simplest ratio of moles p q Empirical formula of magnesium oxide is MgpOq . 03 U3 Chemistry F4(p27-52)csy2p.indd 40 21/12/2022 3:04 PM

MODULE • Chemistry FORM 4 41 UNIT 3 © Nilam Publication Sdn. Bhd. 3 Experiment to Determine Empirical Formula of Copper(II) Oxide LS 3.3.3 Copper(II) oxide + Hydrogen → Copper + Water Set-up of apparatus: Precaution steps: Steps taken Purposes • Hydrogen gas is passed through the glass tube for 10 seconds. • To remove all the air in the glass tube. (The mixture of hydrogen gas and air will cause explosion when lighted). • The flow of hydrogen gas must be continuous throughout the experiment. • To prevent hot copper from reacting with oxygen to form copper(II) oxide again. • The process of heating , cooling and weighing are repeated until a constant mass is obtained. To ensure all copper(II) oxide has changed to copper . Observation: The black colour of copper(II) oxide turns brown . Inference: Copper(II) oxide reacts with hydrogen to produce the brown copper metal . Results: Descriptions Mass (g) Mass of glass tube x Mass of glass tube + copper(II) oxide y Mass of glass tube + copper z Calculation: Elements Cu O Mass (g) z – x y – z Number of mole of atoms z – x 64 y – z 16 Simplest ratio of moles p q Empirical formula of copper(II) oxide is CupOq . Copper(II) oxide powder Glass tube Water Zinc granules Hydrochloric acid (1.0 mol dm–3) Heat PERFORMANCE LEVEL (PL) Mastered Not mastered PL4 Analyse knowledge on mole, chemical formulae and equations in the context of problem solving the natural occurrences or phenomena. 03 U3 Chemistry F4(p27-52)csy2p.indd 41 21/12/2022 3:04 PM

MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 42 UNIT 3 4 Explain why the set-up of apparatus to determine the empirical formula in both experiments are different. (a) • Magnesium is reactive metal (above hydrogen in reactivity series). • Magnesium oxidised easily to form magnesium oxide . (b) • Copper is below hydrogen in the metal reactivity series. • Oxygen in copper(II) oxide can be removed by hydrogen gas to form copper and water. 5 To calculate the empirical formula of a compound, the following table can be used as a guide: Elements Mass of element (g) Number of mole of atom Simplest ratio of moles Calculation steps: (a) Calculate the mass of each element in the compound. (b) Convert the mass of each element to number of mole of atom. (c) Calculate the simplest ratio of moles of atom of the elements. 1 When 11.95 g of metal X oxide is reduced by hydrogen, 10.35 g of metal X is produced. Calculate the empirical formula of metal X oxide. [Relative atomic mass: X = 207, O = 16] Elements X O Mass of element (g) 10.35 1.6 Number of moles of atoms 10.35 207 = 0.05 1.6 16 = 0.1 Ratio of moles 1 2 Simplest ratio of moles 1 2 Empirical formula: XO2 . 2 A certain compound contains the following composition: Na = 15.23%, Br = 52.98%, O = 31.79 % [Relative atomic mass: O = 16, Na = 23, Br = 80] (Assume that 100 g of substance is used) Elements Na Br O Mass of element (g) 15.23 52.98 31.79 Number of moles of atoms 0.66 0.66 1.99 Ratio of moles 1 1 3.01 Simplest ratio of moles 1 1 3 Empirical formula: NaBrO3 . PL3 Exercise LS 3.3.4 PL3 03 U3 Chemistry F4(p27-52)csy2p.indd 42 21/12/2022 3:04 PM

