1. Relative masses and molar mass
Atoms are so small that we compare their masses with a reference: 1/12 of the mass of one atom of carbon-12.
| Quantity | Meaning | Units |
|---|---|---|
| Relative atomic mass, Ar | Weighted mean mass of an atom of an element, relative to 1/12 of the mass of one atom of carbon-12 | None (a ratio) |
| Relative molecular mass, Mr | Mean mass of a molecule, compared with the same carbon-12 standard | None |
| Relative formula mass | Used for ionic compounds (no molecules), found the same way from the formula, e.g. NaCl | None |
| Molar mass, M | Mass of one mole of the substance | g mol−1 |
Mr and molar mass have the same number but are different quantities: Mr(H2O) = 18.0, while M(H2O) = 18.0 g mol−1. Always use the Ar values on the periodic table or data sheet (e.g. Cl = 35.5), not mass numbers.
2. The mole and the Avogadro constant
The mole is the unit of amount of substance (symbol n). It is not a mass. One mole contains exactly 6.02214076 × 1023 specified entities: this is the Avogadro constant, NA, in mol−1. In exams use the value you are given, usually 6.022 × 1023 mol−1.
Always say which entity you are counting: atoms, molecules, ions or electrons. One mole of O2 molecules contains two moles of O atoms.
3. Mass, moles and particles
4. Density and unit conversions
Density is usually given in g cm−3 for liquids. Mass conversions: 1 kg = 1000 g; 1 g = 1000 mg; 1 tonne = 106 g.
5. 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 atoms of each element in one molecule.
Method
- Divide the mass (or percentage) of each element by its Ar.
- Divide every answer by the smallest one.
- If a ratio is close to a simple fraction (such as 1.5 or 1.33), multiply all values by 2 or 3. Don't round 1.5 to 2.
Molecular formula: divide Mr by the empirical formula mass. The answer should be a whole number; multiply the empirical formula by it. For example, an empirical formula CH2O (mass 30.0) with Mr = 180 gives a factor of 6, so C6H12O6.
6. Water of crystallisation
A hydrated salt contains water in its crystal lattice, written as MgSO4·xH2O. Heating drives off the water, leaving the anhydrous salt.
7. Combustion analysis
When a compound containing C, H (and maybe O) burns completely, all its carbon ends up in CO2 and all its hydrogen in H2O. Each CO2 contains 1 C; each H2O contains 2 H. Find oxygen by difference, but only when you know the compound contains just C, H and O.
8. Practical: heating to constant mass
To find x in a hydrate, or the formula of a metal oxide (by heating magnesium in a crucible):
- Weigh the empty crucible and lid, then with the sample.
- Heat. For magnesium, lift the lid now and then to let oxygen in, while keeping the lid on enough to stop magnesium oxide smoke escaping.
- Let it cool, then weigh. Heat, cool and weigh again until two readings agree: constant mass indicates the reaction is complete under these heating conditions (it cannot show that, say, a gentler flame would ever remove all the water).
| Problem | Effect on result |
|---|---|
| Hydrate not heated to constant mass | Mass of water too small, so x is too low |
| Salt decomposes further when heated (losing more than water) | Mass lost too large, so x is too high; not all mass lost is water |
| Magnesium oxide smoke escapes | Mass gained by the solid is too small, so less oxygen appears to have combined |
| Weighing a hot crucible | Air currents make readings unreliable; let it cool first (it is also a burn risk) |
Diagram placeholder
Heating a sample in a crucible
Labels to include:
- Crucible with lid (lid slightly ajar for Mg)
- Pipe-clay triangle
- Tripod
- Bunsen burner
- Balance for weighing between heats
The crucible sits in a pipe-clay triangle on a tripod over a Bunsen burner. It is heated, cooled and weighed repeatedly until the mass stops changing.
9. Significant figures
Give your final answer to the number of significant figures in the data (usually the least precise value), or as the question asks. There is no fixed rule such as "always give Mr to 1 d.p.". Keep unrounded values in your calculator between steps; rounding early can change the final digit.
10. Quick checks
Finesse practice: indicative answers to check your reasoning, not official mark allocations. Ar: H 1.0, C 12.0, N 14.0, O 16.0, Al 27.0, S 32.1, Cu 63.5. NA = 6.022 × 1023 mol−1.
Q1. How many moles are in 4.25 g of NH₃?Show answer
M = 14.0 + 3.0 = 17.0 g mol−1
n = 4.25 ÷ 17.0 = 0.250 mol
Q2. How many oxygen atoms are in 0.0500 mol of Al₂(SO₄)₃?Show answer
Each formula unit has 3 × 4 = 12 O atoms, so n(O) = 0.0500 × 12 = 0.600 mol
N = 0.600 × 6.022 × 1023 = 3.61 × 1023 atoms
Q3. A hydrocarbon is 85.7% C and 14.3% H, with Mr 56.0. Find its molecular formula.Show answer
C: 85.7 ÷ 12.0 = 7.142; H: 14.3 ÷ 1.0 = 14.3
Ratio 1 : 2.00, so empirical formula CH2 (mass 14.0)
56.0 ÷ 14.0 = 4, so C4H8
Q4. 2.50 g of CuSO₄·xH₂O is heated to constant mass, leaving 1.60 g. Find x.Show answer
Water = 0.90 g; n = 0.90 ÷ 18.0 = 0.0500 mol
M(CuSO4) = 63.5 + 32.1 + 64.0 = 159.6; n = 1.60 ÷ 159.6 = 0.01003 mol
x = 0.0500 ÷ 0.01003 = 4.99, so x = 5
Q5. Why must a hydrate be heated to constant mass?Show answer
Repeat heating until there is no further mass loss: under suitable heating conditions this indicates all the water has been driven off. If some water remains, the mass of water lost is too small and x is too low.
11. Sources
Sources and examiner guidance (reviewed 1 October 2026)
- Chemrevise — AQA 1.2 Calculations (amount of substance) revision guide (N. Goalby) — primary content reference (moles, formulae, hydrates, combustion)
- AQA 7405 specification — 3.1.2 Amount of substance (incl. Required Practical 1) — 3.1.2.1 Relative masses, 3.1.2.2 The mole, 3.1.2.4 Empirical and molecular formulae
- AQA 7404/1 examiner report, June 2023 — general: label each substance in calculation steps
Finesse Tuition is not endorsed by AQA or Chemrevise. All explanations and examples here are our own.