Pre-U STPM Chemistry Term 1 CC039342a

ii PREFACE Pre-U STPM Text Chemistry Term 1 is written based on the new syllabus prepared by the Malaysian Examinations Council (MEC). The book is designed and well organised with the following features to help students to understand the concepts taught. v Analysis of STPM Papers (2015 – 2018) Chapter 2015 2016 2017 2018 A B C A B C A B C A B C 1 Atoms, Molecules and Stoichiometry 3 2 1 —2 1 1 2 1 2 Electronic Structure of Atoms 3 1 2 1 —2 1 2 1 3 Chemical Bonding 3 1 —2 2 1 4 1 1 —2 3 1 4 States of Matters 2 2 1 2 1 —2 1 1 5 Reaction Kinetics 2 1 3 3 1 2 6 Chemical Equilibria 1 1 2 1 —2 1 1 —2 2 1 7 Ionic Equilibria and Solubility Equilibria 1 —2 1 2 1 —2 2 1 —2 2 8 Phase Equilibria 1 1 1 1 Total 15 2 3 15 2 3 15 2 3 15 2 3 vi STPM Scheme of Assessment Term of Study Paper Code and Name Type of Test Mark (Weighting) Duration Administration First Term 962/1 Chemistry Paper 1 Written Test Section A 15 compulsory multiple-choice questions to be answered. Section B 2 compulsory structured questions to be answered. Section C 2 questions to be answered out of 3 essay questions. All questions are based on topics 1 to 6. 60 (26.67%) 15 15 30 1— 1 2 hours Central assessment Second Term 962/2 Chemistry Paper 2 Written Test Section A 15 compulsory multiple-choice questions to be answered. Section B 2 compulsory structured questions to be answered. Section C 2 questions to be answered out of 3 essay questions. All questions are based on topics 7 to 13. 60 (26.67%) 15 15 30 1— 1 2 hours Central assessment Third Term 962/3 Chemistry Paper 3 Written Test Section A 15 compulsory multiple-choice questions to be answered. Section B 2 compulsory structured questions to be answered. Section C 2 questions to be answered out of 3 essay questions. All questions are based on topics 14 to 21. 60 (26.67%) 15 15 30 1— 1 2 hours Central assessment 962/5 Chemistry Paper 5 Written Practical Test 3 compulsory structured questions to be answered. 45 (20%) 1— 1 2 hours Central assessment First, Second and Third Terms 962/4 Chemistry Paper 4 School-based Assessment of Practical 13 compulsory experiments and one project to be carried out. 225 to be scaled to 45 (20%) Throughout the three terms School-based assessment vi CHAPTER ATOMS, MOLECULES 1 AND STOICHIOMETRY Concept Map Students should be able to: Fundamental particles of an atom • describe the properties of protons, neutrons and electrons in terms of their relative charges and relative masses; • predict the behaviour of beams of protons, neutrons and electrons in both electric and magnetic fields; • describe the distribution of mass and charges within an atom; • determine the number of protons, neutrons and electrons present in both neutral and charged species of a given proton number and nucleon number; • describe the contribution of protons and neutrons to atomic nuclei in terms of proton number and nucleon number; • distinguish isotopes based on the number of neutrons present, and state examples of both stable and unstable isotopes. Relative atomic, isotopic, molecular and formula masses • define the terms relative atomic mass, Ar , relative isotopic mass, relative molecular mass, Mr , and relative formula mass based on 12C; • interpret mass spectra in terms of relative abundance of isotopes and molecular fragments; • calculate relative atomic mass of an element from the relative abundance of its isotopes or its mass spectrum. The mole and the Avogadro constant • define mole in terms of the Avogadro constant; • calculate the number of moles of reactants, volumes of gases, volumes of solutions and concentrations of solutions; • deduce stoichiometric relationships from the calculations above. Learning Outcomes earning Atoms, Molecules and Stoichiometry Structure of the Atom Isotopes Proton and Nucleon Number Radioactivity • Radioisotopes • Nuclear stability • Half-life • Other nuclear reactions Relative Mass • Relative isotopic mass • Relative atomic mass • Relative molecular mass • Relative formula mass Mass Spectrometer • Determination of relative atomic mass • Mass spectra of molecular species The Mole • Mole and concentration of solutions • Mole and gases Exam Tips Exam Tips CHAPTER 1 2 Chemistry Term 1 STPM 1.1 Fundamental Particles of an Atom 1 An atom consists of three fundamental particles neutron called proton, and electron. 2 The table below lists the year in which the particles were discovered. Particle Year of discovery Discoverer Electron Proton Neutron J.J. Thomson Rutherford Chadwick 1897 1914 1932 3 The properties of the three fundamental particles are listed below: Particle Symbol Mass/kg Relative mass/a.m.u. 1 1 Electron Proton Neutron 1 1834 Charge/C 0 –1 +1 0 Relative charge e or 0 –1e 9.11 × 10–31 –1.6 × 10–19 1.67 × 10–27 +1.6 × 10–19 1.67 × 10–27 p or 1 1 H n or 1 0 n [a.m.u. = atomic mass unit] 4 According to the modern model of the atom, protons and neutrons are found in the nucleus of the atom, and the electrons revolve around the nucleus in the form of an electron cloud called orbital. 5 The size of the atom and the nucleus is: ≈ 10–10 m Electron cloud ≈ 10–15 m Nucleus (containing proton + neutron) 6 The size of the nucleus is about 0.00001 times the size of the atom (or electron cloud). If the diameter of the nucleus is 1 cm, then the diameter of the atom is 100 000 cm or 1 km. 7 Since the electrons are very light by comparison, almost 99% of the mass of an atom is concentrated in the small nucleus. 8 The paths of moving protons, neutrons and electrons through an electric field and a magnetic field are shown below: Other sub-atomic particles such as positron, meson, quark and many others have since being discovered. 1 a.m.u. = 1.66 × 10–27 kg 1 The nucleus is 100 000 times smaller than the atom. + – Electron Proton Neutron • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • ••• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • Electron Magnetic field coming out from the plane of the paper Neutron Proton INFO Atomic Structure STPM Scheme of Assessment Latest STPM scheme of Assessment starting 2012. Concept Map Provides an overall view of the concepts learnt in the chapter. Learning Outcomes A list of subtopics that students will learn in each chapter. Analysis of STPM Papers Shows past yearsʼ STPM questions according to the chapters. QR Code Info/Video Scan QR code using mobile smartphone devices to access related information or videos.

iii Tips Exam Tips Exam CHAPTER 1 4 Chemistry Term 1 STPM The number of protons in Cl and Cl– is the same. Only the number of electrons changes. In a neutral atom, the no. of protons = no. of electrons. Isotopes are atoms having the same number of protons but different number of neutrons. For example, Species No. of protons No. of neutrons No. of electrons 35 17CI 17 18 17 35 17CI– 17 18 18 Example 1.1 Calculate the number of protons, neutrons and electrons in the following species. 1 1H; 12 6C; 238 92U; 16 8O2–; 60 27Co3+ Solution Species No. of protons No. of neutrons No. of electrons H 1 0 1 C 6 6 6 U 92 146 92 O2– 8 8 10 Co3+ 27 33 24 Example 1.2 A sulphur atom has 16 protons, 16 neutrons and 18 electrons. Give the symbol of the atom. Solution The proton number = 16 The nucleon number = 16 + 16 = 32 Since it has two electrons more than protons, the sulphur atom will have a charge of –2. Hence the symbol is 32 16S2–. Isotopes 1 Not all the atoms of a particular element are identical in every aspect. 2 Atoms of a particular element have the same number of protons (hence, the same proton number) and electrons. 3 They may, however, have different number of neutrons in their nucleus (hence, different nucleon number). This means that some atoms of a particular element are heavier than the rest. 4 Atoms having the same proton number but different nucleon number are called isotopes. Isotopes are different atomic species of the same element. 2010/P1/Q41 2015/P1/Q1 2016/P1/Q2 2017/P1/Q6(a)(b) CHAPTER 1 35 Chemistry Term 1 STPM 35 36 37 38 m/e (i) Identify the ions responsible for the peaks with m/e 35, 36, 37 and 38. (ii) There is another peak of relatively low intensity at m/e 18. What species is responsible for this peak? Account for its low intensity. 10 Methyl ethanoate undergoes hydrolysis to produce methanol and ethanoic acid. CH3 COOCH3 + H2 O → CH3 COOH + CH3 OH In an experiment, the isotopically labelled ester is completely hydrolysed in dilute sulphuric acid. CH3 — C 18O CH3 || O After purification, the products of hydrolysis is then analysed in a mass spectrometer. Three of the major peaks are shown below (not to scale). 18 34 60 (a) Identify the species responsible for the three peaks. (b) Deduce which of the two C— O bonds in the ester are broken during hydrolysis. 11 Chlorine consists of two isotopes, 35Cl and 37Cl, in the ratio of 3 : 1. Phosphorus is monoisotopic, 31P. A sample of phosphorous trichloride, PCl3 , is analysed in the mass spectrometer. Apart from lines due to atomic ions, the mass spectrum of the chloride contains 9 lines arranged in 3 groups. (a) Deduce the m/e values for the 9 lines. (b) Identify the ions that are responsible for the lines. (c) Calculate the relative abundance of the various lines within each group. 12 The relative atomic mass of element X is 35.5. (a) Why is the relative atomic mass of X not a whole number? (b) X has two naturally occurring isotopes with nucleon numbers of 35 and 37. Calculate the percentage abundance of the 35X isotope. (c) If X is monoatomic, sketch the mass spectrum of X. 13 Y (an alkane) is 9.5 times heavier than a carbon-12 atom. (a) Identify Y. (b) Calculate the volume of air (measured at s.t.p.) needed for the complete combustion of 0.15 dm3 of X. [Air contains 20% by volume of oxygen.] CHAPTER 1 30 Chemistry Term 1 STPM 1 Which of the following particles would be deflected less than the 3 Li+ ion in a mass spectrometer? A n B e C 1 H+ D 4 He+ 2 35Cl and 37Cl are two isotopes of chlorine. The relative atomic mass of chlorine is 35.5. Which statement is true? A 37Cl is more abundant than 35Cl B 37Cl is radioactive, 35Cl is not C The relative molecular mass of chlorine is 71.0 D 35Cl is more reactive than 37Cl 3 Consider the following conversion: 37Cl + e → 37Cl– Which of the following statements is correct regarding the above? Objective Questions SUMMARY SUMMARY 1 The three fundamental sub-atomic particles are proton, electron and neutron. 2 Proton number is the number of protons in the nucleus of an atom. Nucleon number is the total number of protons and neutrons in the nucleus of an atom. 3 Isotopes are atoms of the same element but with different number of neutrons. 4 Carbon-12 was chosen as the standard for the measurement of relative atomic mass. 5 On the carbon-12 scale, the mass of one atom of C-12 is 12.0000 a.m.u. 6 Relative isotopic mass is the mass of one atom of the isotope relative to 1 12 time the mass of one carbon-12 atom. 7 Relative atomic mass of an element is the average mass of one atom of the element relative to 1 12 time the mass of one carbon-12 atom. 8 Relative molecular mass is the mass of one molecule of the substance relative to 1 12 time the mass of one carbon-12 atom. 9 One mole of any substance is the amount that contains the same number of atoms in exactly 12 g of carbon-12. 10 The number of atoms in exactly 12 g of carbon-12 is 6.02  1023. This number is called the Avogadro’s number or Avogadro’s constant. 1 STPM PRACTICE A The number of protons decreases by one. B The number of neutrons increases by one. C The number of electrons increases by one. D The number of protons decreases by one and the number of electrons increases by one. 4 What is the difference between the relative molecular mass (Mr ) of CO2 and the molar mass of CO2 ? A The Mr has a unit of gram while the molar mass has a unit of g mol–1. B The Mr of CO2 is 44 while the molar mass of CO2 is 44 g mol–1. C The Mr refers to the mass of 1 mol of CO2 while the molar mass is the mass of 1 CO2 molecule. D The Mr is the mass of 1 mol of CO2 divided by the mass of 1 12 time the mass of 1 atom of C-12 while the molar mass of CO2 is the mass of 1 molecule of CO2 multiply by the Avogadro constant. CHAPTER 1 24 Chemistry Term 1 STPM Definition of ‘mole’ (d) It is caused by the presence of a carbon-13 isotope in the molecule of the alcohol. 13CH3 —CH—CH3 + | OH Quick Check 1.4 1 Bromine has two isotopes, 79Br and 81Br in the ratio of 1 : 1. Sketch the mass spectrum showing the peaks formed by the molecular ions. Indicate the relative abundance of the peaks. 2 An element G consists of two isotopes, 34G and 36G. The mass spectrum of a sample of G shows the following major peaks. Mass Relative abundance 68 9 70 x 72 4 (a) Determine the relative abundance of the two isotopes. (b) Calculate the value of x. 3 A mixture of nitrogen (14N2 ) and oxygen (16O2 ) is analysed in the mass spectrometer. Sketch the mass spectrum for the mixture. 4 Sketch the mass spectrum of NO2 when it is analysed in the mass spectrum. Nitrogen consists of 14N and oxygen as 16O and 18O. 1.3 The Mole and the Avogadro's Constant 1 The number of atoms in exactly 12 g of carbon-12 is 6.02  1023. 2 This number is called the Avogadro’s number or Avogadro’s constant. The symbols are NA or L. 3 Based on the carbon-12 scale, one mole of any substance is that amount of substance which contains the same number of particles (atoms, molecules, ions or electrons) as the number of atoms in 12 grams of carbon-12. For example, 1 mol of helium = 6.02  1023 He atoms 1 mol of carbon dioxide = 6.02  1023 CO2 molecules 1 mol of electrons = 6.02  1023 electrons Info Chem An analogy to 'mole' is 'dozen' which represent 12. 2013/P1/Q1, Q16 2015/P1/Q3 2016/P1/Q1, Q18(a) 2017/P1/Q16(d)(e) Notes Provides explanation in a simple way to ease students’ understanding. Quick Check Provides short questions for students to test their understanding of the concepts learnt in the subtopics. Example Gives step-by-step solutions to explain the example. Exam Tips Provides helpful tips for students in answering exam questions. Side Column Notes Provides additional information or pointers to improve students’ understanding. Past-year Questions tagging Enables students to know the topics frequently tested in the exam. Info Chem Provides extra information that relates to the subtopics learnt. Online Quick Quiz Scan QR code for self-assessment at the end of each chapter. Summary Summarises key concepts learnt in the chapter. STPM Practice A variety of examinationtype questions to check students’ understanding of the chapters learnt.

