JPJC 2026 Mole & Stoichiometry Lecture Notes

2026 JC1 H2 Chemistry (9476) 1 Topic 2: The Mole Concept & Stoichiometry Jurong Pioneer Junior College 2026 JC1 H2 Chemistry (9476) Physical Chemistry Topic 2: The Mole Concept and Stoichiometry Learning Outcomes (LOs) / Assessment Objectives Candidates should be able to: (a) define the terms relative atomic, isotopic, molecular and formula mass; (b) define the term mole in terms of the Avogadro constant; (c) calculate the relative atomic mass of an element given the relative abundances of its isotopes; (d) define the terms empirical and molecular formula; (e) calculate empirical and molecular formulae using combustion data or composition by mass; (f) write and/or construct balanced equations; (g) perform calculations, including use of the mole concept, involving (i) reacting masses (from formulae and equations); (ii) volume of gases (e.g. in the burning of hydrocarbons); (iii) volumes and concentrations of solutions; [when performing calculations, candidates’ answers should reflect the number of significant figures given or asked for in the question] (h) deduce stoichiometric relationships from calculations such as those in (g); From Electrochemistry (i) describe and explain redox processes in terms of electron transfer and/or of changes in oxidation number (oxidation state); (j) construct redox equation using the relevant half–equations. Recommended References: 1. Chemistry for Advanced Level by P. Cann & P. Hughes 2. A-level Chemistry by E.N. Ramsden 3. Chemistry in Context by Graham Hill & John Holman

2026 JC1 H2 Chemistry (9476) 2 Topic 2: The Mole Concept & Stoichiometry Section Content LOs Page 1 Introduction 3 2 Relative Masses 2.1 The 12C scale 3 2.2 Definitions (a), (c) 4 3 The Mole Concept 3.1 The Mole and Avogadro Constant (b) 7 3.2 Molar Mass (g) 7 3.3 Molar Volume of Gas (g) 8 3.4 Concentration (g) 10 4 Stoichiometry 4.1 Chemical Equations (f) 13 4.2 Ionic Equations (f) 14 4.3 Limiting Reagents (g) 15 4.4 Percentage Yield (g) 16 5 Empirical and Molecular Formulae 5.1 Definition of Empirical and Molecular Formula (d) 18 5.2 Calculation using Mass Readings (e) 18 5.3 Calculation using % Composition by Mass (e) 19 5.4 Calculation using Combustion Data (e) 19 6 Volumetric Analysis 6.1 Acid-Base titration 24 6.2 Types of Acids and Bases 24 6.3 Common Indicators for Acid-Base Titrations 25 6.4 Back Titration (g) 27 7 Redox 7.1 Oxidation Number (Oxidation State) (i) 29 7.2 Redox Reactions (j) 30 7.3 Balancing Redox Equations 31 7.4 Common Oxidising and Reducing Agents 35 7.5 Redox Titrations (g), (h) 36 Annex A: Commonly used Chemistry Terms 40 Annex B: Summary of Important Stoichiometry Relationships 41 Annex C: Rules for Assigning Oxidation Numbers 41 Annex D: Balancing Redox Equations using Oxidation Number method 42

