AQA Chemistry (8462) 4.3.3.1 · Quantitative
Percentage yield
The equation says you can make 8.0 g. The balance says you made 6.2 g. Where did the rest go, and how do you put a number on the gap?
Triple Science · Higher tier
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8.0 g on paper. 6.2 g in the dish.
A student makes copper sulfate crystals from copper oxide and sulfuric acid. From the balanced equation, the reactants they used can make 8.0 g at most. After filtering, crystallising and drying, the balance reads 6.2 g.
If you had to guess, was the missing 1.8 g destroyed in the reaction, or left behind along the way?
The yield bench · Required practical 1, making a salt
Follow 8.0 g of copper sulfate through the practical.
0 of 5 steps run
React
Warm the dilute sulfuric acid and stir in copper oxide until no more dissolves. The black powder left over is the excess.
Commit first. Does any copper sulfate leave the product at this step?
Think again · weighing
“My yield came out at 108%. My technique must be better than perfect.”
Spot the flaw
SuggestThe working is right: 8.64 g weighed, 8.0 g maximum. Suggest why the percentage yield is over 100%.
Sort the reasons
Less product, faster product, or impossible?
Place each statement.
Tap a card, then tap the box it belongs in. Tap a placed card to take it back.
Manufacturers want a high yield. A low yield wastes raw materials, which raises costs, and the wasted reactants become waste the factory has to deal with. Many raw materials are finite, so less waste is more sustainable. In the Haber process only about 15% of the nitrogen and hydrogen becomes ammonia on each pass, so the unreacted gases are recycled, raising the overall yield to about 98%.
Commit first
CalculateTheoretical yield 8.0 g, actual yield 6.2 g. Which working gives the percentage yield?
Equation
Learn it · chemistry has no equation sheetPercentage yield (%) = (actual yield ÷ theoretical yield) × 100
Both yields in the same unit before you divide.
n = m ÷ Mr → mole ratio → m = n × Mr
How you get the theoretical yield from a mass of reactant.
Where the theoretical yield comes from
On the Higher paper you work out the theoretical yield yourself, from the mass of a reactant and the balanced equation. Find the moles of reactant (n = m ÷ Mr), use the ratio in the equation to get the moles of product, then turn that back into a mass (m = n × Mr). If one reactant is in excess, start from the limiting reactant: the one that is used up.
Command words in this lesson
- Calculate
- Convert, formula, insert, fine-tune, answer. The answer carries the % sign.
- Give
- A reason in a few words, no explanation needed: one mark each.
- Suggest
- Apply what you know to the practical in front of you: where was product lost, or what else was weighed?
- Explain
- Link a cause to its effect on the mass collected, with "so".
Key note · AQA 4.3.3.1 (chemistry only)
Percentage yield
- % yield = (actual ÷ theoretical) × 100.
- Usually less than 100% in practice.
- Reasons: reversible reactions, side reactions, practical losses.
- High yield = less waste, lower cost, more sustainable.
- Higher: theoretical yield from a reactant mass: n = m ÷ Mr, then the mole ratio, then m = n × Mr.
Photograph this card. Tomorrow, cover it and say each line first.
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Not sure which mass goes on top of the fraction?
The yield bench uses model losses chosen to add up to the worked example (8.0 g down to 6.2 g); real losses differ from run to run. It treats the product simply as copper sulfate and does not track water of crystallisation.