Percentage Purity – IGCSE Chemistry Definition

Percentage purity tells you what fraction of a sample is the desired substance, as opposed to impurities. It is a Supplement calculation on the 0620 syllabus and often appears alongside mole calculations on Paper 4, especially in the context of ores, limestone, or impure metal samples.

The 0620 definition

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

When purity matters

In real chemistry, reagents and raw materials are rarely 100% pure. For example:

• Limestone quarried from the ground contains impurities such as sand and clay alongside pure CaCO₃
• Metal ores contain the metal compound plus rock and other minerals
• Manufactured chemicals may contain traces of by-products

If you use an impure sample in a reaction, you must account for the fact that only the pure substance reacts.

Using percentage purity in calculations

Type 1: Finding purity from experimental data

If you react an impure sample and measure the product, you can calculate back to find how much pure reactant was present, then calculate purity.

Type 2: Predicting yield from a known purity

If you know the purity, calculate the mass of pure reactant first, then use stoichiometry to find the expected product.

Worked example

A 25.0 g sample of impure calcium carbonate is heated. The pure CaCO₃ in the sample produces 8.8 g of CO₂.

CaCO₃ → CaO + CO₂ (Mr: CaCO₃ = 100, CO₂ = 44)

Moles of CO₂ = 8.8 / 44 = 0.2 mol

From the equation, 1:1 ratio, so moles of CaCO₃ = 0.2 mol

Mass of pure CaCO₃ = 0.2 x 100 = 20.0 g

Percentage purity = (20.0 / 25.0) x 100 = 80%

Worked exam question

A 5.0 g sample of impure iron is reacted with excess hydrochloric acid. The iron in the sample is 80% pure. Calculate the mass of pure iron in the sample and hence the volume of hydrogen gas produced at RTP. (Ar: Fe = 56, molar gas volume = 24 dm³) Fe + 2HCl → FeCl₂ + H₂ (3)

Mark scheme

Mass of pure Fe = 80/100 x 5.0 = 4.0 g [1]; moles of Fe = 4.0 / 56 = 0.0714 mol; from equation, moles of H₂ = 0.0714 mol [1]; volume = 0.0714 x 24 = 1.71 dm³ (or 1714 cm³) [1]

Common exam mistakes

• Confusing percentage purity with percentage yield. Purity is about the sample before the reaction. Yield is about the product after the reaction.
• Forgetting to account for impurities. If a sample is 80% pure, only 80% of the mass is the reactive substance — using the total mass in mole calculations gives an answer that is too high.
• Putting the values upside down. The mass of pure substance goes on top, the total sample mass on the bottom.