What you'll learn
This revision guide covers two essential quantitative concepts tested in AQA GCSE Chemistry: percentage yield and atom economy. You'll learn how to calculate these values, understand why actual yields differ from theoretical predictions, and evaluate the efficiency and sustainability of chemical processes. These calculations are fundamental for both Paper 1 and Paper 2, particularly in questions about industrial processes and required practicals.
Key terms and definitions
Theoretical yield — the maximum mass of product that could be formed from given reactants, calculated using balanced equations and relative formula masses.
Actual yield — the mass of product actually obtained from a chemical reaction in practice, always measured experimentally.
Percentage yield — the ratio of actual yield to theoretical yield expressed as a percentage; measures how successful a reaction was.
Atom economy — the percentage of reactant atoms that become useful product atoms in a balanced equation; measures how efficiently atoms are used.
By-products — substances formed in a reaction that are not the desired product; lower atom economy reactions produce more by-products.
Sustainable chemistry — designing chemical processes that minimise waste, use fewer resources, and have reduced environmental impact.
Mr (relative formula mass) — the sum of relative atomic masses in a formula; used to convert between moles and mass.
Limiting reactant — the reactant that is completely used up in a reaction, determining the maximum amount of product formed.
Core concepts
Understanding theoretical yield
The theoretical yield represents the maximum possible mass of product you could obtain if every single reactant atom converted perfectly into product with no losses. To calculate theoretical yield:
- Write a balanced equation for the reaction
- Calculate the Mr of reactant and product
- Use the balanced equation to find the mole ratio
- Convert from mass of reactant to moles, then to moles of product, then to mass of product
For example, when heating calcium carbonate (CaCO₃) to produce calcium oxide (CaO), the balanced equation shows the mole ratio:
CaCO₃ → CaO + CO₂
If you start with 10 g of calcium carbonate, you can calculate the theoretical yield of calcium oxide. The mole ratio is 1:1, so one mole of calcium carbonate produces one mole of calcium oxide.
Mr of CaCO₃ = 40 + 12 + (16 × 3) = 100 Mr of CaO = 40 + 16 = 56
Moles of CaCO₃ = 10 ÷ 100 = 0.1 mol Moles of CaO = 0.1 mol (1:1 ratio) Theoretical yield = 0.1 × 56 = 5.6 g
Calculating percentage yield
In real laboratory and industrial processes, you never obtain the theoretical yield. The percentage yield tells you what proportion of the theoretical maximum you actually achieved:
Percentage yield = (actual yield ÷ theoretical yield) × 100
If you performed the calcium carbonate decomposition above and obtained 4.8 g of calcium oxide:
Percentage yield = (4.8 ÷ 5.6) × 100 = 85.7%
This is a typical result for a school laboratory practical.
Reasons for yields being less than 100%
Understanding why reactions don't achieve 100% yield is important for exam questions:
Incomplete reactions — many reactions are reversible or don't go to completion, leaving some reactants unreacted. This is particularly common in organic chemistry.
Side reactions — alternative reactions may occur producing unwanted products. For example, combustion reactions in limited oxygen produce carbon monoxide alongside carbon dioxide.
Practical losses during separation and purification — transferring liquids between containers, filtering precipitates, and purifying products all result in material loss. Some product remains stuck to filter paper or glassware.
Product loss during handling — evaporation of volatile liquids, spillages, and losses during transfer reduce the recovered mass.
You should never claim percentage yields exceed 100% — if calculations suggest this, it indicates measurement errors or impure products containing water or other contaminants.
Understanding atom economy
Atom economy measures the proportion of reactant atoms that end up in useful products rather than waste by-products. It's calculated using:
Atom economy = (Mr of desired product ÷ sum of Mr of all reactants) × 100
For a single product with no by-products, atom economy is 100%. Most industrial processes produce waste, so chemists aim to maximise atom economy for economic and environmental reasons.
Consider two routes to make ethanol:
Route A: Hydration of ethene C₂H₄ + H₂O → C₂H₅OH
Mr of ethanol (desired product) = (2 × 12) + (6 × 1) + 16 = 46 Sum of Mr of all reactants = 28 + 18 = 46 Atom economy = (46 ÷ 46) × 100 = 100%
Route B: Fermentation of glucose C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
Mr of useful product = 2 × 46 = 92 Mr of all reactants = 180 Atom economy = (92 ÷ 180) × 100 = 51.1%
Route A has higher atom economy because all atoms form useful product, while Route B produces carbon dioxide as waste.
Comparing percentage yield and atom economy
These two concepts measure different aspects of reaction efficiency:
| Aspect | Percentage yield | Atom economy |
|---|---|---|
| What it measures | How successful the actual reaction was | How efficient the reaction design is |
| Based on | Experimental results | The balanced equation only |
| Can be improved by | Better technique, optimised conditions | Choosing different reactions |
| Value depends on | Practical skills and equipment | The chemistry of the reaction |
A reaction can have high atom economy but low percentage yield if it's difficult to carry out in practice. Conversely, a well-executed reaction might achieve high percentage yield but have low atom economy if it produces significant waste.
Industrial applications and sustainability
Modern chemical industries consider both percentage yield and atom economy when designing processes:
Economic considerations:
- Low percentage yield means wasting expensive reactants
- Unreacted starting materials may need separating and recycling
- Poor atom economy creates disposal costs for by-products
Environmental considerations:
- By-products may be toxic or polluting
- Extracting and processing raw materials consumes energy and resources
- Waste disposal contributes to environmental damage
The pharmaceutical industry often uses reactions with poor atom economy but works to recycle or find uses for by-products. The fertiliser industry prioritises high atom economy reactions like the Haber process (making ammonia) where unreacted nitrogen and hydrogen are recycled.
