Study Guides
Amount of Substance: The Mole, Empirical Formulae and Gas Calculations
The mole and Avogadro constant, empirical and molecular formulae, hydrated salts, mole calculations from mass, gas volume and concentration, the ideal gas equation, and percentage yield and atom economy, for OCR A Level Chemistry A H432, Module 2.1.3.
- Subject
- Chemistry
- Level
- A LEVELS
- Topic
- Foundations in chemistry
- Author
- Nouman Ahmed
- Updated
Aligned to OCR A Level Chemistry (H432), First assessment 2017 (current specification version 3.1, May 2026). Official specification .
This guide covers 2.1.3, Amount of substance, from Module 2, Foundations in chemistry, of OCR A Level Chemistry A H432.
Before studying this
At GCSE you met the mole as a counting unit for particles and used it in simple reacting-mass calculations. Module 2.1.3 keeps that foundation but extends it substantially: gas volumes and the ideal gas equation, formulae determined from experimental data, hydrated salts, and the two calculations used throughout the rest of the course to judge how “efficient” a reaction is — percentage yield and atom economy. Module 2 is assessed directly in all three written components: Component 01 (Periodic table, elements and physical chemistry – Modules 1, 2, 3 and 5), Component 02 (Synthesis and analytical techniques – Modules 1, 2, 4 and 6), and Component 03 (Unified chemistry – all six modules). It also reappears constantly in later modules, so it’s worth being completely fluent in before moving on.
Syllabus coverage
OCR A LEVEL CHEMISTRY A H432 — Module 2.1.3, Amount of substance
This section covers: the terms amount of substance, mole, the Avogadro constant, molar mass and molar gas volume; empirical and molecular formulae, including calculating them from mass or percentage composition; the terms anhydrous, hydrated and water of crystallisation, and calculating the formula of a hydrated salt; calculations using amount of substance in mol involving mass, gas volume, and solution volume and concentration; the ideal gas equation pV = nRT; the use of stoichiometric relationships in calculations; calculating percentage yield and atom economy, including the sustainability benefits of a high atom economy; and the techniques and procedures required during experiments requiring the measurement of mass, volumes of solutions and gas volumes (Practical Activity Group PAG1).
The mole and the Avogadro constant
Amount of substance is measured in moles (mol). One mole of any substance contains the same number of particles as one mole of any other substance — the Avogadro constant, Nₐ = 6.02 × 10²³ per mole.
- Molar mass (units g mol⁻¹) is the mass of one mole of a substance — numerically equal to its relative formula mass.
- Molar gas volume (units dm³ mol⁻¹) is the volume occupied by one mole of any gas at a given temperature and pressure; at room temperature and pressure this is taken as 24 dm³ mol⁻¹ (the exact value is given on OCR’s data sheet in the exam).
OCR accepts both the classical (carbon-12-based) and the revised (Avogadro-constant-based) definitions of the mole — this distinction reflects a 2018 update to the international definition and won’t affect how you use the mole in calculations.
Empirical and molecular formulae
The empirical formula is the simplest whole-number ratio of atoms of each element in a compound. The molecular formula is the actual number of each type of atom in one molecule — always a whole-number multiple of the empirical formula.
Worked example. A compound is 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass, and has a relative molecular mass of 180. Find its molecular formula.
Assume 100 g of compound, so the masses equal the percentages.
| Element | Mass (g) | ÷ Aᵣ | Moles | ÷ smallest | Ratio |
|---|---|---|---|---|---|
| C | 40.0 | ÷12 | 3.33 | ÷3.33 | 1 |
| H | 6.7 | ÷1 | 6.7 | ÷3.33 | 2 |
| O | 53.3 | ÷16 | 3.33 | ÷3.33 | 1 |
Empirical formula: CH₂O (mass = 30). Since the molecular mass (180) is 6 times the empirical formula mass (30), the molecular formula is C₆H₁₂O₆ (glucose).
Hydrated salts and water of crystallisation
A hydrated salt has a fixed number of water molecules built into its crystal structure (water of crystallisation); an anhydrous salt has none. The formula of a hydrated salt is found the same way as an empirical formula, treating H₂O as one of the “elements” in the ratio calculation — this is a very common practical-based calculation, usually from mass-loss data on heating.