iv 337 Chemistry Term 1 STPM STPM Model Paper (962/1) STPM Model Paper (962/1) Section A [15 marks] Bahagian A [15 markah] Answer all questions in this section. Jawab semua soalan dalam bahagian ini. 1 If n is the number of atoms in 12 g of 12C, what is the number of atoms in 32 g of oxygen? Jika n adalah bilangan atom dalam 12 g 12C, berapakah bilangan atom dalam 32 g oksigen? A 1 2 n C 2n B n D 2.7n 2 A solution contains 20.8 g dm–3 of XCl2 . When react with excess of silver nitrate solution, 28.7 g of silver chloride is formed. What is the relative atomic mass of X? [Relative atomic mass Ag = 108, Cl = 35.5] Suatu larutan mengandungi 20.8 g dm–3 XCl2 . Apabila bertindak balas dengan larutan argentum nitrat berlebihan, 28.7 g argentum klorida terbentuk. Apakah jisim atom relatif X? [Jisim atom relatif Ag = 108, Cl = 35.5] A 52 C 156 B 104.5 D 208 3 The molecular peaks in the mass spectrum of 1,2-dibromoethane, C2 H4 Br2 , is shown below. Which of the graph is correct? [Bromine consists of two isotopes, 79Br and 81Br, in the ratio of 1 : 1] Puncak-puncak molekul dalam spektrum jisim 1,2-dibrometana, C2 H4 Br2 ditunjukkan di bawah. Graf yang manakah adalah benar? [Bromin terdiri daripada dua isotop 79Br dan 81Br dalam nisbah 1 : 1] A 190 B 186 190 C 186 188 190 D 186 190188 4 In the hydrogen atom, how many electronic transitions are possible between the first seven energy levels? Berapa bilangan peralihan yang mungkin berlaku antara tujuh paras tenaga yang pertama dalam atom hidrogen? A 6 C 21 B 7 D 29 5 Which statement is true of the emission spectrum of hydrogen? Pernyataan yang manakah adalah benar mengenai spektrum pancaran atom hidrogen? A There are more lines in the Lyman series than that in the Balmer series Terdapat lebih banyak garisan pada siri Lyman berbanding dengan siri Balmer B The ionisation energy of hydrogen can be calculated from the line with the longest wavelength 351 Activation energy Tenaga pengaktifan The minimum amount of energy required for a chemical reaction to occur Allotropy Alotropi The phenomenon where an element can exist in more than one form. The different forms of the same element are called allotropes. For example, carbon (diamond, graphite); sulphur (rhombic and monoclinic); oxygen (oxygen and ozone, O3 ) Autocatalysis Autopemangkinan A reaction where one of the products can act as a catalyst for the reaction Avogadro’s law Hukum Avogadro At constant temperature and pressure, equal volume of all gases contains the same number of moles of gas. Avogadro’s number/constant Pemalar/nombor Avogadro The number of atoms in exactly 12.0000 g of carbon-12 Azeotrope Azeotrop A mixture of liquids with a constant boiling point where the composition of the vapour is the same as the composition of the liquid. The composition of the azeotrope remains unchanged on distillation. Boiling point Takat didih The temperature at which the vapour pressure of a liquid is equal to the atmospheric pressure. Bond energy Tenaga ikatan The amount of heat energy required to break a covalent bond in one mole of gaseous molecules. Boyle’s law Hukum Boyle The volume of a fixed mass of gas at constant temperature is inversely proportional to its pressure. Bronsted-Lowry acid Asid Bronsted-Lowry A proton donor Bronsted-Lowry base Bes Bronsted-Lowry A proton acceptor Buffer capacity Kapasiti penimbal The amount of acid or base that needs to be added to a buffer solution to change its pH by one unit. Buffer solution Larutan penimbal A solution whose pH does not change significantly when a small amount of acid or base is added to it. Catalyst Mangkin A substance that alters the rate of a chemical reaction without itself being consumed Charles’ law Hukum Charles The volume of a fixed mass of gas at constant pressure is directly proportional to its absolute temperature. Common ion effect Kesan ion sepunya The shift in equilibrium caused by addition of a compound having an ion similar to one of the ions formed from the dissolved substance. Compound Sebatian A type of matter that is made up of more than one type of atoms 357 Chemistry Term 1 STPM Chapter 1 Atoms, Molecules and Stoichiometry Quick Check 1.1 1 (a) 222 86Rn (b) 234 91Pa (c) 38 20Ca (d) 212 82Pb (e) 208 81 Tl; 208 82Pb 2 228 90Y Quick Check 1.2 1 39.985 2 (a) 1.083 times (b) 12C : 13C = 9.9 : 0.1 3 12.0038 Quick Check 1.3 1 110.32 2 (a) 3 (b) 206 82M; 207 82M; 208 82M (c) It is caused by 206M2+ ion 3 (a) 22 21 20 (b) 20.18 (c) 1.68 times Quick Check 1.4 1 162 160 158 Ratio: 158 : 160 : 162 = 1 : 2 : 1 2 (a) 3 : 2 (b) 12 3 [Not to scale] 28 1814 32 4 14 16 18 30 32 46 48 50 14 16 18 30 32 46 48 50 N+ 16O+ 16O+ N16O+ N18O+ N16O16O+ N16O18O+ N18O2 O+ Quick Check 1.5 1 (a) 714.15 g (b) 28.88 g (c) 3369.8 g 2 (a) 23.46 g (b) 8.56 g (c) 1.15 g 3 (a) 5.56  104 (b) 3.34  1025 Quick Check 1.6 1 (a) CaCO3 + 2HCl → CaCl2 + H2 O + CO2 (b) 2.00 g (c) 80.0% 2 2 3 1.95 mol dm–3 Revision Exercise 1 Objective Questions 1 D 2 C 3 C 4 B 5 B 6 A 7 C 8 B 9 C 10 C 11 D 12 B 13 D 14 D 15 C 16 A 17 B 18 D 19 D 20 B 21 B 22 B 23 C 24 B 25 D 26 B 27 D 28 D 29 C 30 B 31 D 32 A Structured and Essay Questions 1 (a) Number of protons in the nucleus of an atom. (b) Total number of protons and neutrons in the nucleus of an atom. (c) Atoms of the same element but with different nucleon number. 2 Refer to section isotopes. 3 (a) Mole ratio N : O = 30.4 14 : 69.9 16 = 2.1 : 4.4 = 1 : 2 Empirical formula is NO2 (b) The molecular peak = 92 (NO2)n = 92 46n = 92 n = 2 Molecular formula is N2 O4 (c) m/e (46) = [NO2 ]+ Or [N2 O4 ]2+ m/e (94) = [N16O18O]+ 4 (a) Refer to text (b) Refer to text (c) (i) 28 29 30 29.8 1.5 1.0 m/e (ii) Ar = (28  29.8) + (29  1.5) + (30  1.0) (29.8 + 1.5 + 1.0) = 28.11 (iii) 28.11 12 = 2.34 times ANSWERS 382 Chemistry Term 1 STPM Index A Acid-base indicator 273 Activated complex 188 Activation energy 187 Allotropy 158 Arrhenius acid 254 Arrhenius base 254 Arrhenius collision theory 186 Atomic number 3 Atomic orbital, shape 49 Aufbau’s principle 51 Autocatalysis 199 Avogadro’s constant 24 Avogadro’s law 137 Azeotrope 324 B Base line, mass spectrum 21 Bohr’s model, Hydrogen atom 41 Bond energy 104 length 104 π-bond 80 σ-bond 80 Boyle’s law 132 Bronsted-Lowry acid 255 Bronsted-Lowry base 255 Buffer capacity 287 Buffer solution 280 C Catalyst 194 Charles’ law 135 Colligative property 169 Common ion effect 293 Compressibility factor, of gas 145 Conjugate acid-base pair 257 Convergence frequency, Line spectrum 44 Convergence limit, Line spectrum 44 Coordinate bond 105 Covalent bond 70 Critical point 164 D Dalton’s law of partial pressure 140 d-orbital 51 Dynamic equilibrium 223 E Electrolyte 253 Equivalent point 275 F Fractional distillation 316 H Half-life 203 Hund’s rule 52 Hybridisation 80 Hydrogen bond 116 I Ideal gas equation 139 Ideal solution 309 Ionic bonding 65 Isotope 4 Isotopic abundance 5 K Kinetic theory of gases 132 L Law of mass action 224 STPM Model Paper A model paper that follows the latest STPM exam format is provided for practice. Glossary Gives a list of important terms to ease the students’ understanding of their meanings. Answers Complete answers are provided. Index Provides a list of terms to enable easy and direct access to the subtopics.

