Percentage Purity Calculations

Percentage purity calculations determine what proportion of a sample is the desired substance. In practice, chemicals are rarely 100% pure, and knowing the purity is essential for accurate stoichiometry in industrial and laboratory contexts.

The method

1. Identify the mass of the pure substance in the sample.
2. Identify the total mass of the sample (pure substance + impurities).
3. Apply the percentage purity formula.

When the mass of pure substance is not given directly, you may need to calculate it from a reaction (using reacting masses methods).

The formula

percentage purity = (mass of pure substance / total mass of sample) x 100

Where:

• mass of pure substance is in grams (g)
• total mass of sample is in grams (g)

Worked examples

Example 1: Direct calculation

Question: A 10 g sample of limestone contains 8.5 g of calcium carbonate. Calculate the percentage purity.

Working:

• Percentage purity = (8.5 / 10) x 100 = 85%

Answer: 85%

Examiner note: When both masses are given directly, this is a single-step calculation. Show the substitution.

Example 2: Finding mass of pure substance from a reaction

Question: A 20 g sample of impure zinc was reacted with excess hydrochloric acid. 0.25 mol of hydrogen gas was collected. Calculate the percentage purity of the zinc. (Ar: Zn = 65)

Working:

• Balanced equation: Zn + 2HCl -> ZnCl2 + H2
• Mole ratio: 1 Zn : 1 H2
• Moles of Zn that reacted = 0.25 mol
• Mass of pure Zn = 0.25 x 65 = 16.25 g
• Percentage purity = (16.25 / 20) x 100 = 81.25%

Answer: 81.25%

Examiner note: The hydrogen produced tells you how much zinc actually reacted. The impurities did not react with the acid.

Example 3: Using gas volume to find purity

Question: 5 g of impure calcium carbonate was heated. 960 cm3 of carbon dioxide gas was collected at room temperature and pressure. If 1 mol of gas occupies 24 dm3 at RTP, calculate the percentage purity. (Ar: Ca = 40, C = 12, O = 16)

Working:

• Balanced equation: CaCO3 -> CaO + CO2
• Convert volume: 960 cm3 = 0.96 dm3
• Moles of CO2 = 0.96 / 24 = 0.04 mol
• Mole ratio: 1 CaCO3 : 1 CO2
• Moles of CaCO3 = 0.04 mol
• Mr of CaCO3 = 100
• Mass of pure CaCO3 = 0.04 x 100 = 4 g
• Percentage purity = (4 / 5) x 100 = 80%

Answer: 80%

Examiner note: This combines gas volume calculations with purity. The volume of gas produced tells you the moles of pure reactant.

Example 4: Impure reactant in a reacting masses problem

Question: A sample of iron ore contains 80 g of iron(III) oxide and the rest is impurities. The total sample mass is 100 g. What mass of iron could be obtained? (Ar: Fe = 56, O = 16)

Working:

• Percentage purity = (80 / 100) x 100 = 80%
• Balanced equation: Fe2O3 + 3CO -> 2Fe + 3CO2
• Mr of Fe2O3 = 160
• Moles of Fe2O3 = 80 / 160 = 0.5 mol
• Mole ratio: 1 Fe2O3 : 2 Fe
• Moles of Fe = 0.5 x 2 = 1 mol
• Mass of Fe = 1 x 56 = 56 g

Answer: 56 g

Examiner note: Use only the mass of pure substance (80 g) in the reacting masses calculation, not the total sample mass (100 g).

Example 5: Reverse problem - finding mass of impure sample needed

Question: A factory needs 28 g of calcium oxide. The limestone it uses is 90% pure calcium carbonate. What mass of limestone is needed? (Ar: Ca = 40, C = 12, O = 16)

Working:

• Balanced equation: CaCO3 -> CaO + CO2
• Mr of CaO = 56, Mr of CaCO3 = 100
• Moles of CaO needed = 28 / 56 = 0.5 mol
• Moles of CaCO3 = 0.5 mol (1:1 ratio)
• Mass of pure CaCO3 = 0.5 x 100 = 50 g
• Mass of limestone = mass of pure CaCO3 / (purity / 100)
• Mass of limestone = 50 / 0.9 = 55.6 g

Answer: 55.6 g

Examiner note: Working backwards: first find the mass of pure substance needed, then divide by the fraction purity to find the total sample mass required.

Common mistakes

1. Confusing purity with yield. Purity is about the sample composition. Yield is about how much product you obtained from a reaction. They are different calculations.
2. Using total sample mass in stoichiometry. Only the mass of the pure substance should be used in mole calculations. The impurities do not react.
3. Putting values the wrong way round. Mass of pure substance goes on top, total sample mass on the bottom.
4. Forgetting to convert the fraction to a percentage. Always multiply by 100 at the end.

When this appears

Percentage purity is Extended tier content and appears on Paper 4. It pairs naturally with percentage yield and builds on reacting masses. See the stoichiometry topic for context.