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Percentage purity: finding the pure part

Find the pure substance in an impure sample from gas or titration data, calculate percentage purity, and combine purity with percentage yield.

  • SEC G3 Pure Chemistry 2027

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An impure sample contains the substance you want and other substances. Percentage purity asks how much of the total sample mass belongs to the desired substance.

Pure substance within the whole sample

Percentage purity = (mass of desired pure substance)/(total sample mass) × 100%

For example, 8.00 g of calcium carbonate in a 10.0 g sample gives 8.00/10.0 × 100 = 80.0% purity. The remaining 2.00 g is other material. You do not assign one molar mass to the whole mixture.

This comparison applies to a starting reactant or a collected product. It differs from percentage yield, which compares actual product with the theoretical maximum.

Finding the pure part from reaction data

When the pure mass is unknown, use a reaction that measures the desired substance:

  1. Calculate moles from the measured gas, titre or precipitate.
  2. Use the balanced equation to find moles of the desired substance in the sample.
  3. Convert those moles to its mass.
  4. Divide by the whole sample mass, then multiply by 100%.

This works only if the measurement is attributable to the desired substance. The questions below assume that impurities are inert, the desired substance reacts completely, and the measured gas is collected without loss. Another acid-reacting metal in a zinc sample, or another carbonate in a calcium carbonate sample, could also produce gas and invalidate that simple inference.

For a titration of an aliquot, scale the amount in the measured portion back to the amount in the whole prepared solution before finding the sample’s purity. An aliquot is a measured portion of a solution.

Conversions you will need

Use gas-volume calculations for a measured gas and solution concentration and titration for a titre. Calculate reacting amounts in moles and use the equation ratio before converting to mass.

Guided calculation: a titre measures part of a sample

Guided practice 1

Sodium carbonate purity from a titration

About 8 min

Problem

A 5.30 g impure sodium carbonate sample is dissolved and made up to 250.0 cm³ of solution. A 25.0 cm³ aliquot requires 25.0 cm³ of 0.200 mol dm⁻³ HCl for complete neutralisation:

Na₂CO₃(aq) + 2HCl(aq) → 2NaCl(aq) + H₂O(l) + CO₂(g)

Find the sample’s percentage purity. Use M(Na₂CO₃) = 106 g mol⁻¹. Assume the sodium carbonate dissolves completely, the solution is thoroughly mixed, and the impurities do not react with HCl.

Distinguish the aliquot from the whole sample

Hints

Hint 1: use the reacting ratio

Convert the HCl titre to moles, then divide by two to find sodium carbonate in the aliquot.

Hint 2: scale to the total solution

The aliquot is one-tenth of the total solution volume. Scale its sodium carbonate amount before comparing with the 5.30 g sample.

View solution step by step

  1. Find acid amount

    Method

    Use the HCl concentration and titre in dm³.

    Reason

    The titre measures the amount of acid reacting with the aliquot.

    Working

    n(HCl) = 0.200 × 0.0250 = 0.00500 mol.

  2. Find carbonate in the aliquot

    Method

    Divide the acid amount by two.

    Reason

    One mole of sodium carbonate reacts with two moles of HCl.

    Working

    n(Na₂CO₃)_aliquot = 0.00500/2 = 0.00250 mol.

  3. Scale to the whole sample

    Method

    Multiply by total solution volume divided by aliquot volume.

    Reason

    The mixed solution has the same concentration throughout.

    Working

    n(Na₂CO₃)ₜₒₜₐₗ = 0.00250 × 250.0/25.0 = 0.0250 mol; pure mass = 0.0250 × 106 = 2.65 g.

  4. Calculate purity

    Method

    Compare pure carbonate mass with the entire sample mass.

    Reason

    Purity describes the original 5.30 g sample, not just the aliquot.

    Working

    Percentage purity = (2.65/5.30) × 100 = 50.0%.