v Analysis of STPM Papers (2015 – 2018) Chapter 2015 2016 2017 2018 A B C A B C A B C A B C 1 Atoms, Molecules and Stoichiometry 3 2 1 —2 1 1 2 1 2 Electronic Structure of Atoms 3 1 2 1 —2 1 2 1 3 Chemical Bonding 3 1 —2 2 1 4 1 1 —2 3 1 4 States of Matters 2 2 1 2 1 —2 1 1 5 Reaction Kinetics 2 1 3 3 1 2 6 Chemical Equilibria 1 1 2 1 —2 1 1 —2 2 1 7 Ionic Equilibria and Solubility Equilibria 1 —2 1 2 1 —2 2 1 —2 2 8 Phase Equilibria 1 1 1 1 Total 15 2 3 15 2 3 15 2 3 15 2 3

vi STPM Scheme of Assessment Term of Study Paper Code and Name Type of Test Mark (Weighting) Duration Administration First Term 962/1 Chemistry Paper 1 Written Test Section A 15 compulsory multiple-choice questions to be answered. Section B 2 compulsory structured questions to be answered. Section C 2 questions to be answered out of 3 essay questions. All questions are based on topics 1 to 6. 60 (26.67%) 15 15 30 1 1 —2 hours Central assessment Second Term 962/2 Chemistry Paper 2 Written Test Section A 15 compulsory multiple-choice questions to be answered. Section B 2 compulsory structured questions to be answered. Section C 2 questions to be answered out of 3 essay questions. All questions are based on topics 7 to 13. 60 (26.67%) 15 15 30 1 1 —2 hours Central assessment Third Term 962/3 Chemistry Paper 3 Written Test Section A 15 compulsory multiple-choice questions to be answered. Section B 2 compulsory structured questions to be answered. Section C 2 questions to be answered out of 3 essay questions. All questions are based on topics 14 to 21. 60 (26.67%) 15 15 30 1 1 —2 hours Central assessment 962/5 Chemistry Paper 5 Written Practical Test 3 compulsory structured questions to be answered. 45 (20%) 1 1 —2 hours Central assessment First, Second and Third Terms 962/4 Chemistry Paper 4 School-based Assessment of Practical 13 compulsory experiments and one project to be carried out. 225 to be scaled to 45 (20%) Throughout the three terms School-based assessment

vii CONTENTS Chapter • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 1 ATOMS, MOLECULES AND STOICHIOMETRY 1 1.1 Fundamental Particles of an Atom 2 1.2 Relative Atomic, Isotopic, Molecular and Formula Masses 13 1.3 The Mole and the Avogadro’s Constant 24 Summary 30 STPM Practice 1 30 QQ 35 Chapter • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 2 ELECTRONIC STRUCTURE OF ATOMS 36 2.1 Electronic Energy Levels of Atomic Hydrogen 37 2.2 Atomic Orbitals: s, p and d 48 2.3 Electronic Configurations 51 2.4 Classification of Elements in the Periodic Table 55 Summary 59 STPM Practice 2 60 QQ 63 Chapter • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 3 CHEMICAL BONDING 64 3.1 Ionic Bonding 65 3.2 Covalent Bonding 70 3.3 Metallic Bonding 107 3.4 Intermolecular Forces: van der Waals Forces and Hydrogen Bonding 110 Summary 124 STPM Practice 3 125 QQ 130 Chapter • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 4 STATES OF MATTERS 131 4.1 Gases 132 4.2 Liquids 147 4.3 Solids 152 4.4 Phase Diagrams 161 Summary 174 STPM Practice 4 175 QQ 180 Chapter • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 5 REACTION KINETICS 181 5.1 Rate of Reaction 182 5.2 Rate Law 191 5.3 The Effect of Temperature on Reaction Kinetics 193 5.4 The Role of Catalysts in Reactions 194 5.5 Order of Reactions and Rate Constants 201

viii Summary 213 STPM Practice 5 214 QQ 220 Chapter • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 6 CHEMICAL EQUILIBRIA 221 6.1 Reversible Reaction and Dynamic Equilibrium 222 6.2 Law of Mass Action and Equilibrium Constant 224 6.3 Deduce Equilibrium Constant, Kc and Kp 225 6.4 Calculation Involving Equilibrium Constant 228 6.5 Nitrogen Dioxide and Stratospheric Ozone 232 6.6 Factors Affecting Equilibrium System and Le Chatelier's Principle 233 6.7 Equilibrium and Temperature 240 6.8 Industrial Processes 243 Summary 245 STPM Practice 6 246 QQ 251 Chapter • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 7 IONIC EQUILIBRIA AND SOLUBILITY EQUILIBRIA 252 7.1 Electrolytes 253 7.2 Acids and Bases 254 7.3 Conjugate Acid-base Pair 257 7.4 Relative Strength of BrØnsted-Lowry Acids and Bases 259 7.5 Acid-base Titration 273 7.6 Buffer Solution 280 7.7 Important Buffers in Human Blood 290 7.8 Solubility Equilibria: Sparingly Soluble Salts 291 7.9 Calculating Ksp from Solubility and Vice Versa 292 7.10 Common Ion Effect 293 7.11 Predicting Precipitation 296 7.12 Solubility Equilibria and Water Softening 298 Summary 300 STPM Practice 7 301 QQ 305 Chapter • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 8 PHASE EQUILIBRIA 306 8.1 Miscible Liquids 307 8.2 Mixture of Two Miscible Liquids 308 8.3 Fractional Distillation 316 8.4 Non-ideal Solutions 319 8.5 Fractional Distillation under Reduced Pressures 330 Summary 332 STPM Practice 8 333 QQ 336 • STPM Model Paper (962/1) 337 • Appendix 343 • Glossary 351 • Answers 357 • Index 382

CHAPTER ATOMS, MOLECULES 1 AND STOICHIOMETRY Concept Map Students should be able to: Fundamental particles of an atom • describe the properties of protons, neutrons and electrons in terms of their relative charges and relative masses; • predict the behaviour of beams of protons, neutrons and electrons in both electric and magnetic fields; • describe the distribution of mass and charges within an atom; • determine the number of protons, neutrons and electrons present in both neutral and charged species of a given proton number and nucleon number; • describe the contribution of protons and neutrons to atomic nuclei in terms of proton number and nucleon number; • distinguish isotopes based on the number of neutrons present, and state examples of both stable and unstable isotopes. Relative atomic, isotopic, molecular and formula masses • define the terms relative atomic mass, Ar , relative isotopic mass, relative molecular mass, Mr , and relative formula mass based on 12C; • interpret mass spectra in terms of relative abundance of isotopes and molecular fragments; • calculate relative atomic mass of an element from the relative abundance of its isotopes or its mass spectrum. The mole and the Avogadro constant • define mole in terms of the Avogadro constant; • calculate the number of moles of reactants, volumes of gases, volumes of solutions and concentrations of solutions; • deduce stoichiometric relationships from the calculations above. Learning earning Outcomes Atoms, Molecules and Stoichiometry Structure of the Atom Proton and Isotopes Nucleon Number Radioactivity • Radioisotopes • Nuclear stability • Half-life • Other nuclear reactions Relative Mass • Relative isotopic mass • Relative atomic mass • Relative molecular mass • Relative formula mass Mass Spectrometer • Determination of relative atomic mass • Mass spectra of molecular species The Mole • Mole and concentration of solutions • Mole and gases

Exam Tips CHAPTER 1 2 Chemistry Term 1 STPM 1.1 Fundamental Particles of an Atom 1 An atom consists of three fundamental particles called proton, neutron and electron. 2 The table below lists the year in which the particles were discovered. Particle Year of discovery Discoverer Electron Proton Neutron J.J. Thomson Rutherford Chadwick 1897 1914 1932 3 The properties of the three fundamental particles are listed below: Particle Symbol Mass/kg Relative mass/a.m.u. 1 1 Electron Proton Neutron 1 1834 Charge/C 0 Relative charge –1 +1 0 e or 0 –1e 9.11 × 10–31 –1.6 × 10–19 1.67 × 10–27 +1.6 × 10–19 1.67 × 10–27 p or 1 1 H n or 1 0 n [a.m.u. = atomic mass unit] 4 According to the modern model of the atom, protons and neutrons are found in the nucleus of the atom, and the electrons revolve around the nucleus in the form of an electron cloud called orbital. 5 The size of the atom and the nucleus is: ≈ 10–10 m Electron cloud ≈ 10–15 m Nucleus (containing proton + neutron) 6 The size of the nucleus is about 0.00001 times the size of the atom (or electron cloud). If the diameter of the nucleus is 1 cm, then the diameter of the atom is 100 000 cm or 1 km. 7 Since the electrons are very light by comparison, almost 99% of the mass of an atom is concentrated in the small nucleus. 8 The paths of moving protons, neutrons and electrons through an electric field and a magnetic field are shown below: Other sub-atomic particles such as positron, meson, quark and many others have since being discovered. 1 a.m.u. = 1.66 × 10–27 kg 1 The nucleus is 100 000 times smaller than the atom. + – Electron Proton Neutron • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • ••• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • Electron Magnetic field coming out from the plane of the paper Neutron Proton INFO Atomic Structure

Exam Tips Exam Tips CHAPTER 1 3 Chemistry Term 1 STPM 9 Neutrons, being neutral, are not affected by the electric or magnetic field. 10 Electrons, which are lighter, are deflected more than the heavier protons and in the opposite direction. Proton Number and Nucleon Number 1 Every atom can be characterised by the numbers of protons, neutrons and electrons it contains. These are represented by two sets of numbers: proton number and nucleon number. 2 The proton number (symbol Z) is the total number of protons in the nucleus of an atom. This is also equal to the number of electrons in the atom. 3 The nucleon number (symbol A) is the total number of protons and neutrons in the nucleus of an atom. [Nucleons are nuclear particles. Electrons are not nucleons]. 4 All atoms of a particular element have the same number of protons. This number is the basic property of the element. 5 For example, the proton number of sodium is 11. This means that any neutral atom in the universe having a proton number of 11 is a sodium atom. 6 The relationship between proton number and nucleon number is given by: A = Z + N where N = number of neutrons 7 The proton number and nucleon number of an atom are written beside the symbol of the element as shown: 8 For example, 12 6 C; 24 12Mg; 238 92U; 56 26Fe 9 When a neutral atom loses an electron (or more than one electron), a positive ion (cation) is formed. In a cation, the number of electrons is less than the number of protons. Atom cation + electrons For example, Species No. of protons No. of neutrons No. of electrons 23 11Na 11 12 11 23 11Na+ 11 12 10 10 When an atom gains electrons, a negative ion (anion) is formed. In an anion, the number of electrons is more than the number of protons. Atom + electrons anion Nucleon number A Proton number Z X Atoms of different elements have different number of sub-atomic particles. Proton number is also known as atomic number. Nucleon number is also known as mass number. The number of protons and electrons in a neutral atom determines the properties of the atom. Number of neutrons = nucleon – proton number number The number of protons in Na and Na+ is the same. Only the number of electrons changes. 2014/P1/Q1

Exam Tips CHAPTER 1 4 Chemistry Term 1 STPM The number of protons in Cl and Cl– is the same. Only the number of electrons changes. In a neutral atom, the no. of protons = no. of electrons. Isotopes are atoms having the same number of protons but different number of neutrons. For example, Species No. of protons No. of neutrons No. of electrons 35 17CI 17 18 17 35 17CI– 17 18 18 Example 1.1 Calculate the number of protons, neutrons and electrons in the following species. 1 1H; 12 6C; 238 92U; 16 8O2–; 60 27Co3+ Solution Species No. of protons No. of neutrons No. of electrons H 1 0 1 C 6 6 6 U 92 146 92 O2– 8 8 10 Co3+ 27 33 24 Example 1.2 A sulphur atom has 16 protons, 16 neutrons and 18 electrons. Give the symbol of the atom. Solution The proton number = 16 The nucleon number = 16 + 16 = 32 Since it has two electrons more than protons, the sulphur atom will have a charge of –2. Hence the symbol is 32 16S2–. Isotopes 1 Not all the atoms of a particular element are identical in every aspect. 2 Atoms of a particular element have the same number of protons (hence, the same proton number) and electrons. 3 They may, however, have different number of neutrons in their nucleus (hence, different nucleon number). This means that some atoms of a particular element are heavier than the rest. 4 Atoms having the same proton number but different nucleon number are called isotopes. Isotopes are different atomic species of the same element. 2010/P1/Q41 2015/P1/Q1 2016/P1/Q2 2017/P1/Q6(a)(b)

Exam Tips CHAPTER 1 5 Chemistry Term 1 STPM 5 Most naturally occurring elements consist of a mixture of isotopes. For example, there are three types of hydrogen atoms. Isotope Name No. of protons No. of neutrons No. of electrons 1 1 H Protium 1 0 1 2 1 H Deuterium 1 1 1 3 1 H Tritium 1 2 1 Protium Proton Electron Neutron Deuterium Tritium 6 The two isotopes of chlorine are 35 17Cl and 37 17Cl. Isotope No. of protons No. of neutrons No. of electrons 35 17CI 17 18 17 37 17CI 17 20 17 7 Since all chlorine atoms must have a proton number of 17, the two isotopes of chlorine can also be represented by 35Cl and 37Cl or Cl-35 and Cl-37. The proton number is omitted in this type of representation. 8 The relative amount of each isotope of an element is called the relative abundance or isotopic abundance. It can be expressed in terms of percentage or ratio. 9 For example, the relative abundance of 35Cl and 37Cl in naturally occurring chlorine is 75% and 25%, or they are in the ratio of 3 : 1. 10 The relative abundance of the isotopes of iron is given below. Isotope Relative abundance (%) 54Fe 56Fe 57Fe 58Fe 5.8 91.6 2.2 0.4 11 Isotopes of an element have the same: (a) number of protons (proton number), (b) number of electrons, (c) electronic configuration, (d) chemical properties. The hydrogen atom (1 1H) is the only atom that does not have any neutron. 37 17 Cl is heavier than 35 17 Cl. Deuterium is sometimes represented by D. Tritium is sometimes represented by T.

