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4.4 The Yield of a Chemical Reaction

GCSE Chemistry · AQA · Chapter 4: Chemical Calculations

A free GCSE Chemistry lesson for AQA (8462) from OBA School: percentage yield, why yields are less than 100%, and empirical formulae from masses or percentages. Watch the video, read the explanation and worked examples, then try the revision questions. You can work through the full lesson, with every video, practice quizzes and printable worksheets with mark schemes, in our free online course.

What you will learn

  • Calculate the percentage yield of a reaction from the actual and maximum yields
  • Give reasons why the actual yield is almost always less than the theoretical yield
  • (Higher tier) use a balanced equation to calculate the theoretical yield of a product
  • Work out an empirical formula from reacting masses or percentage composition

Key words

  • Yield: The amount of product obtained from a reaction.
  • Theoretical (maximum) yield: The mass of product you would get if all of the limiting reactant turned into product and none was lost. Calculated from the balanced equation.
  • Actual yield: The mass of product you really collect.
  • Percentage yield: The actual yield as a percentage of the theoretical yield.
  • Empirical formula: The simplest whole-number ratio of atoms of each element in a compound.

Percentage yield

Even careful chemists rarely collect all the product a reaction could make. The percentage yield compares what you got with the most you could have got.

percentage yield = mass of product actually made ÷ maximum theoretical mass of product × 100

A percentage yield of 100% means no product was lost. A yield of 0% means no product was collected. A value over 100% means the product was not pure or not dry (for example, still wet with water).

Worked example: percentage yield from heating limestone

Heating 50.0 g of calcium carbonate could make a maximum of 28.0 g of calcium oxide. A student collects 22.4 g.
percentage yield = 22.4 ÷ 28.0 × 100 = 80%

Answer: 80%

Why yields are less than 100%

  • The reaction may not go to completion – for example, it is reversible, so some products turn back into reactants.
  • Some product is lost when it is separated from the reaction mixture – for example, left on filter paper or in the beaker, or spilt when transferring.
  • Some reactants may react in unexpected ways – side reactions make other products instead of the one you want.
  • The reactants may not be pure – for example, a rock or ore that contains other minerals. The calculation assumes it is 100% pure, so less product forms than predicted.
  • Some product is lost in handling – it sticks to apparatus, is left behind as dust, or evaporates.

Industry wants high yields: less waste, less reactant needed, lower costs and fewer resources used. Choosing a reaction with a high yield helps make a process more sustainable. Using as little energy as possible also matters: it saves money, conserves fuels and reduces pollution such as carbon dioxide.

Calculating the theoretical yield (Higher tier)

Use the reacting-mass method from 4.2–4.3 to find the maximum mass of product, then compare it with the actual yield.

Worked example: burning magnesium

2Mg + O₂ → 2MgO. A student burns 12 g of magnesium and collects 17 g of magnesium oxide.
moles Mg = 12 ÷ 24 = 0.50 mol, so moles MgO = 0.50 mol
theoretical yield = 0.50 × 40 = 20 g
percentage yield = 17 ÷ 20 × 100 = 85%

Answer: 85%

Worked example: an industrial yield in tonnes

Aluminium is extracted by electrolysis: 2Al₂O₃ → 4Al + 3O₂. A plant uses 204 tonnes of aluminium oxide and collects 97.2 tonnes of aluminium.
Mr Al₂O₃ = 102, so 2 × 102 = 204 parts by mass of Al₂O₃ give 4 × 27 = 108 parts by mass of Al
so 204 tonnes could make a maximum of 108 tonnes (you can work in tonnes – the ratio of masses is the same)
percentage yield = 97.2 ÷ 108 × 100 = 90%

Answer: 90%

Worked example: working backwards from a percentage yield

A reaction has a theoretical yield of 40 g and a percentage yield of 70%.
actual yield = 70 ÷ 100 × 40 = 28 g

Answer: 28 g

Empirical formulae

The empirical formula shows the simplest whole-number ratio of atoms. Ethane is C₂H₆, but its empirical formula is CH₃. To find an empirical formula from experimental data:

  • 1; What to do Write the mass (or %) of each element; Example: 75% C, 25% H by mass C: 75 H: 25
  • 2; What to do Divide each by its Ar; Example: 75% C, 25% H by mass C: 75 ÷ 12 = 6.25 H: 25 ÷ 1 = 25
  • 3; What to do Divide by the smallest answer; Example: 75% C, 25% H by mass C: 6.25 ÷ 6.25 = 1 H: 25 ÷ 6.25 = 4
  • 4; What to do Write the formula (multiply up if you get a half, e.g. 1 : 1.5 → 2 : 3); Example: 75% C, 25% H by mass CH₄

Worked example: an oxide of magnesium

2.4 g of magnesium combines with 1.6 g of oxygen.
Mg: 2.4 ÷ 24 = 0.10 O: 1.6 ÷ 16 = 0.10 ratio 1 : 1

Answer: MgO

Exam tip: Use the Ar of single atoms (O = 16), not O₂ = 32, when finding an empirical formula.

Common mistakes

  • Dividing the wrong way round: it is actual ÷ theoretical, so the answer can never be more than 100% for a pure, dry product.
  • Using the mass of a reactant as the theoretical yield. Work out the maximum mass of product first.
  • Saying the yield is low because “atoms were destroyed”. Mass is conserved; product is lost or not made.
  • Rounding too early in empirical formula questions – 1 : 1.48 is really 1 : 1.5, which becomes 2 : 3.

Revision questions

Try these, then check your answers below each one.

Question 1. A reaction could make a maximum of 12.0 g of product. Only 9.0 g is collected. Calculate the percentage yield.

Answer: 75 %

Working: 9.0 ÷ 12.0 × 100 = 75%.

Question 2. Which of these is a reason why the percentage yield of a reaction is less than 100%? (Choose from: Some product is left behind when it is filtered, Some atoms are destroyed during the reaction, The reactants were weighed on a balance, The product has a high atom economy.)

Answer: Some product is left behind when it is filtered

Working: Product is often lost during separation (on filter paper, in the beaker). Atoms are never destroyed, and atom economy is a different idea.

Question 3. A compound contains 2.4 g of carbon and 0.6 g of hydrogen. (Ar: H = 1, C = 12)<br/>What is its empirical formula?

Answer: CH3

Working: C: 2.4 ÷ 12 = 0.20; H: 0.6 ÷ 1 = 0.60. Ratio 1 : 3, so CH₃.

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