Percentage purity questions show up almost every year in the NSC Physical Sciences Paper 2, usually inside a Quantitative Aspects of Chemical Change context. The good news: once you know the four-step method, every version of this question — whether it’s marble chips reacting with acid, or an impure metal reacting to lose mass — follows the same logic.
What “percentage purity” actually means
A sample is rarely 100% pure. If you react an impure sample of calcium carbonate (CaCO₃) with excess hydrochloric acid (HCl), only the CaCO₃ in the sample reacts — the impurities don’t. So the mass of gas produced tells you how much pure CaCO₃ was actually present.
Percentage purity is simply:
% purity = (mass of pure substance ÷ mass of impure sample) × 100
The whole challenge of these questions is finding that “mass of pure substance” — and that’s where the mass-loss method comes in.
The four-step method
- Find the mass of gas lost. This is usually initial mass − final mass (a mass-loss/conductometric-style setup), or it may be given directly.
- Convert mass of gas to moles using n = m/M.
- Use the balanced equation’s mole ratio to find moles, then mass, of the pure reactant that produced that gas.
- Divide by the mass of the impure sample and multiply by 100.
Worked example
A 5,00 g sample of impure calcium carbonate is reacted with excess dilute hydrochloric acid. The mass of the flask and contents decreases by 1,76 g due to CO₂ gas escaping. Calculate the percentage purity of the CaCO₃ sample.
Balanced equation:
CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g)
Step 1 — mass of CO₂ lost:
m(CO₂) = 1,76 g (given as the mass loss)
Step 2 — moles of CO₂:
n(CO₂) = m/M = 1,76 ÷ 44 = 0,04 mol
Step 3 — moles and mass of pure CaCO₃:
Mole ratio CaCO₃ : CO₂ = 1 : 1, so n(CaCO₃) = 0,04 mol
m(CaCO₃) = n × M = 0,04 × 100 = 4,00 g
Step 4 — percentage purity:
% purity = (4,00 ÷ 5,00) × 100 = 80,0%
Common mistakes that cost marks
- Dividing by the mass of pure product instead of the mass of the impure sample in the final step.
- Using the wrong molar mass — always check whether the question wants M(CaCO₃) = 100 g/mol or M(CO₂) = 44 g/mol at each stage.
- Forgetting the mole ratio when it isn’t 1:1 — always check the balanced equation before assuming a direct swap.
- Rounding too early. Carry extra decimal places through Steps 1–3 and only round your final answer.
- Mixing up “mass of gas lost” with “mass of gas produced” — in a sealed system they’re the same, but in an open system (like this one) the mass loss is the mass of gas that escaped.
Quick reference table
| Step | What to do | Formula |
|---|---|---|
| 1 | Find mass of gas lost | m = initial mass − final mass |
| 2 | Convert to moles | n = m ÷ M |
| 3 | Apply mole ratio, find pure mass | m(pure) = n(pure) × M(pure) |
| 4 | Calculate % purity | % purity = (m(pure) ÷ m(sample)) × 100 |
Once you’re comfortable with this four-step flow, percentage purity questions become one of the more predictable — and mark-friendly — parts of the Quantitative Aspects of Chemical Change topic. Try the worked example again from memory, then attempt a past-paper question using the same steps.
Physics with Mr Eke — CAPS-aligned Physical Sciences content for Grade 10–12.