CHAPTER 1 6 Chemistry Term 1 STPM Nuclei of radioisotopes which are not stable will undergo spontaneous disintegration. Madam Curie was awarded the nobel prize for her work on radioactivitiy. Three main types of radioactive radiations/ particles. 12 Isotopes of an element have different: (a) number of neutrons (nucleon number), (b) mass, (c) density, (d) molecular speed. Example 1.3 Complete the following table. Species Proton number Nucleon number No. of protons No. of neutrons No. of electrons Net charge 90Sr2+ 38 82Br– 35 47 36 39K 19 23Na+ 23 11 +1 Solution Species Proton number Nucleon number No. of protons No. of neutrons No. of electrons Net charge 90Sr2+ 38 90 38 52 36 +2 82Br– 35 82 35 47 36 –1 39K 19 39 19 20 19 0 23Na+ 11 23 11 12 10 +1 Stable and Unstable Isotopes 1 The nuclei of certain isotopes are unstable (they may contain either too many protons or too many neutrons) and will disintegrate spontaneously to form nucleus of smaller atoms (known as daughter nuclei) with the emission of high energy radiations or particles. 2 The unstable isotopes are called radioactive isotopes. The process of spontaneous disintegration is called radioactive decay. (The term radioactivity was suggested by Madam Curie who discovered the natural phenomenon). 3 Since the reaction involves only the nucleus, it is called a nuclear reaction. 4 Essentially, there are three main types of radioactive radiations/ particles emitted by radioactive isotopes when they undergo radioactive decay. They are the alpha-particle, the beta-particle and gamma radiation.

Exam Tips Exam Tips CHAPTER 1 7 Chemistry Term 1 STPM α-particles, β-particles, γ-radiation Radioisotopes Daughter nuclei + 5 Some isotopes decay by emitting all the three types of radiations/particles, but there are some radioactive isotopes which emit α or β particles only. 6 The decay continues until a stable isotope is formed. α-Particles 1 An alpha particle is made up of 2 protons and 2 neutrons: Neutrons Protons An α-particle It has the composition of the nucleus of a helium atom. 2 However, it differs from a helium atom in that it has no electrons and carries a charge of +2. It is represented by the symbol 4 2 He2+. 3 α-particles are positively charged, and have very high ionising power. 4 The positive charge on the α-particles causes them to be deflected towards the negative plate of an electric field. 5 The large size of the α-particles reduces its penetrating power. A stream of α-particles is stopped by a thin sheet of paper. 6 When a nucleus disintegrates by emission of an α-particle, the number of proton (proton number) decreases by two and consequently a new element is formed. The nucleon number of the nuclei decreases by 4 units. 7 An example of α-decay is plutonium-240, 240Pu: 240 94 Pu 4 2He + 236 92 U β-Particles 1 β-particles are high energy electrons moving at a speed comparable to the speed of light (3.0  108 m s–1). 2 A β-particle is produced when a neutron in the nucleus splits to form a proton and an electron. The electron is ejected from the nucleus as β-particle. Alpha particles are represented as 4 2He2+. α-particles have short penetrating range. β-particles are ejected from the nucleus. The proton numbers and nucleon numbers on both sides of the nuclear equation must be balanced. Examples of stable and unstable isotopes: Stable isotope Unstable isotope 1 H 3 H 12C, 13C 14C 16O 15O 23Na 24Na – 235U, 238U

Exam Tips Exam Tips CHAPTER 1 8 Chemistry Term 1 STPM Neutron → proton + electron (β-particle) 1 0n → 1 1H + 0 –1 e n p + e 3 The negatively charged β-particles are deflected towards the positive plate in an electric field. Since the β-particle is lighter than the α-particle, its angle of deflection is larger. 4 Due to its small mass and high velocity, the penetrating power of β-particles is higher than that of α-particles. β-particles are stopped by a thin sheet of aluminium. 5 Due to its very small size and high energy, it can travel further. 6 β-decay produces an extra proton in the nucleus. The proton number increases by one while the nucleon number remains unchanged. A new element is thus produced. 7 An example of β-decay is caesium-137: 137 55 Cs → 0 –1 e + 137 56 Ba γ-Radiation 1 After emitting α or β-particles, the nucleus of the new element is sometimes left in an excited state with excess energy. This excess energy is radiated in the form of γ-radiation. 2 γ-radiation, like X-ray, is high-energy electromagnetic radiation with very short wavelength (10–10 to 10–12 m). 3 Emission of γ-radiation does not affect the proton number or the nucleon number. For example, Pu* → Pu + γ (Excess energy) 4 The ionising and penetrating power of γ-radiation is very much higher than those of α-particles and β-particles. 5 γ-radiation is stopped by a thick block of lead. 6 γ-radiation is neutral and is not affected by electric or magnetic fields. Summary α particle β particle γ ray helium nucleus (+2 charge) electron (–1 charge) gamma (photon) β-particles have longer penetrating range. The nucleon number of both species is the same. No change in the proton number and nucleon number. γ-radiation has very long penetrating range. Info Chem Pu is plutonium-94.

CHAPTER 1 9 Chemistry Term 1 STPM 1 The properties of α, β and γ particles/radiation are summarised in the table below: Particle/Radiation Alpha (α) Beta (β) Gamma (γ) Nature Resembles nucleus of helium atom High energy electron Electromagnetic radiation Composition 2 protons and 2 neutrons Electron Photon Symbol 4 2He 0 –1e γ Relative mass/a.m.u. 4 1 1834 0 Relative charge +2 –1 0 Relative penetrating power 1 100 10 000 2 The path of moving a, β and γ particles/radiation through an electric field is shown below: Lead block Radioactive substance _ + α γ β 3 The relative penetrating power of the three particles/radiation is shown below: α-particle β-particle γ-radiation Paper Aluminium foil Thick block of lead

CHAPTER 1 10 Chemistry Term 1 STPM 1 Balance the following nuclear equations and identify the daughter nuclei. (a) 226Ra → α + A (d) 216Po → α + D (b) 234Th → β + B (e) 212Bi → α + E → β + F (c) 38K → β + C 2 Thorium-232 (232Th) disintegrates by α-particle emission to give an element W, which then loses a β-particle to give element X, which also loses a β-particle to give element Y. What are the proton number and nucleon number of element Y? The neutron: proton ratio determines the stability of an atom. Nuclear Stability 1 Isotopes may be classified as stable or unstable. Stable isotopes do not undergo spontaneous radioactive decay. 2 There are about 280 naturally occurring stable isotopes. Examples are 16O and 12C. 3 The graph below shows a plot of the number of neutrons against the number of protons for all stable isotopes. 140 120 100 80 60 40 20 20 40 60 80 100 0 0 Number of neutrons Number of protons Belt of stability 4 The stable isotopes form a band known as the ‘stability belt’. Isotopes whose points lie outside this belt (either with too many neutrons or too many protons) are unstable and will undergo spontaneous radioactive decay. Relative Stability of Radioisotopes: Half-life 1 The stability of a radioisotope is measured by its half-life, T1 —2 . n —p = 1 Quick Check 1.1

CHAPTER 1 11 Chemistry Term 1 STPM 2 The half-life of a radioisotope is the time taken for the amount of the isotope or its activity (disintegration per second) to decrease to half its original value. 3 The half-life is independent of the amount of radioactive substance present and is not influenced by catalyst or changes in temperature or pressure. 4 For example, the half-life of radium-228 is 6.7 years. This means that the amount of radium-228 will decrease to half its original amount every 6.7 years. Let’s say the original amount of Ra-228 is C g: C → 6.7 years C 2 → 6.7 years C 4 → 6.7 years C 8 → 6.7 years Thus, the longer the half-life of a radioisotope, the more stable it is and vice versa. 5 Half-lives of radioactive isotopes can range from millionth of a second to millions of years as shown in the table below. Isotope Half-life 232Th 1.4  1010 years 14C 5700 years 137Cs 27 years 224Ra 3.7 days 212Bi 60.5 mins 216Po 0.13 seconds 212Po 3  10–7 seconds 6 Radioactive decay is a first order reaction. The half-life is given by the equation: T 1 —2 = ln 2 k where k = decay constant 7 Both T 1 —2 and k are characteristics of the radioisotope. Other Nuclear Reactions 1 The nucleus of some isotopes can disintegrate to form new species when they are bombarded with high energy particles. 2 There are two main types of nuclear reactions: nuclear fission and nuclear fusion. Nuclear Fission 1 When the nucleus of uranium-235 atoms is bombarded with high-energy neutrons, it breaks down into nuclei of molybdenum-95 and lanthanum-129, according to the equation: 235 92U + 1 0 n → 95 42Mo + 139 57La + 2 1 0 n + 7 0 –1 e 2 Tremendous amount of energy in the form of heat is released with the reaction. Definition of half-life The half-life of a particular radioisotope is a constant. The longer the half-life, the more stable the radioisotope. ln = loge Fission – breaking Fusion – joining

Exam Tips Exam Tips CHAPTER 1 12 Chemistry Term 1 STPM Nuclear Fusion 1 Under extreme conditions of temperature and pressure, two deuterium atoms can combine to produce a helium atom according to the equation: 22 1H → 3 2He + 1 0n 2 The energy released in the above reaction is 3.89 × 1011 J per 4 g of deuterium. This amount of energy is equivalent to the energy released on burning 1.18 × 1010 g of coal. 3 The following fusion reaction occurs in the Sun which generates the heat produced from the Sun. 41 1 H → 4 2He + 2 0 1 e Positron Uses of Radioactive Isotopes and Nuclear Reactions 1 Radioactive isotopes have many practical applications due to the fact that radioactivity is easily detected using a Geiger-Muller counter. 2 Radioactive isotopes can be used as tracers. If a radioactive isotope is incorporated into a molecule, then the path of the molecule can be followed using suitable detecting apparatus. 3 Tracers such as 46Sc and 51Cr are used to monitor the movement of silt and mud in rivers. 4 The rate of certain biological process can be determined by adding radioactive samples to the substance. When the tracers are taken in by an organism, the path or the biological process can be followed. 5 Leaks in deep underground pipe-lines can be detected by injecting a small amount of radioactive substance and measuring the activity from the surface. 6 Radioactive isotopes are also used to sterilise medical equipments and food, measure the thickness of metal, and treat cancerous tumours. 7 Radioactive decay is used in carbon dating to estimate the age of archaeological artifacts containing carbon. 8 The energy from nuclear fission and nuclear fusion can be harnessed to produce electricity in a nuclear plant. Carbon Dating 1 A small percentage of carbon dioxide in our atmosphere contains radioactive isotope 14C which is formed between the reaction of the cosmic ray and 14N in the atmosphere. 14 7N + 1 0n → 14 6C + 1 1H Carbon dating Production of electricity Atmospheric CO2 is a mixture of 12CO2 and 14CO2 . Nuclear fusion needs very high temperature for it to occur. Geiger-Muller counter Use as tracers Sterilisation of medical equipments 24Na is used in medicine to study the blood circulatory system. Due to its short halflife, it poses no long-lasting effect on the subject. Info Chem A positron is identical to an electron except it has a charge of +1.

