Percentage Yield Formula: How to Calculate the Yield of a Reaction
The percentage yield of a reaction is how much product you actually got, as a percentage of how much you should have got:
Actual yield is what you weighed at the end. Theoretical yield is what the balanced equation says you should have got if nothing went wrong. Both must be in the same units — both in grams, or both in moles.
A student expected 12.5 g of copper sulfate and weighed 9.8 g
9.8 ÷ 12.5 = 0.784
0.784 × 100 = 78.4
percentage yield = 78.4%
That division is the easy half. The half that costs marks is the theoretical yield, because the question almost never hands it to you — you have to work it out from the balanced equation, and that means moles, mole ratios and, very often, finding the limiting reactant first.
Most pages on this topic show you the division and skip the hard part. This one does the whole journey, from a balanced equation to a percentage, with every number worked through. It also answers four questions almost nobody covers:
- What do you do when your yield comes out above 100%? It happens in real labs, it is a real exam question, and almost no revision page mentions it.
- How do you work backwards? A factory needs 500 kg of product at 82% yield — how much do you start with?
- What is the difference between percentage yield and atom economy? They are both percentages, both about efficiency, and they measure completely different things.
- Which of the four “percentages” is the question asking for? Yield, atom economy, purity and error all look alike under exam pressure.
The two numbers, in one picture
Everything on this page is about two numbers and one division. Get the two numbers right and the division takes five seconds.
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Actual yield
What you measured. The mass on the balance after you dried and weighed your product. A real number from a real experiment. The question gives it to you.
Theoretical yield
What you should have got if every atom behaved and nothing was lost. You calculate it from the balanced equation. The question almost never gives it to you.
That asymmetry is the whole difficulty of this topic. Students who lose marks here almost never get the division wrong — they get the theoretical yield wrong, usually by forgetting the mole ratio or by using the wrong reactant.
Percentage yield questions are really mole questions wearing a disguise.
One rule before we start, because it causes silent errors: both numbers must be in the same units. Grams on top and grams underneath, or moles on top and moles underneath. Mixing grams with moles gives an answer that looks plausible and is completely wrong, and because it looks plausible nobody notices.
How to find the theoretical yield
This is the part the question is really testing. Five steps, the same five every time.
- Write the balanced equation. If it is not balanced, the mole ratio in step four will be wrong and so will everything after it.
- Turn the reactant mass into moles. moles = mass ÷ Mr. If the question gives you two reactant masses, do this for both.
- Find the limiting reactant. Only needed when you were given two or more amounts. Divide each reactant’s moles by its big number in the equation; the smallest answer is the limiting one.
- Use the mole ratio to get moles of product. Read the big numbers straight off the balanced equation.
- Turn moles of product back into a mass. mass = moles × Mr. That mass is your theoretical yield.
Then, and only then, divide the actual yield by it and multiply by 100.
The words “in excess” are the question quietly telling you to skip step three. A reactant in excess cannot be the limiting one, so the other reactant decides the theoretical yield. “Burned in excess oxygen”, “added excess acid”, “with an excess of magnesium” — all the same instruction.
Worked example 1: magnesium oxide
Start with the simple shape, where one reactant is in excess and there is no limiting reactant to find.
4.8 g of magnesium is burned in excess oxygen. The magnesium oxide collected weighs 7.2 g. Calculate the percentage yield. (Mr: Mg = 24.0, MgO = 40.0)
Step 1 — the balanced equation
2Mg + O₂ → 2MgO
Step 2 — moles of magnesium
4.8 ÷ 24.0 = 0.20 mol
Step 3 — skipped, oxygen is in excess
Step 4 — mole ratio is 2 : 2, which is 1 : 1
moles of MgO = 0.20 mol
Step 5 — back to a mass
0.20 × 40.0 = 8.0 g ← theoretical yield
Finally, the percentage
7.2 ÷ 8.0 × 100
percentage yield = 90.0%
Sense check: 7.2 is a bit less than 8.0, so the answer must be a bit under 100%. It is. If you had got 111% you would have divided the wrong way round.
