Percentage Yield and Percentage Purity
Percentage yield and purity for IGCSE Chemistry 0620 Extended: both formulae, theoretical yield from moles, why yields fall below 100%, with marks.
Published by IGCSEChemistry.com.my
Chemistry teaching team: K. S. Tan (15+ years teaching IGCSE Chemistry) and Ms Yash (10+ years teaching IGCSE Chemistry) and Ms Kartini (15+ years teaching IGCSE Chemistry).
Mapped to Cambridge IGCSE Chemistry 0620 (2026–2028). Last updated 2026-08-04.
Real reactions short-change you: run a preparation that should give 10 g and the filter paper holds 7.4 g. Percentage yield measures that gap, percentage purity measures the related one between a sample and the genuine substance inside it, and both are Supplement calculations that Paper 4 attaches to salt preparations and organic syntheses for 3-4 marks. Each is one fraction: the whole skill is knowing which mass goes on top, and computing the theoretical mass with the three-step mole method.
Percentage yield (Supplement)
Percentage yield = (actual yield ÷ theoretical yield) × 100.
The actual yield is the measured mass of product obtained. The theoretical yield is the maximum mass the balanced equation predicts, always the output of a mole calculation: moles of starting material, ratio, mass of product. Then divide and multiply by 100.
Why below 100%? Mark schemes accept: the reaction does not go to completion (or is reversible); product is lost during transfer, filtration or crystallisation; side reactions consume reactants. Pick one and say it specifically. “Some is lost” without a stage named is borderline, and “human error” scores nothing.
| Scenario | Yield interpretation |
|---|---|
| Actual < theoretical | Normal: incomplete reaction or losses |
| Actual = theoretical | 100%: essentially never achieved in practice |
| Actual > theoretical | Impossible for pure product: sample is wet or impure |
Percentage purity (Supplement)
Percentage purity = (mass of pure substance ÷ mass of impure sample) × 100.
The impure sample mass is given in the question; the pure mass usually comes out of a back-calculation. A classic structure: an impure sample of a carbonate is reacted with excess acid, the CO2 is measured, and moles of gas reveal how much real carbonate was present. Purity also has a fast experimental check: a pure substance has a sharp melting point, while impurities lower and broaden it, which links this page to topic 12’s purification methods and the melting-point facts from changes of state.
Keep the two fractions straight: yield compares product obtained against product possible; purity compares true substance against total sample. Both fractions must come out at or below 1, which is a built-in sanity check.
Worked exam question
An impure sample of limestone has a mass of 6.0 g. When reacted with excess hydrochloric acid, it produces 1.2 dm³ of carbon dioxide at r.t.p. Only the calcium carbonate in the limestone reacts: CaCO3 + 2HCl → CaCl2 + H2O + CO2. Calculate the percentage purity of the limestone. [Mr of CaCO3 = 100] (4)
Model answer: n(CO2) = 1.2 ÷ 24 = 0.050 mol (1). Ratio CO2 : CaCO3 = 1 : 1, so n(CaCO3) = 0.050 mol (1). Mass of pure CaCO3 = 0.050 × 100 = 5.0 g (1). Percentage purity = (5.0 ÷ 6.0) × 100 = 83.3% (1).
Mark-by-mark: mark 1 is the gas-to-moles conversion using 24 dm³ at r.t.p. Mark 2 is the (here gentle) 1:1 ratio, still expected on paper. Mark 3 converts to the pure mass via Mr. Mark 4 is the fraction the right way up, ×100, answer to 3 significant figures. A candidate who divides 6.0 by 5.0 reports 120% purity, an impossibility that should send you back up the working.
The mistakes that cost marks
- Inverted fractions. Yield over 100% or purity over 100% is the symptom; swap numerator and denominator and recheck. Pure mass and theoretical yield sit on top only in their opposite formulae; learn each fraction with a worked example, not as abstract words.
- Theoretical yield taken as the reactant mass. The denominator of a yield calculation is the calculated product mass, which requires the mole ratio first.
- Vague loss explanations: “some escaped”, “human error”. Name a stage, for example “product lost when the solution was filtered” or “the reaction did not go to completion”.
- Mixing the two quantities: answering a purity question with the yield formula. Read which comparison the question makes: obtained-vs-possible (yield) or pure-vs-sample (purity).
How examiners want it phrased
| Typical student wording | Accepted mark-scheme wording |
|---|---|
| ”We didn’t get all of it" | "The reaction is incomplete / product was lost during filtration and crystallisation" |
| "The rock is 83% good" | "Percentage purity = (5.0 g ÷ 6.0 g) × 100 = 83.3%" |
| "Yield is what you get over what you should get" | "Percentage yield = (actual yield ÷ theoretical yield) × 100" |
| "More than 100%, great result!" | "An apparent yield above 100% indicates the product is impure or wet” |
The Malaysia note
Yield and purity questions cluster in the second half of Paper 4, where Malaysian candidates in the Oct/Nov series (sitting mocks, trials and the real exam within a 6-8 week window) are most fatigued, and inverted fractions spike accordingly. A memorised pair of worked examples beats a memorised pair of formulae, because the example carries the orientation with it. We rehearse exactly this kind of late-paper resilience in lessons; the 1-hour trial lesson will show you how a Chemistry specialist structures it.
Test yourself
All three are Supplement level. Keep each fraction the right way up, then click to check.
Q1 (3 marks). Zinc oxide reacts with dilute sulfuric acid: ZnO + H2SO4 → ZnSO4 + H2O. A student reacts 4.05 g of zinc oxide with excess acid and obtains 6.44 g of zinc sulfate. Calculate the percentage yield. [Mr: ZnO 81, ZnSO4 161]
Show answer
• n(ZnO) = 4.05 ÷ 81 = 0.0500 mol [1] • Ratio 1 : 1, so theoretical yield = 0.0500 × 161 = 8.05 g [1] • Percentage yield = (6.44 ÷ 8.05) × 100 = 80.0% [1]
Q2 (2 marks). Suggest two reasons why the student’s yield is less than 100%.
Show answer
• The reaction did not go to completion / is incomplete [1] • Product was lost during transfer/filtration/crystallisation (accept: side reactions formed other products) [1]
Q3 (2 marks). A 2.0 g sample of impure sodium chloride is found to contain 1.71 g of pure sodium chloride. Calculate the percentage purity, and explain why an apparent purity above 100% is impossible.
Show answer
• Percentage purity = (1.71 ÷ 2.0) × 100 = 85.5% [1] • The mass of pure substance cannot exceed the mass of the whole sample, so the fraction cannot be greater than 1 [1]
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