Percentage purity is one of those topics that looks simple on paper but trips learners up in the exam room — usually because the question hides an extra step before you can even use the formula. This guide walks through the concept from the ground up and works through four increasingly demanding examples.
What is percentage purity?
No sample of a chemical is ever perfectly pure in practice. A sample of limestone, for instance, is mostly calcium carbonate, but it also contains sand, clay, and other minerals. Percentage purity tells you what fraction of a sample is actually the substance you’re interested in.
The formula is:
% purity = (mass of pure substance ÷ mass of impure sample) × 100
The tricky part is rarely the formula itself — it’s working out the mass of pure substance when the question only gives you a reaction or a titration to work with. That’s where stoichiometry comes in.
Example 1: Straightforward purity
Question: A sample of impure NaCl has a mass of 25 g. Analysis shows it contains 21,25 g of pure NaCl. Calculate the percentage purity of the sample.
Solution:
% purity = (21,25 ÷ 25) × 100 = 85%
This is the only step you get for free — most exam questions build in a calculation before you reach this point.
Example 2: Finding the mass of pure substance from a reaction
Question: A 12,5 g sample of impure calcium carbonate (CaCO₃) is reacted with excess hydrochloric acid. The reaction produces 0,10 mol of CO₂ gas. Calculate the percentage purity of the CaCO₃ sample.
Solution:
Step 1 — Write the balanced equation:
CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g)
Step 2 — Use the mole ratio. CaCO₃ : CO₂ = 1 : 1, so moles of CaCO₃ reacted = 0,10 mol.
Step 3 — Convert to mass using M(CaCO₃) = 100 g·mol⁻¹:
m = n × M = 0,10 × 100 = 10 g
Step 4 — Apply the purity formula:
% purity = (10 ÷ 12,5) × 100 = 80%
Example 3: Working backwards — finding the mass of impure sample needed
Question: A learner needs 5,3 g of pure Na₂CO₃ for an experiment, but only has access to an impure sample that is 75% pure. What mass of the impure sample must be weighed out?
Solution:
Rearrange the purity formula to solve for the mass of impure sample:
mass of impure sample = (mass of pure substance ÷ % purity) × 100
= (5,3 ÷ 75) × 100 = 7,07 g
This variation catches learners who only ever memorise the formula one way round. Always ask what quantity the question is actually asking for before substituting numbers.
Example 4: Purity via titration
Question: A 2,00 g sample of impure NaOH is dissolved in water and made up to 250 cm³. A 25 cm³ portion of this solution is titrated against 0,100 mol·dm⁻³ HCl, requiring 32,5 cm³ to reach the endpoint. Calculate the percentage purity of the NaOH sample.
Solution:
Step 1 — Moles of HCl used:
n(HCl) = c × V = 0,100 × 0,0325 = 3,25 × 10⁻³ mol
Step 2 — Mole ratio. NaOH + HCl → NaCl + H₂O, so n(NaOH) in the 25 cm³ portion = 3,25 × 10⁻³ mol.
Step 3 — Scale up to the full 250 cm³ solution (×10):
n(NaOH) total = 3,25 × 10⁻³ × 10 = 3,25 × 10⁻² mol
Step 4 — Convert to mass using M(NaOH) = 40 g·mol⁻¹:
m = 3,25 × 10⁻² × 40 = 1,30 g
Step 5 — Apply the purity formula:
% purity = (1,30 ÷ 2,00) × 100 = 65%
This is the level of question typically reserved for the final marks in an NSC Paper 2 stoichiometry section — it combines titration calculations, mole ratios, and the purity formula in a single problem.
Common mistakes to avoid
- Forgetting to convert volumes from cm³ to dm³ before using c = n/V.
- Using the mass of the impure sample where the mass of pure substance is required, or vice versa.
- Skipping the mole ratio step and assuming a 1:1 relationship without checking the balanced equation.
- Forgetting to scale up from an aliquot (a portion of the solution) to the full volume prepared, as in Example 4.
Quick reference
% purity = (mass of pure substance ÷ mass of impure sample) × 100
Whenever a question doesn’t hand you the mass of pure substance directly, look for a reaction, a titration, or any other route that lets you calculate moles first — then convert to mass, and only then apply the formula above.