CHAPTER 1 13 Chemistry Term 1 STPM 2 The radioactive carbon dioxide is taken in by plant during photosynthesis. In turn, plants are eaten by animals. Hence, all living tissues contain a certain amount of radioactive 14C isotope. 3 The 14C isotope is a beta-emitter and decays via the following process: 14 6C → 14 7N + 0 –1e The loss of 14C through decay is constantly replenished by the production of 14C in the atmosphere. Eventually, a dynamic equilibrium is established, and the amount of 14C in living organism remains almost constant. 4 When a plant or animal dies, its intake of 14C terminates. Over time the amount of 14C in the dead tissue decreases. Hence, by comparing the activity (rate of disintegration) of 14C per gram of the dead tissue with that of a living tissue, the age of the tissue can be determined. 5 Radioactive decay is a first order reaction. The integrated form of the rate equation is: ln N = –kt + ln No where N = Activity at time t No = Activity at t = 0 6 The following example shows how the age of a piece of dead wood is determined. Example 1.4 The activity of a piece of dead wood is 3.4 disintegration per minute per gram of carbon. Given that the activity of a living specimen is 16.8 disintegration per minute per gram carbon. Calculate the age of the specimen. [k = 1.2  10–4 year–1] Solution Using the equation: ln N = –kt + ln No ln(3.4) = –(1.2  10–4)t + ln(16.8) ∴ t = 13 313 years 1.2 Relative Atomic, Isotopic, Molecular and Formula Masses 1 The relative mass of an atom is expressed in atomic mass unit (a.m.u.). Carbon dating cannot be used on non-organic substances such as rocks and metals. 2008/P1/Q1 2009/P1/Q1 2018/P1/Q18(a)

CHAPTER 1 14 Chemistry Term 1 STPM 2 The relative mass of an atom is an indication of how heavy one atom of an element is compared to an atom of another element. 3 The first reference atom chosen was hydrogen, because it is the lightest. It was later changed to oxygen. 4 In 1961, the International Union of Pure and Applied Chemistry (IUPAC) formally adopted the most abundant isotope of carbon, 12 6C, as standard for measurement of atomic mass. 12C was assigned a mass of exactly 12 atomic mass unit (a.m.u.). This is known as the C-12 scale. Mass of a C-12 atom = 12 a.m.u. ∴ 1 a.m.u. = 1 12 × the mass of a C-12 atom For example, an atom which is 3.5 times heavier than a 12C atom would have a relative mass of (3.5 × 12) = 42 a.m.u. That means, the atom is 42 times heavier than the mass of ( 1 12 × the mass of a C-12 atom). 5 12C was chosen as the standard for the measurement of relative atomic mass because it is a solid and is easily available. Relative Isotopic Mass 1 The relative isotopic mass is defined as the mass of one atom of the isotope relative to 1 12 times the mass of one atom of the C-12 isotope. 2 The relative isotopic mass can also be defined as 12 multiply by the number of times one atom of the isotope is heavier than one atom of carbon-12. Relative isotopic mass = Mass of one atom of the isotope 1 12  the mass of one C-12 atom or = Mass of one atom of the isotope  12 The mass of one C-12 atom 3 For example, an atom of the isotope 40Ca is 3.34 times heavier than one atom of C-12. The relative isotopic mass of the isotope = 3.34  12 = 40.08 4 The relative isotopic mass of an isotope is approximately the same as its nucleon number. For example, Relative abundance: 12C : 13C = 99 : 1 Definition of relative isotopic mass Alternative definition

CHAPTER 1 15 Chemistry Term 1 STPM Isotope Nucleon number Relative isotopic mass (a.m.u.) 4 He 4 4.003 1 H 1 1.008 19F 19 18.99 127I 127 126.910 35CI 35 34.970 37CI 37 36.965 5 Hence, relative isotopic masses can be substituted with the nucleon numbers in normal calculations. Relative Atomic Mass (Ar ) 1 Most elements contain a mixture of isotopes with different relative abundance. For example, Isotope Relative abundance (%) 75.0 35Cl 25.0 37Cl 1.5 204Pb 23.6 206Pb 22.6 207Pb 52.3 208Pb 2 Hence, the relative atomic mass of an element is the weighted average of all the isotopic masses in a sample of the element. 3 The relative atomic mass (Ar) of an element is the weighted average mass of one atom of the element relative to 1 12 times the mass of a C-12 atom. Relative atomic mass = Average mass of one atom of the element 1 12  the mass of one C-12 atom or = Average mass of one atom of the element The mass of one C-12 atom  12 Example 1.5 Chlorine consists of two isotopes 35Cl and 37Cl with relative abundance of 75% and 25%. Calculate the relative atomic mass of chlorine. Solution Ar = (35  75) + (37  25) = 35.5 100 The nucleon number of the atoms can be used in place of relative isotopic mass. Definition of relative atomic mass Exam Tips Exam Tips There are atoms of chlorine with relative masses of 35 and 37. However, there are no atoms with a relative mass of 35.5. 35.5 refers to the average mass of the two isotopes, taking into consideration their relative abundance. 2014/P1/Q2

CHAPTER 1 16 Chemistry Term 1 STPM Relative Molecular Mass (Mr ) 1 The relative molecular mass of a molecular species (elements or compounds) is the mass of one molecule of the substance relative to 1 12 time the mass of one carbon-12 atom. Relative molecular mass = Mass of one molecule of the substance  the mass of one C-12 atom or = Mass of one molecule of the substance  12 The mass of one C-12 atom 1 12 2 The relative molecular mass is equal to the sum of all the relative atomic mass of all the atoms in one molecule of the substance. 3 For example, The Mr of chlorine (Cl2 ) = (35.5  2) = 71.0 The Mr of methane (CH4 ) = 12 + (4  1) = 16 The Mr of carbon dioxide (CO2 ) = 12 + (16  2) = 44 Relative Formula Mass 1 Ionic compounds do not exist as discrete molecules. They consist of an infinite array of opposite charge ions. 2 For ionic compounds, the term relative formula mass is used instead of relative molecular mass. However, these two terms are sometimes interchangeable. Relative formula mass = Mass of one formula unit of the compound 1 12  the mass of one C-12 atom or or = Mass of one formula unit of the compound  12 The mass of one C-12 atom 3 The relative formula mass of an ionic compound is the sum of the relative atomic mass of all the atoms in one formula unit of the compound. The mass of an ion is taken to be the same as the mass of its neutral atom. 4 For example, Relative formula mass of Na+Cl– = 23.0 + 35.5 = 58.5 Relative formula mass of CuSO4 = 63.5 + 32.1 + 4(16.0) = 159.6 Definition of relative molecular mass Definition of relative formula mass

CHAPTER 1 17 Chemistry Term 1 STPM Quick Check 1.2 1 Calculate the relative atomic mass of argon from the following data: Isotope 36Ar 38Ar 40Ar % composition 0.34 0.07 99.59 2 Carbon consists of two isotopes 12C and 13C with relative masses of 12.000 and 13.003 respectively. The relative atomic mass of graphite is 12.011. (a) How many times is one atom of 13C heavier than one 12C atom? (b) Calculate the relative abundance of the two isotopes of carbon. 3 Prior to 1961, a physical atomic mass scale was used based on the 16O scale, where one 16O atom has a mass of 16.0000 atomic mass unit. However, on the 12C scale, the relative mass of 16O is 15.9949. What would be the relative mass of 12C on the 16O scale? Mass Spectrometer 1 The mass spectrometer is an instrument used to determine the relative masses of atoms. 2 The mass spectrometer is used to: (a) determine the relative isotopic mass. (b) determine the relative abundance of the isotopes. (c) determine the relative atomic mass of elements. (d) determine the relative molecular mass of molecular substances. (e) determine the structural formulae of compounds. Determination of Relative Atomic Mass 1 A simplified diagram of the mass spectrometer is shown below: Vaporisation chamber Ionisation chamber + Heated filament – Pump Accelerator Magnetic field Detector Positive ions Amplifier Recorder 2008/P1/Q2 2017/P1/Q16(c) 2016/P1/Q18(b) 2018/P1/Q1, Q18(b) Exam Tips Exam Tips The pump is used to remove air particles and unionised particles which otherwise would hinder the movement of the positive ions. Also, the air molecules might get ionised and produce unwanted peaks in the mass spectrum. VIDEO Mass Spectrometer

CHAPTER 1 18 Chemistry Term 1 STPM 2 The mass spectrometer is made up of five important components: (a) Vaporisation (b) Ionisation (c) Acceleration (d) Deflection (e) Detection 3 The sample of element (if it is a solid or liquid) is first vaporised in the vaporising chamber. 4 The gaseous atoms are then bombarded with high energy electrons released from the surface of a heated cathode where they are ionised to form positive ions. M(g) high-energy electron → M+(g) + e [The voltage of the ionising chamber is adjusted so that only unipositive ions are produced]. 5 The positive ions, with different masses, are then accelerated by an electric field with high negative potentials across parallel plates. The positive ions are focused into a narrow beam when they emerge from the accelerator. 6 The positive ions enter a magnetic field where they are deflected according to their mass charge ratios. 7 Ions with lower mass charge ratios are deflected more than those with higher mass charge ratios. The ions with different mass are thus separated. Magnetic field Ion with low Ion with high mass charge mass charge 8 The ions are detected by the ion detector, and the data ( mass charge ratio and their relative abundance) are transferred to the computer where a mass spectrum is plotted. 9 In practice, the detector is kept in a fixed position, and the strength of the magnetic field is varied gradually so that ions with different mass charge arrive at the detector at different time. Production of positive ions Deflection of the positive ions

CHAPTER 1 19 Chemistry Term 1 STPM 10 A sample of a mass spectrum is shown below: Relative abundance m1 a b c m2 m3 Mass Charge Since the charge on the ions is +1, the mass charge ratio is equal to the relative mass of the isotope. 11 The relative atomic mass of the element is calculated as follows: Ar = (m1  a) + (m2  b) + (m3  c) (a + b + c) Example 1.6 The mass spectrum of magnesium is as follows: 79% m/e 10% 11% 24 25 26 (a) Write the symbols for the isotopes of magnesium. (b) Calculate the relative atomic mass of magnesium. Solution (a) The proton number of magnesium is 12. ∴ The isotopes are: 24 12Mg; 25 12Mg; 26 12Mg (b) Ar = (24  79) + (25  10) + (26  11) 100 = 24.32 Calculating the relative atomic mass from the mass spectrum.

CHAPTER 1 20 Chemistry Term 1 STPM Quick Check 1.3 1 The mass spectrum (not to scale) of an element E is shown below. 6.3 112 25.0 65.8 108 109 111 1.6 Calculate the relative atomic mass of E. 2 A monatomic element (proton number 82) is analysed in the mass spectrometer. The mass spectrum (not to scale) is shown below. 103 206 207 208 (a) How many naturally occurring isotopes does the element contain? (b) Write the symbol for the isotopes of the element. (c) Account for the line with mass of 103. 3 Neon has three isotopes, 20Ne, 21Ne and 22Ne in the ratio of 9.1 : 0.02 : 0.88. (a) Sketch the mass spectrum of neon. (b) Calculate the relative atomic mass of neon. (c) How many times is one atom of neon heavier than one atom of carbon-12? Fragmentation of the molecule Mass Spectra of Molecular Species 1 The mass spectra of molecular species are more complicated than those of elements because they contain more than one atom in their structures. 2 When a molecule is bombarded with high energy electrons in the ionising chamber, the molecule might break to form fragments. 3 For example, when a sample of water is analysed in the mass spectrometer, the following will occur: H2 O(g) + high-energy electron → [H2 O]+(g) + 2e The [H2 O]+ is called the molecular ion or parent ion. 4 However, the H2 O molecule can also break to form fragmented ions. H2 O(g) + e* → H+(g) + OH+(g) + 3e H2 O(g) + e* → 2H+(g) + O+(g) + 4e 2010/P1/Q1 2015/P1/Q2

CHAPTER 1 21 Chemistry Term 1 STPM 5 The mass spectrum of water would consist of two sets of lines, one from the molecular ion and the other from the fragmented ions as shown below: Relative abundance Mass spectrum of water Mass to charge ratio (m/e) 1 16 17 18 Molecular ion The ions responsible for the peaks are: Mass Ion 1 H+ 16 O+ 17 OH+ 18 H2 O+ 6 The relative molecular mass of a molecular species is given by the peak or line with the highest mass or m/e ratio (not necessarily the peak with the highest abundance). 7 For example, the mass spectrum of a compound is shown below: Relative intensity (m/e) 10 15 20 25 30 35 40 45 100 80 60 40 20 0 The base line Molecular ion The relative molecular mass of the compound is 46. The other lines are caused by fragment ions. 8 For organic compounds, sometimes a very small peak with mass equal to (Mr + 1) unit is also found in the spectrum. This is known as the (M + 1) peak. It is caused by the presence of a carbon-13 isotope in the molecule. [Carbon consists of two isotopes 12C : 13C = 99 : 1] H—O—H → H+ + [O—H]+ + 2e Info Chem The line with the highest abundance is called the 'base line'.