2026 JC1 H2 Chemistry (9476) 3 Topic 2: The Mole Concept & Stoichiometry 1 INTRODUCTION Stoichiometry is one of the most important subjects in general chemistry. It is the study of the quantitative relationships or ratios between two or more substances undergoing a physical change or chemical change (chemical reaction). Jeremias Benjaim Richter defined stoichiometry in 1792 as the science of measuring quantities or mass ratios of chemical elements. The word derives from the Greek words: stoicheion (meaning "element") and metron (meaning "to measure"). Review of terms used in chemistry (a) Atom is the smallest part of an element which can ever exist or take part in chemical reactions. (b) Element is a pure substance that only contains one type of atoms having the same chemical properties, e.g. helium, He (monatomic); hydrogen, H 2 (diatomic); ozone, O3 (triatomic). (c) Isotopes are atoms of the same element (i.e. same proton number) having different number of neutrons. For instance, 35Cl and 37Cl are isotopes of chlorine. Isotopes have similar chemical properties but different physical properties. (d) Compound is a pure substance that consists of two or more elements chemically combined in fixed proportions. Unlike a mixture , a compound cannot be separated into its components through physical means. (e) Molecule is an electrically neutral group of two or more atoms held together by covalent bonds, e.g. bromine, Br2 (diatomic molecule); sulfur dioxide, SO2 (triatomic molecule). (f) Ion is an electrically charged species, e.g. sulfide ion, S 2– (monoatomic ion); sulfite ion, SO32– (polyatomic ion); ammonium ion, NH4+ (polyatomic ion). 2 RELATIVE MASSES 2.1 THE 12C SCALE Since the masses of the atoms and molecules are very small, it is not practical to deal with their actual masses. (E.g., mass of one H atom is equivalent to 1.6710−27 kg). Instead, it is more convenient to compare their masses relative to that of another atom, known as the standard. The most abundant isotope of carbon, 12C is the chosen standard. In this relative atomic mass scale, a 12C atom is assigned a mass of exactly 12 units. Hence, we say the relative mass of a particle is defined as the number of times the particle is as heavy as 1 12 the mass of a 12C atom. Note that for the following definitions, the relative masses have no units as they are not actual masses, but relative to the arbitrary standard.

2026 JC1 H2 Chemistry (9476) 4 Topic 2: The Mole Concept & Stoichiometry 2.2 DEFINITIONS • Relative isotopic mass of an element is defined as the ratio of the mass of one atom of a single isotope of the element to 1 12 the mass of an atom of 12C. = 12 mass of one atom of an isotope of an elementrelative isotopic mass 1 of the mass of one atom of C12 (no units) • Relative atomic mass (symbol Ar) of an element is defined as the ratio of the weighted average mass of one atom of the element to 12 1 the mass of an atom of 12C. =r 12 weighted average mass of one atom of an element 1 of the mass of one atom of C12A (no units) FOR YOUR INFORMATION: When given two values (e.g. 35 and 37), simple averaging assumes that both values have even weightage (50 % contribution to the final value). Thus the math will work out to be: +   =  +  = =      1 1 35 37simple average 35 37 362 2 2 Weighted averaging takes into account that both values do not have the same percentage of contribution to the final value. If the value 35 contributes 75 % to the average value while 37 contributes the remaining 25 %, then the math will work out to be:    =  +  =      75 25weighted average 35 37 35.5100 100 Note: • Because of the concept of weighted average, relative atomic mass is rarely a simple whole number because of the existence of isotopes of varying abundance (see Table 1). Isotope Relative isotopic mass Percentage relative abundance Relative atomic mass 1H 2H 1.0 2.0 99.98 0.02 1.0 12C 13C 12.0 13.0 98.89 1.11 12.0 35Cl 37Cl 35.0 37.0 75.5 ≈ 75 24.5 ≈ 25 35.5 79Br 81Br 79.0 81.0 50.5 ≈ 50 49.5 ≈ 50 80.0 Table 1

2026 JC1 H2 Chemistry (9476) 5 Topic 2: The Mole Concept & Stoichiometry • Simplifying the above equation: ( )=r relative isotopic mass relative abundanceA • Ar (when quoting from the Periodic Table) should be to 1 decimal place. Example 1: The element gallium has two isotopes 69Ga and 71Ga, in the ratio of 3:2. Calculate the Ar of gallium. Ar of Ga    =  +  =       69.832 69 713 +2 3 +2 Example 2: Copper exists in two isotopic forms, 63Cu and 65Cu. The relative atomic mass of Cu is 63.5. Calculate the relative abundance of the two isotopes of Cu. Let the relative abundanc