MODULE • Chemistry FORM 4 43 UNIT 3 © Nilam Publication Sdn. Bhd. 3 2.08 g of element X combines with 4.26 g of element Y to form a compound with formula XY3. Calculate the relative atomic mass of element X. [RAM: Y = 35.5] Elements X Y Mass of element (g) 2.08 4.26 Number of moles of atom 2.08 x 4.26 35.5 = 0.12 Simplest ratio of moles 1 3 4 2.07 g of element Z reacts with bromine to form 3.67 g of a compound with empirical formula ZBr2. Calculate the relative atomic mass of element Z. [RAM: Br = 80] Elements Z Br Mass of element (g) 2.07 3.67 – 2.07 = 1.6 Number of moles of atoms 2.07 z 1.6 80 = 0.02 Simplest ratio of moles 1 2 z = relative atomic mass of Z Mol Z Mol Br = 1 2 2.07 z 0.02 = 1 2 z = 207 5 The empirical formula of compound X is CH2 and relative molecular mass is 56. Determine the molecular formula of compound X. [Relative atomic mass: H = 1; C = 12] (12 + 2)n = 56 n = 56 14 = 4 Molecular formula = (CH2)4 = C4H8 6 2.58 g of a hydrocarbon contains 2.16 g of carbon. The relative molecular mass of the hydrocarbon is 86. [Relative atomic mass: H = 1; C = 12] (i) Calculate the empirical formula of the hydrocarbon. Elements C H Mass of element (g) 2.16 2.58 – 2.16 = 0.42 Number of moles of atoms 0.18 0.42 Ratio of moles 1 2 1 3 = 7 3 Simplest ratio of moles 3 7 Empirical formula = C3H7 PL3 x = relative atomic mass of X Mol X Mol Y = 1 3 2.08 x 0.12 = 1 3 x = 52 PL3 PL3 PL3 03 U3 Chemistry F4(p27-52)csy2p.indd 43 21/12/2022 3:04 PM

MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 44 UNIT 3 (ii) Determine the molecular formula of the hydrocarbon. (3 × 12 + 7 × 1)n = 86 n = 86 43 = 2 Molecular formula = (C3H7)2 = C6H14 Formula % composition by mass of an element = Total RAM of the element in the compound RMM/RFM of compound × 100% Example Calculate the percentage composition by mass of nitrogen in the following compounds: [Relative atomic mass: N = 14, H = 1, O = 16, S = 32, K = 39] (i) (NH4)2SO4 (ii) KNO3 %N = 2 × 14 132 × 100% %N = 14 101 × 100% = 21.2% = 13.9% 1 Chemical formula of an ionic compound comprising of the ions Xm+ and Yn– is obtained by exchanging the charges on each ion. The formula obtained will be XnYm. LS 3.3.5 2 Example: (i) Sodium oxide Ion Na+ O2– Charges +1 –2 Exchange of charges 2 1 Simplest ratio 2 1 Number of combining ions 2 Na+ O2– Formula Na2O (ii) Copper(II) nitrate (iii) Zinc oxide Cu2+ 1 NO3 – 2 +2 –1 ➾ Cu(NO3)2 (Simplest ratio) Percentage Composition by Mass of an Element in a Compound LS 3.3.5 Chemical Formula for Ionic Compounds (Simplest ratio) Zn2+ 2 1 O2– 2 1 +2 –2 ➾ ZnO 03 U3 Chemistry F4(p27-52)csy2p.indd 44 21/12/2022 3:04 PM