CHAPTER 1 22 Chemistry Term 1 STPM Example 1.7 The mass spectrum of methane, CH4 is as follows: 1 1 141312 5 16 m/e Give the formulae of the ions responsible for the peaks. Solution m/e value Ion 1 H+ 12 C+ 13 CH+ 14 CH2 + 15 CH3 + 16 CH4 + Example 1.8 The mass spectrum of a sample of chlorine gas is shown below: Intensity (%) 30 100 90 80 70 60 50 40 30 20 10 0 35 40 45 50 55 60 65 70 75 80 (a) Give the ions responsible for the lines. (b) Calculate the relative atomic mass of chlorine. Solution (a) m/e Ion 35 35Cl+ 37 37Cl+ 70 35Cl—35Cl+ 72 35Cl—37Cl+ 74 37Cl—37Cl+

CHAPTER 1 23 Chemistry Term 1 STPM Using the mass spectrum to determine the structural formula of a compound (b) The relative abundance of 35Cl and 37Cl is 75% : 25% ∴ Ar = (35  75) + (37  25) 100 = 35.5 Example 1.9 The mass spectrum (not to scale) of an alcohol with the molecular formula of C3 H8 O is shown below. 15 17 4543 60 61 (a) Draw the possible structures of the alcohol. (b) Give the ions that are responsible for the peaks with mass 15, 17, 43, 45 and 60. (c) Suggest a structural formula of the alcohol. (d) What species is responsible for the peak at mass 61? Solution (a) The two isomers are: CH3 CH2 CH2 OH 1-propanol CH3 —CH—CH3 2-propanol | OH (b) Mass 15 17 43 45 60 Ion CH3 + OH+ CH3 CH2 CH2 + or CH3 C+H | CH3 CH2 CH2 OH+ or CH3 C+H | OH CH3 CH2 CH2 OH+ or CH3 CHCH3 | OH+ (c) Since there is no peak corresponding to the following ‘signature’ fragments: CH3 CH2 CH2 OH → CH3 CH2 + + CH2 OH+ 29 31 The alcohol cannot be CH3 CH2 CH2 OH. The structure of the alcohol is: CH3 —CH—CH3 | OH Exam Tips Exam Tips The line corresponding to m e = 29 usually indicates the ethyl radical, C2 H5 .

CHAPTER 1 24 Chemistry Term 1 STPM Definition of ‘mole’ (d) It is caused by the presence of a carbon-13 isotope in the molecule of the alcohol. 13CH3 —CH—CH3 + | OH Quick Check 1.4 1 Bromine has two isotopes, 79Br and 81Br in the ratio of 1 : 1. Sketch the mass spectrum showing the peaks formed by the molecular ions. Indicate the relative abundance of the peaks. 2 An element G consists of two isotopes, 34G and 36G. The mass spectrum of a sample of G shows the following major peaks. Mass Relative abundance 68 9 70 x 72 4 (a) Determine the relative abundance of the two isotopes. (b) Calculate the value of x. 3 A mixture of nitrogen (14N2 ) and oxygen (16O2 ) is analysed in the mass spectrometer. Sketch the mass spectrum for the mixture. 4 Sketch the mass spectrum of NO2 when it is analysed in the mass spectrum. Nitrogen consists of 14N and oxygen as 16O and 18O. 1.3 The Mole and the Avogadro's Constant 1 The number of atoms in exactly 12 g of carbon-12 is 6.02  1023. 2 This number is called the Avogadro’s number or Avogadro’s constant. The symbols are NA or L. 3 Based on the carbon-12 scale, one mole of any substance is that amount of substance which contains the same number of particles (atoms, molecules, ions or electrons) as the number of atoms in 12 grams of carbon-12. For example, 1 mol of helium = 6.02  1023 He atoms 1 mol of carbon dioxide = 6.02  1023 CO2 molecules 1 mol of electrons = 6.02  1023 electrons Info Chem An analogy to 'mole' is 'dozen' which represent 12. 2013/P1/Q1, Q16 2015/P1/Q3 2016/P1/Q1, Q18(a) 2017/P1/Q16(d)(e)

CHAPTER 1 25 Chemistry Term 1 STPM 4 One mole of substance not only represents a specific number, it also represents a specific mass of the substance. The amount of substance that contains the Avogadro’s number of particles is the relative mass of that substance expressed in grams. 5 For example, 1 mol of 24Mg = 24 g of 24Mg = 6.02  1023 atoms. 1 mol of 40Ar = 40 g of 40Ar = 6.02  1023 atoms. 1mol of 16O2 = 32 g of 16O2 = 6.02  1023 molecules. 1 mol of NaOH = 40 g of NaOH = 6.02  1023 ‘NaOH’ units. 1 mol of CO2 = 44 g of CO2 = 6.02  1023 molecules. 6 However, 1 mol of 16O2 contains 2  (6.02  1023) atoms. 1 mol of NaOH contains 2  (6.02  1023) ions. 1 mol of CO2 contains 3  (6.02  1023) atoms. 1 mol of NaCl contains 2  (6.02  1023) ions. Example 1.10 Calculate the mass of the following substances: (a) 0.45 mol of aluminium (b) 2.65  1024 atoms of oxygen (c) 2.30  1022 molecules of carbon dioxide Solution (a) 1 mol of Al = 27.0 g 0.45 mol of Al = 0.45  27.0 = 12.15 g (b) 1 mol of oxygen atoms = 16.0 g 6.02  1023 oxygen atoms = 16.0 g ∴ 2.65  1024 oxygen atoms = 2.65  1024  16.0 g 6.02  1023 = 70.43 g (c) 1 mol of CO2 = 44.0 g 6.02  1023 CO2 molecules = 44.0 g ∴ 2.30  1022 CO2 molecules = 2.30  1022 6.02  1023  44.0 g = 1.68 g The mass of one mole of a substance is the relative mass of the substance expressed in gram.

CHAPTER 1 26 Chemistry Term 1 STPM The most common solvent is water. Example 1.11 Ethanoic acid, C2 H4 O2 , is the active ingredient in vinegar. (a) What is the mass (in g) of 0.85 mol of ethanoic acid? (b) How many moles of ethanoic acid are in 250 cm3 of vinegar that contains 12.5% by mass of ethanoic acid? [Density of vinegar = 1.18 g cm–3] Solution (a) Mass of 1 mol of ethanoic acid = 60 g Mass of 0.85 mol = 0.85  60 = 51.0 g (b) Mass of 250 cm3 of vinegar = 250  1.18 g = 295.0 g Mass of ethanoic acid = 12.5 100  295.0 = 36.88 g Number of moles of ethanoic acid = 36.88 60 = 0.615 mol Quick Check 1.5 1 Calculate the mass of lead that contains (a) 3.45 mol of lead, (b) 8.4  1022 atoms, (c) 9.8  1024 atoms. 2 Calculate (a) the mass of potassium that has the same number of atoms as 24.0 g of calcium, (b) the mass of ethanol that contains the same number of atoms as in 34 g of aluminium oxide, (c) the mass of sodium chloride that has 6 times the number of ions as the number of ions in 0.45 g of aluminium sulphate. 3 Calculate (a) the number of moles of H2 O molecule in 1.0 m3 of water, (b) the number of H2 O molecules in 1 dm3 of water. [Density of water = 1.0 g cm–3] Mole and Concentration of Solutions 1 A solution is a homogeneous mixture of two or more substances. 2 The substance present in smaller quantity is called the solute, and the substance present in larger quantity is called the solvent. 3 For example, in aqueous potassium bromide, potassium bromide is the solute and water is the solvent. 4 Similarly, when solid iodine dissolves in liquid cyclohexane, iodine is the solute and cyclohexane is the solvent. 5 Concentration is the amount of solute present in a fixed quantity of the solution.

CHAPTER 1 27 Chemistry Term 1 STPM 6 Concentration is expressed either in terms of g dm–3 (which is the mass of the solute in 1 dm3 or 1000 cm3 of the solution), or mol dm–3 (which is the number of moles of the solute in 1 dm3 of the solution). 7 Concentration in mol dm–3 is called molar concentration or molarity, and has a symbol of M (pronounced as molar). 8 The relationship between molarity and concentration is: Molarity (mol dm–3) = Concentration (g dm–3) Molar mass of solute (g mol–1) 9 A concentrated solution is one which contains a higher amount of solute per dm3 of solution whereas a dilute solution is one that contains lesser amount of solute per dm3 of solution. 10 The number of moles of solute in a given volume of solution is given by: No. of moles = Volume (in cm3 ) 1000  Molarity (in mol dm–3) or = MV 1000 Example 1.12 What is the concentration in (a) g dm–3 and (b) mol dm–3, of a solution containing 2.46 g of sodium chloride in 250 cm3 of water? Solution (a) Concentration in g dm–3 = 2.46  1000 250 = 9.84 g dm–3 (b) Molar concentration = 9.84 58.5 mol dm–3 = 0.168 mol dm–3 [Molar mass of NaCl = 58.5 g mol–1] Example 1.13 (a) 25.0 cm3 of aqueous 0.10 M NaOH is pipetted into a conical flask. (i) Calculate the number of moles of NaOH in the 25.0 cm3 sample. (ii) What is the molar concentration of NaOH in the 25.0 cm3 sample? Units for concentration

CHAPTER 1 28 Chemistry Term 1 STPM (b) 100.0 cm3 of water is then added to the NaOH solution in the conical flask. (i) What is the concentration of the NaOH in the conical flask? (ii) How many moles of NaOH is present in the diluted NaOH solution? Solution (a) (i) No. of moles = MV 1000 = 0.10  25.0 1000 = 2.50  10–3 mol (ii) The molar concentration is still 0.10 mol dm–3. This can be proved as follows: No. of moles of NaOH in 25.0 cm3 = 2.50  10–3 mol No. of moles of NaOH in 1 dm3 = 1000 25.0  2.50  10–3 = 0.10 mol ∴ Molar concentration = 0.10 mol dm–3 (b) (i) Molar concentration = 1000 25.0 + 100  2.50  10–3 mol dm–3 = 0.020 mol dm–3 (ii) The number of moles of NaOH is still 2.50  10–3 mol. Quick Check 1.6 1 2.50 g of an impure sample of calcium carbonate requires 27.6 cm3 of 1.45 M HCl for complete reaction. (a) Write a balanced equation for the reaction. (b) Calculate the mass of pure calcium carbonate in the sample. (c) What is the percentage purity of the sample of calcium carbonate. 2 25.0 cm3 of a 0.063 M solution containing a base M(OH)n requires 23.60 cm3 of 0.10 M sulphuric acid for complete reaction. Calculate the value of n. 3 Calculate the molar concentration of HCl when 250 cm3 of 2.50 M HCl is added to 400 cm3 of 1.60 M HCl. Mole and Gases 1 1 mole of gas is the amount of gas that contains the same number of gas particles as the number of atoms in 12 g of carbon-12. 2 In other words, 1 mole of gas is the amount of gas that contains 6.02  1023 gas particles. 3 For example, (a) 1 mole of He contains 6.02  1023 He atoms, (b) 1 mole of hydrogen contains 6.02  1023 H2 molecules, (c) 1 mole of CO2 contains 6.02  1023 CO2 molecules. 2010/P2/Q5(a) 2011/P1/Q2 2013/P1/Q19(a) 2017/P1/Q1 2018/P1/Q7

