Percentage Yield & Purity

Summary: Percentage yield compares the actual mass of product obtained to the theoretical mass predicted from stoichiometry. Percentage purity measures how much of a sample is the desired substance. Both are key calculation skills for IGCSE Chemistry. Tags: igcse chemistry calculations yield purity stoichiometry Created: 2026-07-14 Last Updated: 2026-07-14

Percentage Yield

Formula

% Yield = (actual yield / theoretical yield) × 100

• Actual yield = the mass of product you actually obtain from the experiment (measured)
• Theoretical yield = the maximum mass of product predicted by calculation using the balanced equation and reacting masses

Worked Example 1: Simple Yield

50 g of CaCO₃ is heated. 20 g of CaO is obtained. Calculate the percentage yield.

CaCO₃(s) → CaO(s) + CO₂(g)

1. Mr(CaCO₃) = 100; Mr(CaO) = 56
2. Moles of CaCO₃ = 50/100 = 0.50 mol
3. Mole ratio 1:1 → theoretical moles of CaO = 0.50 mol
4. Theoretical mass of CaO = 0.50 × 56 = 28.0 g
5. % Yield = (20.0 / 28.0) × 100 = 71.4%

Worked Example 2: Finding Actual Yield

In a reaction with theoretical yield 15.0 g and percentage yield 80%, what mass is actually obtained?

Actual yield = (80/100) × 15.0 = 12.0 g

Reasons for Yield < 100%

This is a classic exam question (2-3 marks). You must be able to explain WHY the actual yield is usually less than theoretical:

Yield > 100%? (Trick Question)

If yield > 100%, the product is not pure — it contains impurities that add mass. Possible causes:

• Product not fully dried (still contains water/solvent)
• Product contains unreacted reactants or by-products
• Weighing error

Percentage Purity

Formula

% Purity = (mass of pure substance / total mass of impure sample) × 100

Worked Example

A 10.0 g sample of impure CaCO₃ reacts with excess HCl to produce 3.52 g of CO₂. Calculate the percentage purity of the sample.

CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g)

1. Mr(CO₂) = 44; Mr(CaCO₃) = 100
2. Moles of CO₂ produced = 3.52/44 = 0.0800 mol
3. Mole ratio CaCO₃ : CO₂ = 1:1 → moles of CaCO₃ = 0.0800 mol
4. Mass of pure CaCO₃ = 0.0800 × 100 = 8.00 g
5. % Purity = (8.00 / 10.0) × 100 = 80.0%

Combined Yield + Purity Problems

Sometimes both concepts appear in one question:

A 25.0 g sample of impure zinc (90% pure) reacts with excess HCl. The reaction has 85% yield. What mass of H₂ is actually produced?

Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)

1. Mass of pure Zn = (90/100) × 25.0 = 22.5 g
2. Moles of Zn = 22.5 / 65 = 0.346 mol
3. Mole ratio 1:1 → theoretical moles of H₂ = 0.346 mol
4. Theoretical mass of H₂ = 0.346 × 2 = 0.692 g
5. Actual yield = (85/100) × 0.692 = 0.588 g

Water of Crystallisation & Purity

Hydrated salts contain water molecules in their crystal structure (e.g., CuSO₄·5H₂O, MgSO₄·7H₂O). This water contributes mass and must be accounted for in calculations.

Finding x in M·xH₂O

4.99 g of CuSO₄·xH₂O is heated to constant mass. 3.19 g of anhydrous CuSO₄ remains. Find x.

1. Mass of H₂O driven off = 4.99 − 3.19 = 1.80 g
2. Moles of anhydrous CuSO₄ = 3.19 / 159.5 = 0.0200 mol
3. Moles of H₂O = 1.80 / 18 = 0.100 mol
4. Mole ratio CuSO₄ : H₂O = 0.02 : 0.10 = 1 : 5
5. Therefore x = 5 → CuSO₄·5H₂O

Atom Economy (Extension)

While not always examined in IGCSE, atom economy is a related green chemistry concept:

% Atom Economy = (Mr of desired product / Σ Mr of all reactants) × 100

• Measures how efficiently reactants are incorporated into the desired product
• Higher atom economy = less waste, more sustainable

Common Question Types

Type 1: Calculate Percentage Yield (2-3 marks)

• Frequency: ~40% of papers (Stoichiometry topic)
• Given: actual yield + either theoretical yield OR enough data to calculate it
• Method: find theoretical yield from moles → % yield formula

Type 2: Explain Why Yield < 100% (2-3 marks)

• Give 2-3 specific reasons
• Must be relevant to the described experiment
• “Reversible reaction”, “product lost during filtration/crystallisation”, “side reactions”

Type 3: Calculate Purity from Gas Volume or Mass (3-4 marks)

• Given: mass of impure sample + volume/mass of product
• Method: product → moles → moles of pure reactant → mass of pure → % purity

Type 4: Water of Crystallisation (3-4 marks)

• Frequency: ~25% of papers
• Given: mass of hydrated salt + mass after heating
• Method: find mass of H₂O lost → moles ratio → x

Common Mistakes

• Using actual yield instead of theoretical in further calculations: Always work backwards from the product measured
• Forgetting to × 100: The formula is % yield = (actual/theoretical) × 100
• Mixing up yield and purity: Yield = reaction efficiency; purity = sample composition
• Not accounting for water of crystallisation in Mr calculations for hydrated compounds
• Rounding moles too early: Keep at least 3 significant figures until the final answer
• Forgetting that gases must be at RTP when using 24 dm³/mol for volume-to-moles conversion

Key Facts to Memorize

• % Yield = (actual / theoretical) × 100
• % Purity = (mass of pure / total mass) × 100
• Actual yield < theoretical yield is NORMAL (reversible reactions, losses, side reactions)
• Purity is found by reacting the sample and measuring the product formed
• Water of crystallisation: heat hydrated salt → measure mass loss → mole ratio
• Always start from the product when working out purity (the product tells you how much pure reactant was there)

Related Notes

• Relative Masses and Moles — Moles = mass / Mr
• Reacting Masses — Reacting mass calculations using mole ratios
• Empirical and Molecular Formulae — Finding formulae from composition data
• Concentration and Gas Volumes — Moles = concentration × volume; moles = volume / 24 dm³
• Chemical Equations and Calculations — Balancing equations
• Water of Crystallisation — Detailed coverage of hydrated salts
• IGCSE-Chem-Index

Past Paper Sources

• 0620/43 May/June 2019 Q8(f): Calculate % yield of ester from given masses (3 marks)
• 0971/42 Oct/Nov 2022 Q6(d): Explain why yield is less than 100% (2 marks)
• 0620/32 Feb/March 2020 Q5(c): Calculate % purity of limestone from CO₂ released (4 marks)
• 0620/53 May/June 2018 Q3: Water of crystallisation practical — find x in MgSO₄·xH₂O (5 marks)
• 0971/43 Oct/Nov 2021 Q7(e): Combined purity + yield calculation (4 marks)

IGCSE Chemistry (0620/0971) wiki. Core calculation topic — expect yield/purity questions in most papers.