Work back from a gas measurement

Examiner practice 2

Percentage Purity (From Gas Moles)

4 marks

Examination question

10.0 g of impure CaCO₃ reacts with excess dilute HCl to produce 0.080 mol CO₂. Assume all the calcium carbonate reacts, all the gas is collected, and the impurities do not react with the acid. Calculate sample purity using M(CaCO₃) = 100 g mol⁻¹ and CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂. [4 marks]

Work back from gas to the pure solid

View solution step by step

  1. Apply the equation ratio

    1 mark

    Method

    Use the 1:1 ratio from CO₂ to pure CaCO₃.

    Reason

    Only the reacting calcium carbonate produces the measured gas.

    Working

    n(CaCO₃)ₚᵤᵣₑ = 0.080 mol.

  2. Find pure calcium carbonate mass

    1 mark

    Method

    Multiply pure amount by molar mass.

    Reason

    Purity requires mass of pure substance in the numerator.

    Working

    mₚᵤᵣₑ = 0.080(100) = 8.0 g.

  3. Form the purity fraction

    1 mark

    Method

    Divide pure mass by total sample mass.

    Reason

    The denominator is the complete 10.0 g impure sample.

    Working

    8.0/10.0 = 0.80.

  4. Report percentage purity

    1 mark

    Working

    Percentage purity = 0.80(100) = 80%.

Challenge 3

Percentage Purity (From Gas Volume at RTP)

Minimal support

Gas-volume transfer

For Zn(s) + H₂SO₄(aq) → ZnSO₄(aq) + H₂(g), a 6.50 g impure zinc sample reacts with excess dilute sulfuric acid and produces 1.20 dm³ H₂ at RTP. Assume all the zinc reacts, all the hydrogen is collected, and the impurities are inert to the acid. Calculate zinc purity using 24 dm³ mol⁻¹ and Aᵣ(Zn) = 65.

Convert gas volume back to pure metal mass

Hints

Hint 1: start from measured gas
At RTP, n(H₂) = 1.20/24.
Hint 2: work back through the equation
The ratio Zn:H₂ is 1:1; convert zinc moles to pure zinc mass before dividing by 6.50 g.
View solution step by step

  1. Convert hydrogen volume to amount

    Method

    Divide RTP gas volume by 24 dm³ mol⁻¹.

    Reason

    The equation comparison requires moles of hydrogen.

    Working

    n(H₂) = 1.20/24 = 0.0500 mol.

  2. Find pure zinc amount

    Method

    Use the 1:1 zinc-to-hydrogen ratio.

    Reason

    Each mole of reacting zinc produces one mole of hydrogen.

    Working

    n(Zn)ₚᵤᵣₑ = 0.0500 mol.

  3. Find pure zinc mass

    Method

    Multiply the pure zinc amount by 65 g mol⁻¹.

    Reason

    Purity compares pure mass with total sample mass.

    Working

    m(Zn)ₚᵤᵣₑ = 0.0500(65) = 3.25 g.

  4. Calculate purity

    Method

    Divide pure zinc mass by 6.50 g and multiply by 100.

    Reason

    The full impure sample is the denominator.

    Working

    Percentage purity = (3.25/6.50) × 100 = 50.0%.

Try independently: purity before yield

Mind stretcher 1: Use both percentages in the correct orderExtension

A 12.0 g impure magnesium carbonate sample has 75.0% purity. It is heated:

MgCO₃(s) → MgO(s) + CO₂(g)

The impurities do not form magnesium oxide. The actual mass of dry, pure MgO collected is 3.60 g. Calculate its percentage yield. Use molar masses in g mol⁻¹: MgCO₃ = 84, MgO = 40.

Show answer

Pure magnesium carbonate mass = 12.0 × 0.750 = 9.00 g.

The equation ratio is 1:1, so theoretical magnesium oxide mass is:

m_theoretical = 9.00/84 × 40 = 4.285714… g

Keep that unrounded value:

Percentage yield = 3.60/((9.00/84) × 40) × 100% = 84.0%

Purity reduces the reacting mass before calculating the maximum product. Yield compares collected product with that maximum afterwards. Using all 12.0 g as magnesium carbonate would overestimate the maximum and underestimate the yield.

Practise and check

Try yield and purity questions in the chemical calculations topic check. Name the numerator and denominator before choosing a formula, and check that the experimental data justify your pure-substance calculation.

Syllabus and review details

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