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3.2.7Stoichiometry

Percentage Yield and Purity — Cambridge IGCSE Chemistry 0620 Revision Notes & Topical Questions

5 min read 6 parts Lesson 28 of 28 Unit quiz

📚 Cambridge IGCSE Chemistry 0620 Hub / Topical Revision
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Curriculum Equation Key قوانين المول والحسابات الكيميائية - Moles & Stoichiometry FormulasFast Exam Solution
n = m / M_r | N = n × N_A (6.022 × 10²³) | M = n / V | V_gas = n × 22.4 L (STP)
Word Equation & Conditions: عدد المولات = الكتلة بالجرام ÷ الكتلة المولية | عدد الجسيمات = عدد المولات × عدد أفوجادرو
Location & Specifications: كيمياء التحصيلي وثانوي وبكالوريا و Cambridge IGCSE Chemistry.
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Section 1 of 6Theoretical and Actual Yield

Theoretical yield
  • Theoretical yield This is the maximum mass of product predicted from the balanced equation, assuming the reaction goes to completion.
Actual yield
  • Actual yield This is the mass of product really obtained at the end of the experiment, measured on a balance.
Percentage yield
  • Percentage yield Percentage yield = (actual yield ÷ theoretical yield) × 100, and it can never be more than 100% for a pure, dry product.
Units
  • Units Actual and theoretical yields must be in the same units, such as grams, before dividing.

Section 2 of 6Why Yield Is Less Than 100%

Why Yield Is Less Than 100%
Incomplete reaction
  • Incomplete reaction Some reactions are reversible, so reactants and products reach a balance and not all reactants turn into product.
Side reactions
  • Side reactions Some reactants may form other unwanted products, reducing the amount of the desired product.
Losses in handling
  • Losses in handling Product is lost when liquids are transferred, solids are filtered, or crystals are left behind on glassware.
Impure reactants
  • Impure reactants If a reactant is not pure, less of it is available to make the product.

Section 3 of 6Calculating Percentage Yield

1
Write the balanced equation and find the relative formula masses of the reactant and the product.
2
Convert the given mass of reactant to moles using moles = mass ÷ Mr.
3
Use the mole ratio in the equation to find the moles of product expected.
4
Convert these moles to a mass using mass = moles × Mr; this is the theoretical yield.
5
Divide the actual yield by the theoretical yield and multiply by 100.

Section 4 of 6Percentage Purity

Pure substance
  • Pure substance A pure substance contains only one compound, while an impure sample contains other substances as well.
Formula
  • Formula Percentage purity = (mass of pure substance ÷ total mass of sample) × 100.
Example
  • Example If 8.0 g of a sample contains 7.2 g of the pure compound, the purity is 7.2 ÷ 8.0 × 100 = 90%.
Why it matters
  • Why it matters Medicines and foods must be of high purity, so a high yield alone does not show the product is suitable.

Section 5 of 6Worked Example

Question

  • 25.0 g of calcium carbonate is heated strongly: CaCO₃(s) → CaO(s) + CO₂(g).
  • The mass of calcium oxide obtained is 11.2 g.
  • Calculate the percentage yield.
  • (Ar: Ca = 40, C = 12, O = 16)

  1. Mr of CaCO₃ = 40 + 12 + 48 = 100, so moles of CaCO₃ = 25.0 ÷ 100 = 0.250 mol.
  2. The ratio CaCO₃CaO is 1 : 1, so 0.250 mol of CaO is expected.
  3. Mr of CaO = 40 + 16 = 56, so the theoretical yield = 0.250 × 56 = 14.0 g.
  4. Percentage yield = (11.2 ÷ 14.0) × 100 = 80%.

Answer The percentage yield of calcium oxide is 80%.

Section 6 of 6Exam Checklist & Tips

3.2.7 – Key Points for the Exam

  • Theoretical yield The maximum possible mass of product calculated from the equation.
  • Actual yield The mass of product actually obtained in an experiment.
  • Percentage yield Actual yield divided by theoretical yield, multiplied by 100.
  • Percentage purity Mass of pure substance divided by total mass of sample, multiplied by 100.
  • Reversible reaction A reaction in which products can change back into reactants, so it does not go to completion.

Common Mistakes

  • Dividing the theoretical yield by the actual yield instead of the other way round.
  • Mixing units, for example using the actual yield in grams and the theoretical yield in kilograms.
  • Using the mass of the reactant instead of the mass of the product when finding the theoretical yield.
  • Thinking a percentage yield above 100% is acceptable; it usually means the product is wet or impure.

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