MODULE • Chemistry FORM 4 45 UNIT 3 © Nilam Publication Sdn. Bhd. ACTIVITY 1: WRITE THE CHEMICAL FORMULAE AND NAMES OF THE FOLLOWING COMMON COMPOUNDS O2–, Oxide ion CO32–, Carbonate ion SO42–, Sulphate ion Cl–, Chloride ion Br–, Bromide ion I–, Iodide ion OH–, Hydroxide ion NO3 – , Nitrate ion K+ Potassium ion K O2 Potassium oxide K CO 2 3 Potassium carbonate K2SO4 Potassium sulphate KCl Potassium chloride KBr Potassium bromide KI Potassium iodide KOH Potassium hydroxide KNO3 Potassium nitrate Na+ Sodium ion Na O2 Sodium oxide Na CO 2 3 Sodium carbonate Na2SO4 Sodium sulphate NaCl Sodium chloride NaBr Sodium bromide NaI Sodium iodide NaOH Sodium hydroxide NaNO3 Sodium nitrate H+ Hydrogen ion H CO 2 3 Carbonic acid H2SO4 Sulphuric acid HCl Hydrocloric acid HBr Hydrobromic acid HI Hydroiodic acid HNO3 Nitric acid Ag+ Silver ion Ag O2 Silver oxide Ag CO 2 3 Silver carbonate Ag2SO4 Silver sulphate AgCl Silver chloride AgBr Silver bromide AgI Silver iodide AgOH Silver hydroxide AgNO3 Silver nitrate NH4 + Ammonium ion (NH4) CO 2 3 Ammonium carbonate (NH4)2SO4 Ammonium sulphate NH Cl 4 Ammonium chloride NH Br 4 Ammonium bromide NH4I Ammonium iodide NH NO 4 3 Ammonium nitrate Ca2+ Calcium ion CaO Calcium oxide CaCO3 Calcium carbonate CaSO4 Calcium sulphate CaCl2 Calcium chloride CaBr2 Calcium bromide CaI2 Calcium iodide Ca(OH)2 Calcium hydroxide Ca(NO3)2 Calcium nitrate Cu2+ Copper(II) ion CuO Copper(II) oxide CuCO3 Copper(II) carbonate CuSO4 Copper(II) sulphate CuCl2 Copper(II) chloride CuBr2 Copper(II) bromide CuI2 Copper(II) iodide Cu(OH)2 Copper(II) hydroxide Cu(NO3)2 Copper(II) nitrate Mg2+ Magnesium ion MgO Magnesium oxide MgCO3 Magnesium carbonate MgSO4 Magnesium sulphate MgCl2 Magnesium chloride MgBr2 Magnesium bromide MgI2 Magnesium iodide Mg(OH)2 Magnesium hydroxide Mg(NO3)2 Magnesium nitrate Zn2+ Zinc ion ZnO Zinc oxide ZnCO3 Zinc carbonate ZnSO4 Zinc sulphate ZnCl2 Zinc chloride ZnBr2 Zinc bromide ZnI2 Zinc iodide Zn(OH)2 Zinc hydroxide Zn(NO3)2 Zinc nitrate Pb2+ Lead(II) ion PbO Lead(II) oxide PbCO3 Lead(II) carbonate PbSO4 Lead(II) sulphate PbCl2 Lead(II) chloride PbBr2 Lead(II) bromide PbI2 Lead(II) iodide Pb(OH)2 Lead(II) hydroxide Pb(NO3)2 Lead(II) nitrate Al 3+ Aluminium ion Al O2 3 Aluminium oxide Al2(CO3)3 Aluminium carbonate Al2(SO4)3 Aluminium sulphate AlCl3 Aluminium chloride AlBr3 Aluminium bromide AlI3 Aluminium iodide Al(OH)3 Aluminium hydroxide Al(NO3)3 Aluminium nirate LS 3.3.5 PL1 PERFORMANCE LEVEL (PL) Mastered Not mastered PL5 Evaluate knowledge on mole concept, chemical formulae and equations in the context of problem solving and decision-making to carry out a task. 03 U3 Chemistry F4(p27-52)csy2p.indd 45 21/12/2022 3:04 PM

MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 46 UNIT 3 ACTIVITY 2: WITHOUT REFERRING TO THE TABLE IN ACTIVITY 1, WRITE THE CHEMICAL FORMULA OF THE FOLLOWING COMPOUNDS Oxide ion Carbonate ion Sulphate ion Chloride ion Bromide ion Iodide ion Hydroxide ion Nitrate ion Potassium ion K O2 K CO 2 3 K2SO4 KCl KBr KI KOH KNO3 Sodium ion Na O2 Na CO 2 3 Na2SO4 NaCl NaBr NaI NaOH NaNO3 Hydrogen ion H CO 2 3 H2SO4 HCl HBr HI HNO3 Silver ion Ag O2 Ag CO 2 3 Ag2SO4 AgCl AgBr AgI AgOH AgNO3 Ammonium ion (NH4) CO 2 3 (NH4)2SO4 NH Cl 4 NH4 Br NH4 I NH4 NO3 Calcium ion CaO CaCO3 CaSO4 CaCl2 CaBr2 CaI2 Ca(OH)2 Ca(NO3)2 Copper(II) ion CuO CuCO3 CuSO4 CuCl2 CuBr2 CuI2 Cu(OH)2 Cu(NO3)2 Magnesium ion MgO MgCO3 MgSO4 MgCl2 MgBr2 MgI2 Mg(OH)2 Mg(NO3)2 Zinc ion ZnO ZnCO3 ZnSO4 ZnCl2 ZnBr2 ZnI2 Zn(OH)2 Zn(NO3)2 Lead(II) ion PbO PbCO3 PbSO4 PbCl2 PbBr2 PbI2 Pb(OH)2 Pb(NO3)2 Aluminium ion Al O2 3 Al2(CO3)3 Al2(SO4)3 AlCl3 AlBr3 AlI3 Al(OH)3 Al(NO3)3 LS 3.3.5 PL1 03 U3 Chemistry F4(p27-52)csy2p.indd 46 21/12/2022 3:04 PM