Green chemistry principles encourage:
- Reactions with 100% atom economy where possible
- Using renewable feedstocks
- Finding valuable uses for by-products
- Minimising energy requirements
Calculations with excess reactants
In industry, one reactant is often used in excess to ensure the limiting reactant is completely used. When calculating atom economy or theoretical yield with excess reactants:
- Atom economy calculations use the balanced equation (not the actual masses used)
- Theoretical yield calculations depend on the limiting reactant only
- Excess reactants don't affect theoretical yield calculations
If you're given masses of two reactants, calculate moles of each and compare with the balanced equation ratio to identify the limiting reactant. Only this amount determines your theoretical yield.
Worked examples
Example 1: Percentage yield calculation
Question: Copper oxide reacts with sulfuric acid to produce copper sulfate and water. A student reacted 4.0 g of copper oxide with excess sulfuric acid and obtained 8.5 g of copper sulfate crystals. Calculate the percentage yield. (Ar: Cu = 64, O = 16, S = 32, H = 1) [4 marks]
Solution:
Step 1: Write the balanced equation CuO + H₂SO₄ → CuSO₄ + H₂O
Step 2: Calculate Mr values Mr of CuO = 64 + 16 = 80 Mr of CuSO₄ = 64 + 32 + (4 × 16) = 160 ✓
Step 3: Calculate theoretical yield Moles of CuO = 4.0 ÷ 80 = 0.05 mol Moles of CuSO₄ = 0.05 mol (1:1 ratio) ✓ Theoretical yield = 0.05 × 160 = 8.0 g ✓
Step 4: Calculate percentage yield Percentage yield = (8.5 ÷ 8.0) × 100 = 106.25% ✓
Note: This exceeds 100%, indicating the crystals contained water or the measurements were inaccurate. In an exam, state this would suggest experimental error.
Example 2: Atom economy calculation
Question: Calcium carbonate decomposes on heating to produce calcium oxide and carbon dioxide. Calculate the atom economy for producing calcium oxide. (Ar: Ca = 40, C = 12, O = 16) [3 marks]
Solution:
Step 1: Write the balanced equation CaCO₃ → CaO + CO₂ ✓
Step 2: Calculate relevant Mr values Mr of CaO (desired product) = 40 + 16 = 56 Mr of CaCO₃ (all reactants) = 40 + 12 + (3 × 16) = 100 ✓
Step 3: Calculate atom economy Atom economy = (56 ÷ 100) × 100 = 56% ✓
Example 3: Combined calculation
Question: Iron is extracted from iron oxide in a blast furnace using carbon monoxide:
Fe₂O₃ + 3CO → 2Fe + 3CO₂
(a) Calculate the atom economy for producing iron. (Ar: Fe = 56, C = 12, O = 16) [3 marks] (b) A company produces 980 tonnes of iron from 1500 tonnes of iron oxide. Calculate the percentage yield. [3 marks]
Solution:
(a) Mr of useful product (2Fe) = 2 × 56 = 112 ✓ Mr of all reactants = [56 + 56 + (3 × 16)] + [3 × (12 + 16)] = 160 + 84 = 244 ✓ Atom economy = (112 ÷ 244) × 100 = 45.9% ✓
(b) Theoretical yield calculation: Mr Fe₂O₃ = 160, Mr of 2Fe = 112 Theoretical yield = (1500 ÷ 160) × 112 = 1050 tonnes ✓ Percentage yield = (980 ÷ 1050) × 100 ✓ = 93.3% ✓
Common mistakes and how to avoid them
Using actual masses instead of the balanced equation for atom economy — atom economy is a theoretical calculation based solely on the equation, not experimental masses. Always use Mr values from the balanced equation.
Confusing percentage yield with atom economy — percentage yield compares actual results with theoretical predictions (experiment vs calculation), while atom economy compares useful products with all reactants (equation analysis only).
Forgetting to multiply Mr by the balancing number — if the equation shows 2NH₃, the relevant mass is 2 × 17 = 34, not 17. Always check coefficients in the balanced equation.
Calculating atom economy using only one reactant — the denominator must be the sum of Mr for all reactants. For A + B → C, use (Mr of C) ÷ (Mr of A + Mr of B) × 100.
Not identifying the limiting reactant — when given masses of two reactants, calculate which is limiting before finding theoretical yield. The excess reactant doesn't limit the amount of product.
Rounding too early — keep at least 3 significant figures in intermediate steps and round only the final answer to match the data given (usually 2-3 significant figures).
Exam technique for percentage yield and atom economy
Command word "Calculate" requires working shown clearly. Write the formula, substitute values, then calculate. Typically worth 3-4 marks: 1 mark for formula/method, 1-2 marks for correct substitution/working, 1 mark for answer with units.
Show all calculation steps — even if you can do it mentally, write: formula → substituted values → answer. This gains method marks if your final answer is wrong.
Include appropriate units — percentage yield and atom economy are percentages (%), theoretical and actual yields need mass units (g, kg, tonnes). Missing units can lose marks.
Extended response questions linking yield and economy to sustainability require named concepts: economic costs, environmental impact, waste disposal, and specific examples like recycling unreacted materials or toxicity of by-products.
Quick revision summary
Theoretical yield is the maximum product mass calculated from balanced equations and Mr values. Actual yield is measured experimentally and is always lower due to incomplete reactions, practical losses, and side reactions. Percentage yield = (actual ÷ theoretical) × 100 measures reaction success. Atom economy = (Mr of desired product ÷ sum of Mr of all reactants) × 100 measures efficiency of atom use. High atom economy reduces waste and improves sustainability. Both concepts are crucial for evaluating industrial processes economically and environmentally.