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5.1.6Additional lessons

Percentage Yield Calculations — IB Chemistry SL Revision Notes & Topical Questions

5 min read 6 parts Lesson 64 of 98

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Part 1 of 6Yield

5.1.6 Percentage Yield Calculations

Percentage yield measures how efficiently a reaction converts reactants into the desired product.

Theoretical yield
  • The maximum mass of product predicted by the balanced equation from the limiting reactant, assuming complete reaction.
Actual yield
  • The mass of product really obtained, which is measured after purification.
Percentage yield
  • Percentage yield = actual yield ÷ theoretical yield × 100.
Range
  • The percentage yield cannot exceed 100% unless the product is impure or wet.

Part 2 of 6Calculation Steps

1
Write the balanced equation and find the limiting reactant.
2
Calculate the moles of the desired product from the moles of the limiting reactant.
3
Convert to a theoretical mass using the molar mass of the product.
4
Divide the actual mass obtained by the theoretical mass and multiply by 100.
5
State the answer to the appropriate number of significant figures.

Part 3 of 6Why Yields Are Below 100%

Why Yields Are Below 100%
Reversible reactions
  • If the reaction reaches equilibrium, some reactants remain and the reaction does not go to completion.
Side reactions
  • Other products may form from the same reactants, reducing the amount of the desired product.
Losses
  • Product is lost during transfer, filtering, washing and recrystallisation.
Impure reactants
  • Starting materials that are not pure contain less of the reacting substance than expected.

Part 4 of 6Improving Yield

Improving Yield
Reaction conditions
  • Altering temperature, pressure or concentration can favour the desired reaction.
Better technique
  • Careful transfers and washing minimise losses of product.
Excess of cheap reactant
  • Using an excess of an inexpensive reactant can drive the reaction closer to completion.
Industrial value
  • Even a small improvement in percentage yield saves large amounts of raw material on an industrial scale.

Part 5 of 6Worked Example

Question

  • 6.00 g of ethanol (Mr 46.08) reacts with excess ethanoic acid to give ethyl ethanoate (Mr 88.12): C₂H₅OH + CH₃COOH → CH₃COOC₂H₅ + H₂O.
  • 8.50 g of ester is obtained.
  • Calculate the percentage yield.

  1. n(ethanol) = 6.00 ÷ 46.08 = 0.1302 mol.
  2. The ratio ethanolester is 1 : 1, so the theoretical amount of ester is 0.1302 mol.
  3. Theoretical mass = 0.1302 × 88.12 = 11.47 g.
  4. Percentage yield = 8.50 ÷ 11.47 × 100 = 74.1%.

Answer Percentage yield = 74.1%

Part 6 of 6Exam Checklist & Tips

5.1.6 – Key Points for the Exam

  • Theoretical yield The maximum amount of product calculated from the limiting reactant.
  • Actual yield The amount of product obtained experimentally.
  • Percentage yield Actual yield as a percentage of theoretical yield.
  • Reversible reaction A reaction in which products can react to re-form the reactants.

Common Mistakes

  • Calculating the theoretical yield from the excess reactant instead of the limiting reactant.
  • Dividing the theoretical yield by the actual yield, which inverts the ratio.
  • Believing that a yield above 100% means the experiment was perfect, rather than that product is impure.
  • Mixing up moles and grams when comparing yields.

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