Calculating from amount of substance
The same three relationships come up repeatedly, all rearrangements of “amount = quantity ÷ quantity per mole”:
- Mass: moles = mass ÷ molar mass
- Gas volume: moles = gas volume ÷ molar gas volume (at the same temperature and pressure)
- Solution concentration: moles = concentration (mol dm⁻³) × volume (dm³)
Worked example. How many moles of NaOH are in 250 cm³ of a 0.100 mol dm⁻³ solution?
moles = concentration × volume = 0.100 × (250/1000) = 0.0250 mol
The ideal gas equation
For gases not at room temperature and pressure, use:
pV = nRT
where p is pressure in pascals (Pa), V is volume in m³, n is amount in mol, R is the gas constant (given on the data sheet), and T is temperature in kelvin. Getting the units right — especially converting cm³ or dm³ to m³, and °C to K — is the single most common source of error in this calculation.
Percentage yield and atom economy
Two separate ideas, often confused:
- Percentage yield = (actual yield ÷ theoretical yield) × 100. It measures how much product you actually collected compared with the maximum the stoichiometry allows — lost to side reactions, incomplete reactions, or purification losses.
- Atom economy = (molar mass of desired product ÷ sum of molar masses of all reactants) × 100. It measures how much of the mass of all the reactants ends up in the useful product, rather than in waste by-products — a property of the reaction’s stoichiometry itself, not of how carefully you carried it out.
A reaction can have 100% yield of a product but poor atom economy if most of the reactant mass ends up in a wasted by-product — which is why the specification links high atom economy to the sustainability of chemical processes: less waste to dispose of, and less reactant needed overall.
Practical techniques for these calculations (PAG1)
Every calculation in this sub-topic depends on measurements made with a specific technique: mass is measured by difference (weighing a container before and after adding or removing a substance, or before and after heating, to find the change), volumes of solutions are measured using a volumetric flask (to make up a solution of known concentration) or a burette/pipette (to measure out or transfer a precise volume), and gas volumes are measured using a gas syringe or by displacement of water, reading the scale at eye level to avoid parallax error. These techniques and procedures are examined directly, not just the calculations that follow from them.
Common mistakes
- Mixing up percentage yield and atom economy. Yield is about how much of the possible product you collected; atom economy is about how much of the reactant mass becomes useful product, calculated from the balanced equation alone.
- Forgetting to convert units before using pV = nRT. Pressure must be in Pa, volume in m³, and temperature in kelvin — mixing in cm³, dm³ or °C without converting is the most common cause of an answer being wrong by a power of ten.
- Using the wrong mass in an empirical formula calculation. For a hydrated salt, remember the water of crystallisation is included in the ratio as if it were its own “element” (H₂O), not omitted.
- Rounding too early. In multi-step mole calculations, carry full calculator precision through each step and only round the final answer.
Quick revision checklist
- Definitions: amount of substance, mole, Avogadro constant, molar mass, molar gas volume
- Empirical vs molecular formula, and calculating both from % composition
- Hydrated salts and water of crystallisation calculations
- Mole calculations from mass, gas volume, and solution concentration
- The ideal gas equation pV = nRT, with correct SI units
- Percentage yield vs atom economy — what each one actually measures
Related resources
- Atoms, Molecules and Stoichiometry (Cambridge AS & A Level) — the equivalent AS-level content from a different awarding body
- Formulae, Equations and the Mole (Cambridge IGCSE / O Level) — the GCSE-level foundation this page builds on
- OCR A Level Chemistry hub
Written against OCR A Level Chemistry A H432, specification version 3.1 (May 2026), https://www.ocr.org.uk/qualifications/as-and-a-level/chemistry-a-h032-h432-from-2015/, verified 2026-08-18. Always check the current specification for your examination year.
Related resources
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Study Guides
Acids, Bases and Neutralisation: Titration Technique and Calculations
Common acids and alkalis, strong versus weak acid dissociation, neutralisation reactions, standard solution preparation, acid-base titration technique, and titration calculations, for OCR A Level Chemistry A H432, Module 2.1.4.
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Practice Questions
OCR A Level Chemistry: Amount of Substance — Practice Questions
Original exam-style practice questions with full worked answers on moles, the ideal gas equation, titrations and percentage yield for OCR A Level Chemistry.
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Revision Notes
OCR A Level Chemistry: Amount of Substance — Revision Notes
Condensed recall notes on the mole, empirical formulae, titrations, gas volumes, yield and atom economy for OCR A Level Chemistry H432.
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