CHAPTER 1 29 Chemistry Term 1 STPM 4 The mass of 1 mole of gas is the same as the relative atomic mass or relative molecular mass of the gas expressed in grams. For example, (a) 1 mole of He weighs 4.0 g, (b) 1 mole of oxygen gas weighs 32.0 g, (c) 1 mole of carbon dioxide weighs 44.0 g. 5 The volume occupied by a gas varies according to conditions. The volume occupied by a gas depends on: (a) the amount of gas (b) temperature (c) pressure 6 Hence, when stating the volume occupied by a certain amount of gas, the temperature and pressure at which the volume was measured must be stated . 7 Avogadro’s Law states that equal volumes of all gases, under the same conditions, contain the same number of moles. 8 Hence, equal number of moles of any gas, under the same conditions, would occupy the same volume. It does not depend on the nature of the gas. 9 For example, at temperature T and a pressure of P, 0.25 moles of N2 occupies a volume of 4.5 dm3 . This means that, at the same temperature and pressure, 0.25 mol of O2 , CO2 , H2 or SO2 would also occupy a volume of 4.5 dm3 . 10 Experimentally, it was found that 1 mole of any gas would occupy a volume of 22.4 dm3 at 273 K (0 °C) and 101 kPa pressure (1 atm). 11 This volume (22.4 dm3 ) is known as the molar volume of a gas. The temperature of 273 K and 101 kPa is known as the standard temperature and pressure (s.t.p.). Example 1.14 A sample of carbon dioxide contains 0.011 g of the gas. Calculate (a) the number of CO2 molecules in the sample, (b) the volume occupied by the sample at s.t.p. Solution (a) Molar mass of CO2 = 44 g mol–1 No. of CO2 molecules = 0.011 44  (6.02  1023) = 1.51  1020 molecules (b) No. of moles of CO2 = 0.011 44 = 2.50  10–4 mol ∴ Volume of CO2 at s.t.p. = (2.50  10–4)  22.4 dm3 = 0.0056 dm3 = 5.60 cm3 Molar volume of a gas

CHAPTER 1 30 Chemistry Term 1 STPM 1 Which of the following particles would be deflected less than the 3 Li+ ion in a mass spectrometer? A n B e C 1 H+ D 4 He+ 2 35Cl and 37Cl are two isotopes of chlorine. The relative atomic mass of chlorine is 35.5. Which statement is true? A 37Cl is more abundant than 35Cl B 37Cl is radioactive, 35Cl is not C The relative molecular mass of chlorine is 71.0 D 35Cl is more reactive than 37Cl 3 Consider the following conversion: 37Cl + e → 37Cl– Which of the following statements is correct regarding the above? Objective Questions SUMMARY SUMMARY 1 The three fundamental sub-atomic particles are proton, electron and neutron. 2 Proton number is the number of protons in the nucleus of an atom. Nucleon number is the total number of protons and neutrons in the nucleus of an atom. 3 Isotopes are atoms of the same element but with different number of neutrons. 4 Carbon-12 was chosen as the standard for the measurement of relative atomic mass. 5 On the carbon-12 scale, the mass of one atom of C-12 is 12.0000 a.m.u. 6 Relative isotopic mass is the mass of one atom of the isotope relative to 1 12 time the mass of one carbon-12 atom. 7 Relative atomic mass of an element is the average mass of one atom of the element relative to 1 12 time the mass of one carbon-12 atom. 8 Relative molecular mass is the mass of one molecule of the substance relative to 1 12 time the mass of one carbon-12 atom. 9 One mole of any substance is the amount that contains the same number of atoms in exactly 12 g of carbon-12. 10 The number of atoms in exactly 12 g of carbon-12 is 6.02  1023. This number is called the Avogadro’s number or Avogadro’s constant. STPM PRACTICE 1 A The number of protons decreases by one. B The number of neutrons increases by one. C The number of electrons increases by one. D The number of protons decreases by one and the number of electrons increases by one. 4 What is the difference between the relative molecular mass (Mr ) of CO2 and the molar mass of CO2 ? A The Mr has a unit of gram while the molar mass has a unit of g mol–1. B The Mr of CO2 is 44 while the molar mass of CO2 is 44 g mol–1. C The Mr refers to the mass of 1 mol of CO2 while the molar mass is the mass of 1 CO2 molecule. D The Mr is the mass of 1 mol of CO2 divided by the mass of 1 12 time the mass of 1 atom of C-12 while the molar mass of CO2 is the mass of 1 molecule of CO2 multiply by the Avogadro constant.

CHAPTER 1 31 Chemistry Term 1 STPM 5 Each molecule of haemoglobin contains 4 iron atoms. If the percentage by mass of iron in haemoglobin is 0.33%, what is the relative molecular mass of haemoglobin? A 3.4  102 C 1.7  105 B 6.8  104 D 6.8  106 6 10 g of an impure sample of calcium carbonate needs 25.0 cm3 of 1.0 mol dm–3 sulphuric acid for complete reaction. What is the percentage purity of the sample of calcium carbonate? [Assume that the impurities do not react with sulphuric acid] A 25% C 65% B 50% D 80% 7 Which of the following statements is correct regarding isotopes? A They are elements in the same group in the periodic table. B They are atoms with the same number of protons and neutrons. C They occupy the same position in the periodic table. D They have the same mass. 8 An oxide of M has a formula of M2 O. 12.8 g of the oxide on reduction produces 11.2 g of M. Calculate the relative atomic mass of M. A 28 C 84 B 56 D 112 9 The mass spectrum of an element is shown below. 24 48 Which statement is true about the element? A The relative atomic mass of the element is 36 B The element has two isotopes of relative mass 24 and 48 C The line at m/e 24 is caused by a dipositive ion D The element is radioactive 10 Which of the following is the least deflected in a mass spectrometer? A 2 H+ C 7 Li+ B 4 He2+ D 7 Li2+ 11 Carbon (relative atomic mass = 12.001) has two isotopes: 12C and 13C. Which of the following is the approximate composition of the two isotopes? % of 12C % of 13C A 50 50 B 60 40 C 80 20 D 99 1 12 Air bags contain a mixture of sodium azide and potassium nitrate. On collision, nitrogen gas is produced via the following reactions: 2NaN3 (s) → 2Na(s) + 3N2 (g) 10Na(s) + 2KNO3 (s) → K2 O(s) + 5Na2 O(s) + N2 How many moles of nitrogen gas are produced from 0.50 mol of sodium azide? A 0.75 B 0.80 C 1.6 D 2.0 13 23Na is non-radioactive whereas 22Na is radioactive with a half-life of about 2.6 years. Which statement is not correct? A Both isotopes have the same number of electrons B Both isotopes have the same number of protons C Both isotopes have the same number of charged particles D It takes 2.6 years for 22Na to be completely converted to 23Na 14 In which species is the number of neutrons and number of electrons the same? A 9 Be B 23Na+ C 19F D 18O2– 15 A 0.15 g of a dibasic acid requires 32.0 cm3 of 0.10 mol dm–3 sodium hydroxide for complete reaction. What is the relative molecular mass of the acid? A 35.6 B 62.5 C 93.8 D 104.2

CHAPTER 1 32 Chemistry Term 1 STPM 16 20 cm3 of ethyne, C2 H2 , is mixed with 70 cm3 of oxygen and the mixture sparked continuously. The resultant gases are cooled to room temperature. Which of the following is not true about the combustion? A The total volume of the mixture after cooling is 50 cm3 B 20 cm3 of oxygen remains C 20 cm3 of CO2 is produced D The final mixture contains CO2 and O2 only 17 Which of the following contains the largest number of molecules? A 1 g of He C 1 g of NH3 B 1 g of H2 D 1 g of C6 H12O6 18 A mixture of 25 cm3 of ethane and 25 cm3 of butane was completely burned in excess oxygen. What is the volume of carbon dioxide produced? [All measurements are done under identical conditions.] A 50 cm3 C 120 cm3 B 100 cm3 D 150 cm3 19 What is the mass of 6.02  1024 molecules of nitrogen? A 14 g C 140 g B 28 g D 280 g 20 Which of the following peaks would not be found in the mass spectrum of propanone, CH3 — C — CH3 ? || O A 15 C 28 B 19 D 58 21 The chlorine dioxide molecule, ClO2 , consists of 16O, 35Cl and 37Cl isotopes. How many molecular peaks would there be in the mass spectrum of the ClO2 molecule? A 1 C 3 B 2 D 4 22 The mass of one carbon-12 atom is m g. What is the mass of one mole of 23Na? A (6.02  1023)  m g B 23 12  (6.02  1023)  m g C 23 12  m g D 23  (6.02  1023)  m g 23 When 70 cm3 of water is added to 50 cm3 of an aqueous solution of sodium sulphate, Na2 SO4 , the resulting solution has a concentration of 0.50 mol dm–3. What is the original concentration of the sodium sulphate solution? A 0.35 mol dm–3 B 0.50 mol dm–3 C 1.20 mol dm–3 D 2.40 mol dm–3 24 Which of the following is correct? A A mass spectrum provides the evidence of the existence of electrons. B A mass spectrum provides the evidence of the existence of isotopes. C A mass spectrum provides the evidence of the existence of allotropes. D A mass spectrum provides the evidence of the existence of protons. 25 The percentage of nickel in a protein molecule, X, is 0.360%. If a molecule of X consists of three nickel(II) ions, what is the relative molecular mass of X? [Relative atomic mass of Ni = 58.7] A 163 C 163  102 B 489 D 489  102 26 Boron has two isotopes with relative masses of 10 and 11. When 1.00 g of boron reacts with hydrogen, 1.278 g of B2 H6 is produced. What is the relative abundance of the lighter isotope to that of the heavier isotope? A 1 : 1 C 3 : 2 B 1 : 4 D 2 : 3 27 Gallium (Ga) has two isotopes, 69Ga and 71Ga. If the relative atomic mass of Ga is 69.7, what is the percentage abundance of 71Ga? A 65.0 C 45.5 B 54.5 D 35.0 28 Phosphorus trichloride, PCl3 , is formed from 31P, 35Cl and 37Cl. The percentage abundance of 35Cl is 75%. Which statement about the mass spectrum of PCl3 is not correct? A The maximum m/e value is at 142. B The number of peaks for PCl3 + is 4. C The relative abundance of the P35Cl3 + peak to the P37Cl3 + peak is 3 : 1. D The base peak corresponds to the P37Cl3 + peak.

CHAPTER 1 33 Chemistry Term 1 STPM 29 Which statement(s) is/are true of the nuclides of 12 6C and 13 6C? I They have different densities. II 12C is more reactive than 13C. III The mass spectrum shows only two peaks. A I C I and III B II D II and III 30 Urea, (H2 N)2 CO, reacts with water according to the equation: (H2 N)2 CO + H2 O → CO2 (g) + 2NH3 (g) How many moles of gas molecules are released when 0.56 g of urea reacts completely with water? [The molar mass of urea = 60 g mol–1] A 1.87  10–2 C 1.68 B 2.80  10–2 D 3.36 31 Consider the reaction below: C2 H6 + Cl2 → C2 H5 Cl + HCl 4C2 H5 Cl + 4Na + Pb → (C2 H5 )4 Pb + 4NaCl How many moles of HCl will be released for each mole of (C2 H5 )4 Pb produced? A 1 2 C 3 B 1 D 4 32 An element Y with relative atomic mass of 35.5 consists of isotope 35X and 37X. What is the relative abundance of the heavier isotope? A 0.25 C 0.75 B 0.5 D 0.90 Structured and Essay Questions 1 Define the following terms: (a) Proton number (b) Nucleon number (c) Isotope 2 In what ways are the isotopes of an element similar to one another and in what ways do they differ? 3 Compound X has the following composition by mass: N = 30.4%, O = 69.9% The mass spectrum of X (where only certain peaks are shown) is shown below: 16 46 92 94 m/e (a) Determine the empirical formula of X. (b) Determine the molecular formula of X. (c) Identify the ions responsible for the peaks at m/e = 46 and m/e = 94 and account for their relative intensity. 4 (a) Define (i) relative isotopic mass and (ii) relative atomic mass (b) With the aid of a labelled diagram, describe how the relative atomic mass of an element can be determined using the mass spectrometer.