MODULE • Chemistry FORM 4 47 UNIT 3 © Nilam Publication Sdn. Bhd. ACTIVITY 3: WRITE THE CHEMICAL FORMULA AND TYPE OF PARTICLES FOR THE FOLLOWING ELEMENT/COMPOUND Compound/ Element Formula Type of particles Compound/ Element Formula Type of particles Sodium sulphate Na2SO4 Ion Zinc carbonate ZnCO3 Ion Ammonium carbonate (NH4)2CO3 Ion Ammonium carbonate (NH4)2CO3 Ion Magnesium nitrate Mg(NO3)2 Ion Silver chloride AgCl Ion Hydrochloric acid HCl Ion Sulphuric acid H2SO4 Ion Potassium oxide K2O Ion Copper(II) nitrate Cu(NO3)2 Ion Magnesium oxide MgO Ion Hydrogen gas H2 Molecule Lead(II) carbonate PbCO3 Ion Carbon dioxide gas CO2 Molecule Iron(III) sulphate Fe2(SO4)3 Ion Oxygen gas O2 Molecule Magnesium chloride MgCl2 Ion Aluminium sulphate Al2(SO4)3 Ion Zinc sulphate ZnSO4 Ion Lead(II) chloride PbCl2 Ion Silver nitrate AgNO3 Ion Potassium iodide KI Ion Ammonium sulphate (NH4)2SO4 Ion Copper(II) carbonate CuCO3 Ion Zinc oxide ZnO Ion Potasium carbonate K2CO3 Ion Nitric acid HNO3 Ion Sodium hydroxide NaOH Ion Ammonia gas NH3 Molecule Aqueous ammonia NH3(aq) Ion and Molecule Magnesium Mg Atom Ammonium chloride NH4Cl Ion Zinc Zn Atom Nitrogen dioxide gas NO2 Molecule Copper(II) sulphate CuSO4 Ion Sodium chloride NaCl Ion Iodine I2 Molecule Silver Ag Atom Chlorine Cl2 Molecule Bromine Br2 Molecule LS 3.3.5 PL2 03 U3 Chemistry F4(p27-52)csy2p.indd 47 21/12/2022 3:04 PM

MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 48 UNIT 3 1 A chemical equation summarises what happen during chemical reaction. Reactants (Substances take part in the reaction) Products (Substances that are produced) Produce Example: Reaction between zinc powder and hydrochloric acid produce zinc chloride aqueous and hydrogen gas. Identify the reactant and products. • Reactants: Zinc and hydrochloric acid • Products: Zinc chloride and hydrogen gas Write the word equation. • Zinc + Hyrochloric acid ➝ Zinc chloride + Hydrogen gas Write the chemical formula of the reactants and products. List down the number of atoms of each element on both sides of the equation. Zn atom H atom Cl atom Zn + HCl Left 1 1 1 ZnCl2 + H2 Right 1 (Balanced) 2 (Not balanced) 2 (Not balanced) ➝ Balance the number of atoms of each element on both sides of the equation by adjusting the coefficients in front of the formula. LS 3.4.1 Zn atom H atom Cl atom Zn + 2HCl Left 1 2 2 ZnCl2 + H2 Right 1 (Balanced) 2 (Balanced) 2 (Balanced) ➝ Put the state symbols for every reactant and products: Remark: State symbols (s) (I) (g) (aq) Solid Liquid Gas Aqueous Zn(s) + 2HCl(aq) ➝ ZnCl2(aq) + H2(g) Qualitative interpretation of the chemical equation (the reactants and the products) LS 3.4.2 (i) The reactants are solid zinc and hydrochloric acid. (ii) The products are zinc chloride solution and hydrogen gas. Quantitative interpretation of the chemical equation (the coefficient of each formula shows the number of moles of reactants react and products formed) LS 3.4.2 Zn + 2HCl ➝ ZnCl2 + H2 Coefficient 1 2 1 1 • 1 mol of zinc reacts with 2 mol of hydrochloric acid to produce 1 mol of zinc chloride and 1 mol of hydrogen. LS 3.4.1 3.4 CHEMICAL EQUATIONS CS 3.4 03 U3 Chemistry F4(p27-52)csy2p.indd 48 21/12/2022 3:04 PM