Mini Chemistry
Part of Mini EducationSwitch subject

Percentage Yield & Percentage Purity

Percentage yield and purity: use the formulas correctly, and explain why yield/purity are <100% (losses, side reactions, incomplete reaction, impurities).

  • SEC G3 Pure Chemistry 2027
On this page

Learning objectives

  • calculate % yield and % purity.

Percentage yield and percentage purity compare different pairs of quantities: theoretical vs actual, and pure substance vs total sample.

1. Definition

A. Theoretical Yield and Actual Yield

  • Theoretical yield is the maximum amount of product calculated from the balanced equation (using the limiting reactant).
  • Actual yield is the amount of product actually obtained in the lab.

B. Percentage Yield

Percentage yield compares actual yield with theoretical yield:

Percentage yield = (Actual yield)/(Theoretical yield) × 100

C. Percentage Purity

Percentage purity is the fraction of the sample that is the desired pure substance:

Percentage purity = (Mass of pure substance)/(Total mass of impure sample) × 100

2. Key Ideas

A. Yield: Same Substance, Same Units

Actual and theoretical yields must be for the same product and in the same unit (g with g, mol with mol, dm³ with dm³ at the same conditions).

B. Purity: Use Chemistry to Find the Pure Part

You almost never get told “mass of pure substance” directly. You calculate it from reaction data (gas volume, titration, mass of precipitate, etc.).

C. Yield Can Be Over 100% (And That Is a Red Flag)

If you get >100%, you have impure product, wet crystals, incomplete drying, or a calculation/unit mistake.

Recall: Limiting reactant decides theoretical yield

Theoretical yield must be based on the limiting reactant. Revise this step if needed: Limiting Reactant

3. Detailed Explanations

Quick Recall (do not mix these up)
  • Yield compares actual product to theoretical product: %yield = actual/theoretical × 100.
  • Purity compares pure substance to sample mass: %purity = (mass of pure)/(mass of sample) × 100.
  • Theoretical yield comes from the balanced equation (and the limiting reactant).
  • Yield above 100% usually means wet/impure product or a unit mistake.

A. Why Actual Yield Is Usually Lower

  • Reaction does not go to completion (reversible reactions, equilibrium).
  • Side reactions form other products.
  • Loss of product during filtration, transfer, crystallisation, washing, drying.
  • Product remains dissolved (especially in crystallisation).

B. How Purity Is Found From Reaction Data (Standard Pattern)

  1. Use the data to calculate moles (or mass) of product formed.
  2. Convert that to moles (or mass) of the pure substance in the impure sample using the equation.
  3. Use the purity formula.

C. What Purity Is Not

Purity is not “percentage yield”. Purity is about the starting sample; yield is about the product obtained.

4. Common Mistakes

A. Yield Mistakes

  • Using actual yield in moles but theoretical yield in grams.
  • Calculating theoretical yield from the wrong reactant (ignoring limiting reactant).
  • Forgetting to convert cm³ to dm³ in gas/solution questions.
  • Getting >100% and accepting it without explanation.

B. Purity Mistakes

  • Using total sample mass as if it all reacted (it didn’t; impurities might not react).
  • Trying to assign an Mᵣ to the whole impure sample; use the molar mass of the pure reacting substance.

5. Exam Tips

Two-Line Setup That Scores Marks

Write both lines before substituting numbers:

  1. “theoretical yield of product = …” (from stoichiometry)
  2. “percentage yield = actual ÷ theoretical × 100”
  • For purity, write: “mass of pure substance in sample = …” before you calculate the percentage.
  • If you get >100% yield, state the reason (wet/impure product, incomplete drying) instead of ignoring it.

6. Worked Examples

Modelled example 1

Percentage Yield (Direct)

Core

Problem

The theoretical yield of iron is 50.0 g and the actual mass collected is 42.5 g. Calculate the percentage yield.
Study the worked solution

  1. Identify actual and theoretical quantities

    Method

    Use 42.5 g as actual yield and 50.0 g as theoretical yield.

    Reason

    Yield compares collected product with the calculated maximum for the same substance and unit.

    Working

    Actual = 42.5 g; theoretical = 50.0 g.