CHAPTER 1 34 Chemistry Term 1 STPM (c) Naturally occurring silicon consists of three isotopes 28Si, 29Si and 30Si in the relative abundance of 29.8 : 1.5 : 1.0. (i) Use a graph paper to sketch the mass spectrum of silicon. (ii) Calculate the relative atomic mass of silicon. (iii) What is the mass of one silicon atom relative to the mass of one 12C atom? 5 Beams of particles from different sources are passed between two plates which are electrically charged. The 1 H+ ion and an α-particle are deflected as shown. 4° 2° A B (I) (II) (a) State the polarity (+ or –) of plate A. (b) Identify particle (I). Give a reason for your answer. (c) Calculate the angles of deflection when the following particles are passed through the same electric field (i) 2 H+, and (ii) 1 0 n 6 Chlorine has two isotopes 35Cl and 37Cl in the ratio of 3 : 1. Bromine has two isotopes 79Br and 81Br in the ratio of 1 : 1. Assuming that carbon and hydrogen consist of 12C and 1 H atoms only, determine the possible mass/ charge ratio for the molecular peaks in the mass spectrum of the compound C2 H3 Br2 Cl and hence calculate the relative heights of these peaks. 7 (a) What do you understand by radioactive isotope? (b) Consider the following radioactive disintegration of uranium- 235. 235 92U → 2a W →b X →A 211 82Pb →B 207 82Y (i) Identify the isotopes W and X by giving their proton numbers and nucleon numbers. (ii) Identify the particles A and B. (iii) What is the relationship between Pb-211 and Y-207? 8 (a) Define mole. (b) The active ingredient in aspirin is acetylsalicylic acid. O O C CH3 COOH (i) Calculate the mass of one mole acetylsalicylic acid. (ii) How many molecules of acetylsalicylic acid are present in 4.8  10–3 g of acetylsalicylic acid? (iii) A sample of aspirin contains 89.5% by mass of acetylsalicylic acid. Calculate the mass of acetylsalicylic acid in 2.5 g of the sample. 9 (a) Define an isotope in terms of its sub-atomic particles. (b) A sample of hydrogen chloride is analysed in a mass spectrometer and the mass spectrum is obtained as shown.

CHAPTER 1 35 Chemistry Term 1 STPM 35 36 37 38 m/e (i) Identify the ions responsible for the peaks with m/e 35, 36, 37 and 38. (ii) There is another peak of relatively low intensity at m/e 18. What species is responsible for this peak? Account for its low intensity. 10 Methyl ethanoate undergoes hydrolysis to produce methanol and ethanoic acid. CH3 COOCH3 + H2 O → CH3 COOH + CH3 OH In an experiment, the isotopically labelled ester is completely hydrolysed in dilute sulphuric acid. CH3 — C 18O CH3 || O After purification, the products of hydrolysis is then analysed in a mass spectrometer. Three of the major peaks are shown below (not to scale). 18 34 60 (a) Identify the species responsible for the three peaks. (b) Deduce which of the two C— O bonds in the ester are broken during hydrolysis. 11 Chlorine consists of two isotopes, 35Cl and 37Cl, in the ratio of 3 : 1. Phosphorus is monoisotopic, 31P. A sample of phosphorous trichloride, PCl3 , is analysed in the mass spectrometer. Apart from lines due to atomic ions, the mass spectrum of the chloride contains 9 lines arranged in 3 groups. (a) Deduce the m/e values for the 9 lines. (b) Identify the ions that are responsible for the lines. (c) Calculate the relative abundance of the various lines within each group. 12 The relative atomic mass of element X is 35.5. (a) Why is the relative atomic mass of X not a whole number? (b) X has two naturally occurring isotopes with nucleon numbers of 35 and 37. Calculate the percentage abundance of the 35X isotope. (c) If X is monoatomic, sketch the mass spectrum of X. 13 Y (an alkane) is 9.5 times heavier than a carbon-12 atom. (a) Identify Y. (b) Calculate the volume of air (measured at s.t.p.) needed for the complete combustion of 0.15 dm3 of X. [Air contains 20% by volume of oxygen.]

CHAPTER ELECTRONIC STRUCTURE 2 OF ATOMS Concept Map Students should be able to: Electronic energy levels of atomic hydrogen • explain the formation of the emission line spectrum of atomic hydrogen in the Lyman and Balmer series using Bohr’s Atomic Model. Atomic orbitals: s, p and d • deduce the number and relative energies of the s, p and d orbitals for the principal quantum numbers 1, 2 and 3, including the 4s orbitals; • describe the shape of the s and p orbitals. Electronic configuration • predict the electronic configuration of atoms and ions given the proton number (and charge); • define and apply Aufbau principle, Hund’s rule and Pauli exclusion principle. Classification of elements into s, p, d and f blocks in the Periodic Table • identify the position of the elements in the Periodic Table as (a) block s, with valence shell configurations s1 and s2, (b) block p, with valence shell configurations from s2p1 to s2p6, (c) block d, with valence shell configurations from d1s2 to d10s2; • identify the position of elements in block f of the Periodic Table. Learning earning Outcomes Electronic Structure of Atoms Electromagnetic Radiation Electronic Configuration Atomic Orbitals Line Spectrum of Hydrogen Bohr’s Model of Hydrogen Atom Ionisation Energy of Hydrogen • Ionisation energies and electronic shells Electronic Configuration of Ions Modern Atomic Model Filling of Electrons

37 Chemistry Term 1 STPM CHAPTER 2 2.1 Electronic Energy Levels of Atomic Hydrogen Electromagnetic Radiations 1 Electromagnetic radiations consist of radiations/waves such as the gamma radiations, X-rays, ultraviolet radiations, infrared radiations, visible light, microwaves and many others. 2 Each type of radiation is characterised by their wavelengths or frequencies. 3 The wavelength (λ) is defined as the distance between two adjacent peaks or troughs as shown below. It has the unit of meter (m) or nanometer (nm). [1 nm = 1 × 10–9 m] Peak Trough Wavelength (λ) 4 Frequency (f) is the number of waves that passes through a certain point per second. It has a unit of Hertz (Hz) or per second (s–1). The diagram below shows two waves, one of high frequency and one of low frequency. High frequency Low frequency 5 The frequency of a wave is inversely proportional to its wavelength. f ∝ 1 λ Another symbol for frequency is ‘v’.

38 Chemistry Term 1 STPM CHAPTER 2 6 The relationship between the wave length (λ) and the frequency (f) of the electromagnetic radiation is given by: λ f = C [where C = velocity of light = 3.0  108 m s–1] 7 The diagram below shows the various types of electromagnetic radiations. 8 In 1900, Max Planck put forward his quantum theory which states that energy in the form of electromagnetic radiation can be emitted or absorbed in discrete amount (or packets) called quanta (singular: quantum). 9 The magnitude of a quantum depends on the frequency (or wavelength) of the radiation. The energy (in J) associated with a quantum of energy (or a photon) is given by the Planck’s equation: E = hf Where h = Planck’s constant = 6.63  10–34 J s or = 3.99  10–13 kJ mol–1 s Quantum theory Concept of quanta Frequency increasing Wavelength increasing Visible light 110 10–2 10–4 10–6 3 fi 108 3 fi 1010 3 fi 1012 3 fi 1014 10–8 10–10 10–12 10–14 3 fi 1016 3 fi 1018 3 fi 1020 3 fi 1022 Meter 700 400 nm 1 Second X-rays Ultraviolet radiation Infrared radiation Microwaves TV waves Radio waves y-rays Cosmic rays Red Orange Yellow Green Blue Violet Quick Check 2.1 1 A quantum of a particular electromagnetic radiation has energy of 10–18 J. Calculate (a) the energy in kJ mol–1, (b) the frequency, and (c) the wavelength of the radiation. 2 A gamma radiation has a frequency of 1  1024 Hz. Calculate (a) its wavelength, (b) its energy in J, and (c) its energy in kJ mol–1.

39 Chemistry Term 1 STPM CHAPTER 2 The Line Spectrum of Hydrogen 1 When sunlight passes through a prism, the colours of the rainbow are observed. This type of spectrum is known as a continuous spectrum because it contains all the possible wavelengths (or frequencies) of the visible light. Continuous spectrum contains all the wavelengths that make up visible light. 400 450 Wavelength 500 550 600 650 700 750 Red ~ 700 nm ~ 400 nm Violet Sunlight The Continuous Spectrum 2 However, when electricity is passed through a discharge tube filled with hydrogen gas at low pressure, a pink light is emitted. When this light passes through a prism or is observed through a spectroscope a line spectrum is obtained. Exam Tips Exam Tips The spectrum consists of a few coloured lines, each corresponding to a particular wavelength or frequency. (Red: 656.2 nm, blue-green: 486.1 nm, blue-violet: 434.0 nm and violet: 410.0 nm) Gas discharge tube containing hydrogen Slits Prism Violet Blueviolet Bluegreen Red 410.0 nm434.0 nm486.1 nm656.2 nm 2013/P1/Q18(a) 2016/P1/Q3, Q18(c) 2015/P2/Q4 2017/P1/Q2 2018/P1/Q2, Q19(a) INFO Line Spectrum of Atomic Hydrogen

40 Chemistry Term 1 STPM CHAPTER 2 3 Using more sophisticated instruments, lines in other parts of the spectrum were also detected. These lines are in the ultraviolet and the infrared regions. Part of the line spectrum of hydrogen is shown below: Frequency Wavelength Ultraviolet (Lyman series) Visible (Balmer series) Infrared (Paschen series) 4 Each set of lines is known as a series and is named after its discoverer. Region Name of series Ultraviolet Lyman Visible Balmer Infrared Paschen Infrared Bracket Infrared Pfund 5 In each series, the intervals between the lines get smaller towards the high frequency (low wavelength) end of the spectrum and finally merge to form a continuous spectrum which ends at a maximum frequency value. Such spectrum is called a converging spectrum. 6 The line spectra of other elements can also be obtained in the same way. No two elements have identical line spectra. Hence, it is possible to identify the element from its line spectrum. 7 Mathematically, the wavelength of the lines in the emission spectrum of hydrogen can be calculated using the Rydberg’s equation: 1 λ = RH 1 n1 2 – 1 n2 2  where RH = Rydberg’s constant = 1.097  107 m–1 n1 and n2 = Integers n2 can take any values from (n1 + 1), (n1 + 2) .....∞ 8 The values of n1 and n2 for each series are given in the table below: Series Region n1 n2 Lyman Ultraviolet 1 2, 3, 4 .....∞ Balmer Visible 2 3, 4, 5, 6 .....∞ Paschen Infrared 3 4, 5, 6, 7 .....∞ Bracket Infrared 4 5, 6, 7, 8 .....∞ The hydrogen line spectrum Converging spectrum. Exam Tips Exam Tips The Rydbergʼs equation can be used to calculate the wavelengths of the lines in the hydrogen spectrum only. Info Chem Characteristics of the series: • Each series consists of discrete lines with specific frequencies. • The interval between the lines gets smaller towards the high frequency end. • Each series ends with a line with maximum frequency.

41 Chemistry Term 1 STPM CHAPTER 2 9 The first line in the Lyman series corresponds to n1 = 1 and n2 = 2, while the second line corresponds to n1 = 1 and n2 = 3 and so on. 10 The same applies to the other series. Example 2.1 Calculate the wavelengths and frequencies of the following lines in the emission spectrum of hydrogen. (a) First line in the Balmer series. (b) Last line in the Balmer series. Solution (a) Using n1 = 2 and n2 = 3 1 λ = 1.097  107  1 22 – 1 32  1 λ = 1.534  106 m–1 ∴ λ = 6.52  10–7 m or = 652 nm f = C λ = 3.0 × 108 6.52 × 10–7 = 4.60 × 1014 Hz (b) Using n1 = 2 and n1 = ∞ 1 λ = 1.097 × 107  1 22 – 1 ∞2  1 λ = 2.74 × 106 m–1 ∴λ = 3.65  10–7 m or = 365 nm f = C λ = 3.0 × 108 3.65 × 10–7 = 8.22  1014 Hz Bohr's Model of the Hydrogen Atom 1 In order to explain the formation of the line spectrum of hydrogen, Niels Bohr, a Danish physicist, put forward his model of the hydrogen atom in 1913. [He received the Nobel Prize for physics in 1922 for his theory explaining the emission line spectrum of hydrogen]. 2 He postulated that electrons revolve round the nucleus in fixed circular paths of different radii called orbits, much like the planets revolving around the Sun. 3 The energy of the orbits is quantised. Each orbit is represented by an integer, n, which is called the principle quantum number. Concept of orbit: A fixed circular path 2009/P1/Q3 2014/P1/Q18(a) VIDEO Bohr's Model of the Hydrogen Atom