Percentage yield formula and worked examples for A-Level Chemistry
Percentage yield is the ratio of the actual mass of product made in a reaction to the maximum mass you could theoretically make, expressed as a percentage. The formula is: Percentage yield = (actual yield / theoretical yield) × 100. It tells chemists how efficient a reaction is, and it is one of the most reliably tested calculations on AQA A-Level Paper 1 and Paper 3.
This guide covers the formula, how to calculate theoretical yield from a balanced equation, two worked examples, the link to atom economy, and the small errors that lose students easy marks every year.
One formula
Percentage yield = (actual yield / theoretical yield) × 100. Mass or moles works, as long as you use the same unit for both.
Theoretical yield first
Find moles of limiting reactant, scale by the balanced equation, then convert to mass of product.
Atom economy is different
Atom economy measures how much of the reactants ends up in the desired product. Percentage yield measures how well you got there.
The percentage yield formula
The percentage yield formula is short, but each part matters. Percentage yield = (actual yield / theoretical yield) × 100. The actual yield is what you actually made in the lab, measured by weighing or by titration. The theoretical yield is the maximum mass of product the balanced equation predicts if every mole of limiting reactant converted with no losses.
A percentage yield of 100% is impossible in practice. Real reactions lose product to side reactions, incomplete conversions, evaporation, and transfer losses between flasks. A good organic synthesis often runs at 60–80%. Anything below 30% suggests either a problem with the procedure or a reaction that is fundamentally limited.
Mass or moles? Either works You can calculate percentage yield using mass or using moles, as long as you use the same unit on top and bottom. Moles are usually faster because you avoid an extra Mr multiplication, but examiners accept either approach.
Step-by-step method
Every percentage yield calculation follows the same four steps. Write the balanced equation. Find moles of the limiting reactant. Use the stoichiometric ratio to find moles of product. Convert to mass of product (the theoretical yield), then apply the formula.
If the question gives you reactant masses but does not say which is limiting, you have to check both. Calculate moles of each reactant, divide by their stoichiometric coefficients, and the smaller value is your limiting reactant. This is where most marks are lost: Picking the wrong limiting reactant gives a wrong theoretical yield and therefore a wrong percentage yield.
| Step | What to do | Common mistake |
|---|---|---|
| 1 | Write the balanced equation | Forgetting to balance, especially for organic products |
| 2 | Identify the limiting reactant | Assuming the first reactant listed is limiting |
| 3 | Calculate moles of product from the ratio | Using the ratio of reactants, not reactant to product |
| 4 | Convert moles of product to mass using Mr | Using the Mr of the reactant by mistake |
| 5 | Apply (actual / theoretical) × 100 | Forgetting to multiply by 100, or dividing the wrong way around |
Worked example 1: A simple synthesis
A student reacts 4.00 g of magnesium with excess oxygen to form magnesium oxide. They collect 6.20 g of magnesium oxide. Calculate the percentage yield.
Step 1: Balanced equation. 2Mg + O2 → 2MgO.
Step 2: Magnesium is the limiting reactant (oxygen is in excess). Moles of Mg = 4.00 / 24.3 = 0.165 mol.
Step 3: From the 2:2 ratio (which simplifies to 1:1), moles of MgO = 0.165 mol.
Step 4: Theoretical yield in mass = 0.165 × 40.3 = 6.65 g.
Step 5: Percentage yield = (6.20 / 6.65) × 100 = 93.2%.
A yield of 93.2% is high but realistic for a clean reaction with magnesium burning in oxygen. The 6.8% loss is likely from magnesium oxide smoke escaping during the burn.
Worked example 2: An organic synthesis
A student reacts 12.0 g of ethanol (C2H5OH, Mr = 46.0) with excess sodium dichromate and dilute sulfuric acid to make ethanoic acid (CH3COOH, Mr = 60.0). They collect 11.5 g of ethanoic acid. Calculate the percentage yield.
Step 1: The relevant equation simplifies to C2H5OH + 2[O] → CH3COOH + H2O, a 1:1 ratio of ethanol to ethanoic acid.
Step 2: Ethanol is limiting. Moles of ethanol = 12.0 / 46.0 = 0.261 mol.
Step 3: Moles of ethanoic acid (1:1 ratio) = 0.261 mol.
Step 4: Theoretical yield = 0.261 × 60.0 = 15.65 g.
Step 5: Percentage yield = (11.5 / 15.65) × 100 = 73.5%.
Organic yields of 70–80% are normal. The losses come from over-oxidation to side products, evaporation of the volatile organic compound, and product lost when transferring between flasks during the distillation stage.
Why percentage yield is never 100%
Four reasons explain why real yields never hit 100%. Reactions may not go to completion if they are reversible. Side reactions can convert reactants into unwanted products. Product can be lost during purification (filtration, distillation, recrystallisation). The reactants themselves may not be 100% pure.
An AQA examiner question often asks students to suggest two reasons for the actual yield being lower than the theoretical yield. Pick from those four and you cover almost every accepted mark scheme answer. Avoid vague answers like "experimental error" or "human error", which do not get the mark.
Percentage yield versus atom economy Percentage yield measures how much of the predicted product you actually made. Atom economy measures how much of the reactant mass ends up as the desired product (the rest goes to by-products). A reaction can have 100% atom economy but a low yield if the procedure is wasteful, or 30% atom economy but a high yield. The two metrics measure different things.
The atom economy formula for comparison
Atom economy = (Mr of desired product / sum of Mr of all products) × 100. Or equivalently, (Mr of desired product / sum of Mr of all reactants) × 100, since mass is conserved.
Atom economy is a property of the reaction itself, not the procedure. Addition reactions usually have 100% atom economy because there is only one product. Substitution and elimination reactions have lower atom economy because they produce by-products that are not the target molecule. This is why green chemistry favours addition reactions where possible.
Common mistakes that cost easy marks Identifying the wrong limiting reactant. Using the Mr of the reactant instead of the product. Calculating moles using mass / atomic number instead of mass / Mr. Forgetting to multiply by 100. Confusing percentage yield with atom economy. Reporting yields to too many significant figures: Three sig figs is the convention.
What examiners look for in 6-mark questions
On a 6-mark percentage yield question, the marks usually break down as: 1 mark for the balanced equation (if not given), 1 mark for moles of the limiting reactant, 1 mark for moles of product using the correct ratio, 1 mark for theoretical yield in mass, 1 mark for the percentage yield calculation, and 1 mark for the answer to the correct number of significant figures.
Writing out every step is non-negotiable. Even if your final answer is wrong, you can pick up 4 or 5 marks on a 6-mark question with clear working. A correct final answer with no working can only score the final mark. Show your method, line by line, with units.
Key facts to memorise for the exam
- Percentage yield = (actual yield / theoretical yield) × 100
- Always start by writing the balanced equation
- Identify the limiting reactant before calculating theoretical yield
- Use moles, not mass, for the stoichiometric ratio
- Multiply by 100 at the end to convert the ratio to a percentage
- Atom economy = (Mr of desired product / sum of Mr of products) × 100
- Real yields are 60–95% for clean reactions, lower for complex syntheses
- Report final answers to 3 significant figures with the % symbol
