Amount of Substance: Moles, the Ideal Gas Equation and Atom Economy

This guide covers 3.1.2, Amount of substance, from the Physical chemistry section of OxfordAQA International AS and A-level Chemistry 9620. It follows Atomic structure (3.1.1) and supplies the calculation toolkit – moles, gas laws, formulae and balanced-equation arithmetic – used throughout the rest of the course.

Relative atomic mass and relative molecular mass

You need to be able to define relative atomic mass (Ar) and relative molecular mass (Mr), both expressed in terms of ¹²C. A specific naming convention applies: the term relative formula mass is used specifically for ionic compounds, rather than relative molecular mass, because ionic compounds do not exist as discrete molecules.

The mole and the Avogadro constant

You need to understand the Avogadro constant as the number of particles in one mole, and apply the mole concept to electrons, atoms, molecules, ions, formulas and equations, as well as to concentration of a substance in solution (measured in mol dm⁻³). You need to be able to carry out calculations using the Avogadro constant, using mass of substance together with Mr and amount in moles, and using concentration, volume and amount of substance in a solution. Note that the specification explicitly states you will not be expected to recall the value of the Avogadro constant itself – it will be provided, so your calculation method matters more than memorising the constant.

The ideal gas equation

You need to use the ideal gas equation, pV = nRT, with all variables expressed in SI units, in calculations. As with the Avogadro constant, you will not be expected to recall the value of the gas constant R – it will be given. This means the actual skill being tested is unit consistency: converting volumes to m³, pressures to Pa, and temperatures to kelvin before substituting into the equation, since an otherwise correct method will produce a wrong answer if any variable is left in the wrong units.

Empirical and molecular formula

You need to know that empirical formula is the simplest whole-number ratio of atoms of each element in a compound, while molecular formula is the actual number of atoms of each element in a compound, and understand the relationship between the two. You need to be able to calculate empirical formula from data giving composition by mass or percentage by mass, and calculate molecular formula from the empirical formula together with relative molecular mass – the molecular formula is always a whole-number multiple of the empirical formula, found by dividing the given Mr by the empirical formula’s own mass.

Balanced equations and associated calculations

You need to be able to write balanced equations (both full and ionic) for reactions studied, and balance equations for unfamiliar reactions when reactants and products are specified. Building on this, you need to use balanced equations to calculate masses, volumes of gases, percentage yields, percentage atom economies, and concentrations and volumes for reactions in solutions.

Percentage atom economy has a specific formula you need to apply:

atom economy = (molecular mass of desired product ÷ sum of molecular masses of all reactants) × 100

The specification also expects you to understand the economic, ethical and environmental advantages, for both society and industry, of developing chemical processes with high atom economy – a process that wastes less starting material as by-product is generally cheaper to run and produces less unwanted waste.

Required Practical 1

This subtopic carries the specification’s Required Practical 1: making up a volumetric solution and carrying out a simple acid-base titration. This practical links directly to the mole-and-concentration calculations above, since a titration’s whole purpose is determining an unknown concentration or amount using the stoichiometry of a balanced equation.

A worked example

A hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass, and has a relative molecular mass of 42. Assuming 100 g of the compound: 85.7 g carbon ÷ 12 = 7.14 mol, and 14.3 g hydrogen ÷ 1 = 14.3 mol. Dividing both by the smaller value (7.14) gives a ratio of 1 : 2.003, so the empirical formula is CH₂, with empirical formula mass 12 + 2 = 14. Dividing the given Mr by the empirical formula mass: 42 ÷ 14 = 3, so the molecular formula is (CH₂)₃ = C₃H₆. This two-stage process – composition data to empirical formula, then empirical formula plus Mr to molecular formula – is the exact method the specification expects you to apply to any similar dataset.

How to approach it

Since neither the Avogadro constant nor the gas constant R needs to be memorised, spend your revision time on method and unit consistency instead – particularly converting to SI units before using pV = nRT, which is the single most common source of numerical errors in this subtopic. Practise the empirical-to-molecular-formula workflow as a strict two-step process: find the empirical formula from composition data first, then only afterwards use the given Mr to find the whole-number multiplier. For atom economy questions, be careful to identify the correct “desired product” from a reaction that may produce more than one product, since atom economy is calculated relative to that one target product’s mass, not the total mass of everything formed. Finally, revise the balanced-equation calculation chain as one connected skill rather than separate topics: writing a correct equation, using it to find a mole ratio, and converting between moles and mass, gas volume or solution concentration are all one continuous method that Required Practical 1 exercises directly.

Official syllabus

OxfordAQA, International AS and A-level Chemistry (9620) specification, Version 4.3, for International AS and A-level exams May/June 2020 onwards: official specification PDF, section 3.1.2 “Amount of substance”. Verified 2026-09-02.