1. Balancing equations
Balance the atoms of every element, and the total charge on each side, by changing the coefficients (big numbers in front). Never change subscripts: that changes the substance. Add state symbols: (s), (l), (g), (aq).
2. Ionic equations
Write the full equation, split only dissolved strong electrolytes into ions, then cancel the spectator ions that appear unchanged on both sides. Don't split solids, water, gases or weak acids.
3. The mole bridge
Every reacting-quantity calculation follows the same route:
The known or wanted quantity can be a mass (n = m/M), a solution (n = cV) or a gas (pV = nRT). Label each step with the substance it refers to.
4. Limiting reagents
The limiting reagent is used up first and fixes how much product forms. To find it, divide each reactant's moles by its coefficient; the smallest value is limiting. Don't just compare masses or raw moles.
5. Percentage yield and purity
Calculate the theoretical yield from the limiting reagent. In the example above, if 15.1 g of AlCl3 was collected, yield = 15.1 ÷ 17.8 × 100 = 84.8%.
Yields are below 100% because of incomplete reactions, side reactions, reversible reactions reaching equilibrium, and product lost in transfer or purification. A yield above 100% suggests the product is wet or impure.
For example, a 2.50 g sample containing 2.10 g CaCO3 is 84.0% pure.
Percentage yield vs conversion: yield compares product made with the maximum possible; percentage conversion compares reactant used up with reactant supplied. They are not the same thing.
6. Atom economy
Multiply each M by its coefficient in the balanced equation. By conservation of mass, you can use the total M of all products as the denominator instead.
An addition reaction with only one product, such as CH2=CH2 + H2O → C2H5OH, has an atom economy of 100%.
Why it matters, and its limits
High atom economy means less waste, fewer raw materials and less disposal. But it ignores the actual yield, excess reagents, solvents and catalysts. It also says nothing about energy use, toxic chemicals or renewable feedstocks, so a high atom economy does not automatically make a process sustainable.
7. 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, Ca 40.1.
Q1. Write the ionic equation for AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq).Show answer
Na+ and NO3− are spectators: Ag+(aq) + Cl−(aq) → AgCl(s)
Q2. What mass of CO₂ forms when 5.00 g CaCO₃ decomposes completely?Show answer
n(CaCO3) = 5.00 ÷ 100.1 = 0.04995 mol = n(CO2) (1 : 1)
mass = 0.04995 × 44.0 = 2.20 g
Q3. 2.80 g N₂ reacts with 0.900 g H₂: N₂ + 3H₂ → 2NH₃. Which is limiting, and what mass of NH₃ forms?Show answer
n(N2) = 2.80 ÷ 28.0 = 0.100 → ÷ 1 = 0.100
n(H2) = 0.900 ÷ 2.0 = 0.450 → ÷ 3 = 0.150. N2 is limiting.
n(NH3) = 2 × 0.100 = 0.200 mol
mass(NH3) = 0.200 × 17.0 = 3.40 g
Q4. In Q3, 2.55 g of NH₃ is collected. What is the percentage yield?Show answer
2.55 ÷ 3.40 × 100 = 75.0%
Q5. Find the atom economy for H₂ in CH₄ + H₂O → CO + 3H₂.Show answer
Desired: 3 × 2.0 = 6.0. Reactants: 16.0 + 18.0 = 34.0.
6.0 ÷ 34.0 × 100 = 17.6%
8. Sources
Sources and examiner guidance (reviewed 1 October 2026)
- Chemrevise — AQA 1.2 Calculations (amount of substance) revision guide (N. Goalby) — primary content reference (equations, yield, atom economy)
- AQA 7405 specification — 3.1.2 Amount of substance (incl. Required Practical 1) — 3.1.2.5 Balanced equations and associated calculations
- AQA 7404/2 examiner report, June 2023 — Q07.3: yield must compare theoretical and actual mass of the same product
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