Calculating Percentage Yield and Theoretical Mass Revision Notes | GCSE AQA Higher Chemistry

1. Why Reactions Never Give Perfect Results

You mix 10g of reactants expecting 8g of product. You get 6g. What went wrong? Nothing — this is completely normal in chemistry.

Real reactions never give the perfect amount predicted by equations. Some product gets lost during filtering, some reactant doesn't react completely, and side reactions steal your materials. This is where percentage yield becomes crucial — it tells you how efficient your reaction actually was.

A 100% yield means you got exactly what the equation predicted. A 75% yield means you got three-quarters of the expected amount. Anything above 90% is considered excellent in most reactions.

2. The Percentage Yield Formula

The formula looks straightforward, but there's a catch. You need two values: the actual mass (clear and systematic — you weigh it) and the theoretical mass (harder — you calculate it from the equation). Let's tackle the calculation step first.

3. Calculating Theoretical Mass from Equations

Before you can find percentage yield, you need the theoretical mass. This requires three steps: balance the equation, find the mole ratio, then convert back to mass. Here's the method that works every time.

Worked example: Calculating Theoretical Mass

Calculate the theoretical mass of water produced when 4g of hydrogen reacts with excess oxygen. 2H₂ + O₂ → 2H₂O

1. $2H_2 + O_2 \rightarrow 2H_2O$ — 2 moles H₂ produce 2 moles H₂O (1:1 ratio)
2. Moles of $H_2 = \frac{4g}{2 \text{ g/mol}} = 2 \text{ mol}$ — Using molar mass of H₂ = 2 g/mol
3. Moles of $H_2O = 2 \text{ mol}$ (1:1 ratio) — From balanced equation: 2 mol H₂ → 2 mol H₂O
4. Mass = $2 \text{ mol} \times 18 \text{ g/mol} = 36g$ — Using molar mass of H₂O = 18 g/mol

Answer: 36g of water (theoretical mass)

Notice how we used the mole ratio from the balanced equation. This is the key step that students often miss — the coefficients in the equation tell you exactly how many moles of each substance react together.

4. Complete Percentage Yield Calculations

Now we can combine both skills. You'll calculate the theoretical mass, compare it to the actual mass obtained, then find the percentage yield. This is exactly what you'll face in exam questions.

Worked example: Complete Percentage Yield Calculation

6g of magnesium reacts with excess hydrochloric acid to produce hydrogen gas. The actual mass of hydrogen collected is 0.4g. Calculate the percentage yield. Mg + 2HCl → MgCl₂ + H₂

1. Moles of Mg = $\frac{6g}{24 \text{ g/mol}} = 0.25 \text{ mol}$ — Using Ar(Mg) = 24
2. Moles of $H_2 = 0.25 \text{ mol}$ (1:1 ratio) — From equation: 1 mol Mg → 1 mol H₂
3. Theoretical mass = $0.25 \times 2 = 0.5g$ — Using Mr(H₂) = 2 g/mol
4. $\% \text{ Yield} = \frac{0.4g}{0.5g} \times 100 = 80\%$ — Actual ÷ Theoretical × 100

Answer: 80% yield

An 80% yield is actually quite good for this type of reaction. Some hydrogen probably escaped during collection, and the magnesium might not have been completely pure. These factors always reduce the actual yield below the theoretical maximum.

5. Why Yields Are Less Than 100%

Perfect yields are impossible in real chemistry. Understanding why helps you explain low percentage yields in exam questions and suggests ways to improve reactions in industry.

Incomplete reactions: Not all reactant molecules collide with enough energy to react. Some starting material always remains unreacted, especially in reversible reactions that reach equilibrium.

Side reactions: Reactants might form unwanted products instead of your target compound. These competing reactions steal your starting materials and reduce the main product yield.

Practical losses: Product gets lost during filtering, transferring between containers, or washing. Some remains stuck to glassware or dissolves in washing solvents.

6. Exam Technique and Common Mistakes

Percentage yield questions follow predictable patterns in exams. Master the method and avoid the traps that catch most students. Here's your strategy for full marks.

Mark allocation pattern: Expect 1 mark for the balanced equation, 1 mark for moles calculation, 1 mark for theoretical mass, and 1 mark for the final percentage yield. Show all working to secure method marks even if your final answer is wrong.

Exam precision note: Exam tip: show the balanced equation, calculate the theoretical amount first, then use percentage yield = (actual mass / theoretical mass) × 100., then link it to evidence, data, or an equation. A high-scoring GCSE answer names the particles involved, states what changes, and justifies the conclusion using correct scientific language. Where calculations are used, show each step with units and keep significant figures sensible. Finally, check that terminology matches AQA wording so your answer is unambiguous and mark-scheme friendly.

Percentage yield compares what was actually made with the maximum theoretical amount. A low percentage yield can happen because the reaction is reversible, because side reactions produce other products, or because some product is lost during separation steps such as filtering, washing, or transferring. Students should always balance the equation first, then calculate the theoretical mass before applying the percentage-yield formula.