  2. Form the yield percentage

    Method

    Divide actual by theoretical and multiply by 100.

    Reason

    The actual amount is the achieved fraction of the maximum.

    Working

    Percentage yield = (42.5/50.0) × 100 = 85.0%.

Guided practice 2

Percentage Yield (Theoretical Yield From Stoichiometry)

About 8 min

Problem

For 2Mg(s) + O₂(g) → 2MgO(s), 4.80 g Mg reacts with excess oxygen and produces 7.20 g MgO. Calculate percentage yield. Use Aᵣ: Mg = 24, O = 16.

Find the maximum before comparing

Hints

Hint 1: calculate theoretical product
n(Mg) = 4.80/24 and the equation gives a 1:1 Mg:MgO amount ratio.
Hint 2: compare like masses
M(MgO) = 40 g mol⁻¹; convert theoretical moles to grams before using actual/theoretical.
View solution step by step

  1. Calculate magnesium amount

    Method

    Divide magnesium mass by molar mass.

    Reason

    The balanced equation relates reactants and products in moles.

    Working

    n(Mg) = 4.80/24 = 0.200 mol.

  2. Find theoretical magnesium oxide

    Method

    Use the 2:2 amount ratio and convert product amount to mass.

    Reason

    Excess oxygen means magnesium determines the maximum product.

    Working

    n(MgO) = 0.200 mol; m_theoretical = 0.200(40) = 8.00 g.

  3. Calculate percentage yield

    Method

    Divide actual 7.20 g by theoretical 8.00 g.

    Reason

    Both values now refer to the same product in the same unit.

    Working

    Percentage yield = (7.20/8.00) × 100 = 90.0%.

Common misconception 3

Yield Is Not Purity

Find and correct the mistake

Learner response

A 10.0 g impure calcium carbonate sample contains 8.0 g pure CaCO₃. A learner calculates (8.0/10.0) × 100 = 80% and calls it percentage yield. Explain why the label is wrong and state the correct comparison.

Name the numerator and denominator

Correct name

View solution step by step

  1. Identify the quantities

    Method

    Recognise 8.0 g as pure substance and 10.0 g as total impure sample.

    Reason

    Neither value is an actual or theoretical product yield.

    Working

    Pure part / total sample = 8.0/10.0.

  2. Correct the interpretation

    Method

    Name the calculation percentage purity.

    Reason

    Purity measures the desired substance within a sample; yield measures actual product against a theoretical maximum.

    Working

    Percentage purity = (8.0/10.0) × 100 = 80%.

Examiner practice 4

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. Calculate sample purity using Mᵣ(CaCO₃) = 100 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 5

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 produces 1.20 dm³ H₂ at RTP. 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%.

7. Mind Stretchers

Mind stretcher 1: Combine Purity and Yield (Two Different Percentages)Extension

An impure sample of magnesium carbonate, MgCO₃, of mass 12.0 g is heated: MgCO₃(s) → MgO(s) + CO₂(g) The sample has 75% purity of MgCO₃. The actual mass of MgO collected is 3.60 g. Calculate the percentage yield of MgO. (Use Mᵣ(MgCO₃) = 84, Mᵣ(MgO) = 40.)

Show Answer

Mass of pure MgCO₃ = 0.75 × 12.0 = 9.00 g.

n(MgCO₃) = 9.00/84 = 0.107 mol; From equation: MgCO₃ → MgO (1:1); m(MgO)_theoretical = 0.107 × 40 = 4.28 g
Percentage yield = 3.60/4.28 × 100 = 84.1%

Mind stretcher 2: Spot the Error (Over 100% Yield)Extension

A student calculated a percentage yield of 112% for crystallising CuSO₄ * 5H₂O. State the most likely reason and the correct action.

Show Answer

Most likely reason: crystals were not dried properly (wet crystals contain water), or product contains impurities.

Correct action: dry the crystals (e.g., blot between filter papers, leave to dry), then re-weigh and recalculate. Do not accept >100% without explanation.

8. Quiz

Quiz Time!

Ready to test your knowledge? Try the quiz to practice separating yield vs purity without mixing them up.

Go to Quiz Page