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1.2.4Physical Chemistry

Reaction Yields — AQA AS Chemistry Revision Notes & Topical Questions

5 min read 6 parts Lesson 11 of 102

📚 AQA AS Chemistry Hub / Topical Revision
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Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
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Part 1 of 6Types of Yield

Theoretical yield
  • Theoretical yield The maximum mass of product that could form, assuming complete reaction and no losses, calculated from the equation.
Actual yield
  • Actual yield The mass of pure product really obtained in the experiment.
Percentage yield
  • Percentage yield Percentage yield = (actual yield ÷ theoretical yield) × 100, and it cannot be above 100% for a pure product.
Purpose
  • Purpose Percentage yield shows how efficiently the reactants have been turned into the desired product.

Part 2 of 6Calculating Percentage Yield

1
Write the balanced equation and find the moles of the limiting reactant.
2
Use the mole ratio to find the moles of product that would form if the reaction went to completion.
3
Convert the moles of product into a theoretical mass using mass = moles × Mr.
4
Divide the actual mass obtained by the theoretical mass.
5
Multiply by 100 and give the answer to a suitable number of significant figures.

Part 3 of 6Why Yields Are Below 100%

Why Yields Are Below 100%
Incomplete reaction
  • Incomplete reaction Reversible reactions do not go to completion, so some reactant is left unreacted at equilibrium.
Side reactions
  • Side reactions Reactants may form unwanted by-products, using up material that would have made the desired product.
Transfer losses
  • Transfer losses Product is lost during transfers, filtration, washing and separation, because some remains on the glassware.
Impure reactants
  • Impure reactants If the reactants are impure, the real amount of reacting substance is smaller than calculated.

Part 4 of 6Improving Yield

Excess reagent
  • Excess reagent A cheap reagent used in excess helps to use up the more valuable one.
Careful technique
  • Careful technique Rinsing apparatus and reducing transfers keeps more product.
Conditions
  • Conditions Changing temperature, pressure or reaction time can increase the yield of some reactions.
Recycling
  • Recycling Unreacted starting material can be recovered and fed back into the process in industry.

Part 5 of 6Worked Example

Question

  • 10.0 g of calcium carbonate is heated: CaCO₃ → CaO + CO₂.
  • 4.20 g of calcium oxide is obtained.
  • Calculate the percentage yield (Mr: CaCO₃ = 100.1, CaO = 56.1).

  1. n(CaCO₃) = 10.0 ÷ 100.1 = 0.0999 mol.
  2. The ratio CaCO₃CaO is 1 : 1, so theoretical n(CaO) = 0.0999 mol.
  3. Theoretical mass of CaO = 0.0999 × 56.1 = 5.60 g.
  4. Percentage yield = 4.20 ÷ 5.60 × 100 = 74.9 %.

Answer Percentage yield = 74.9 %

Part 6 of 6Exam Checklist & Tips

1.2.4 – Key Points for the Exam

  • Theoretical yield The maximum mass of product that could form from the amounts of reactants used.
  • Actual yield The mass of product actually obtained.
  • Percentage yield Actual yield divided by theoretical yield, multiplied by 100.
  • Limiting reactant The reactant that is used up first and so limits the amount of product.

Common Mistakes

  • Using the mass of the excess reactant to calculate the theoretical yield.
  • Writing the percentage yield as theoretical ÷ actual, which gives a value above 100%.
  • Ignoring the mole ratio in the balanced equation when moving from reactant to product.
  • Saying yield is lower because mass is lost; atoms are conserved, and the product is lost in practice.

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