Do not compare 7.2 g with 4.8 g. They are different substances. Magnesium oxide is heavier than the magnesium it came from, because the oxygen joined on. Dividing product mass by reactant mass gives 150%, which is nonsense, and it is the single most common wrong answer on this topic.
The limiting reactant, and why it decides everything
When a question gives you two reactant masses, it is testing whether you know that one of them runs out first. The one that runs out first is the limiting reactant, and it alone sets the theoretical yield. Once it is gone the reaction stops, however much of the other reactant is left sitting in the flask.
Here is the trap. You cannot just pick whichever reactant you have less of, because the equation does not use them in equal amounts.
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The method is three words long: divide by the number. Take each reactant’s moles, divide by the big number in front of it in the balanced equation, and the smallest answer is the limiting reactant. It works every time, with any number of reactants.
Worked example 2: ammonia
28 g of nitrogen is reacted with 9.0 g of hydrogen to make ammonia. 22 g of ammonia is collected. Calculate the percentage yield. (Mr: N2 = 28.0, H2 = 2.0, NH3 = 17.0)
Step 1 — the balanced equation
N₂ + 3H₂ → 2NH₃
Step 2 — moles of each reactant
N₂ : 28 ÷ 28.0 = 1.00 mol
H₂ : 9.0 ÷ 2.0 = 4.50 mol
Step 3 — divide each by its equation number
N₂ : 1.00 ÷ 1 = 1.00 ← smaller, so LIMITING
H₂ : 4.50 ÷ 3 = 1.50
Step 4 — mole ratio from the LIMITING one. N₂ : NH₃ is 1 : 2
moles of NH₃ = 1.00 × 2 = 2.00 mol
Step 5 — back to a mass
2.00 × 17.0 = 34.0 g ← theoretical yield
Finally, the percentage
22 ÷ 34.0 × 100 = 64.7
percentage yield = 64.7%
Now look at what happens if you pick the wrong reactant. Hydrogen is the one you have more moles of, so it feels like the generous one, and plenty of students use it:
Using hydrogen — wrong
4.50 ÷ 3 × 2 = 3.00 mol
3.00 × 17.0 = 51.0 g
22 ÷ 51.0 × 100 = 43.1%
Hydrogen never ran out, so it could never limit anything.
Using nitrogen — right
1.00 ÷ 1 × 2 = 2.00 mol
2.00 × 17.0 = 34.0 g
22 ÷ 34.0 × 100 = 64.7%
Nitrogen ran out first, so it set the ceiling.
A twenty-one percentage point difference, from one decision made in the first thirty seconds. This is why step three is not optional.
Why yield is never 100%
Exam questions ask you to explain a low yield, not just calculate it, and a vague answer like “some was lost” rarely scores. There are four proper reasons, and good answers name which one applies to the experiment in front of them.
| Reason | What is actually happening | When to use it |
|---|---|---|
| The reaction is reversible | It reaches equilibrium and stops before all the reactant has turned into product. Some product turns back | Any equation written with ⇆. The Haber process is the classic one |
| Side reactions | Some of your reactant made something else instead. The atoms went somewhere, just not where you wanted | Organic synthesis especially, where several products are possible |
| Losses in handling | Product left on the filter paper, stuck to the beaker, lost in transfers, or washed away during purification | Any practical with filtering, washing or recrystallising |
| Impure reactants | You weighed 10 g but only 9 g of it was the substance, so there was less to react than you thought | When the question mentions purity, or an old or damp reagent |
In a written answer, name the reason and then tie it to the method. “Some product was lost when the mixture was filtered, because crystals stayed on the filter paper” scores where “some was lost” does not.
What if your yield is over 100%?
Almost no revision page mentions this, and it is a standard exam question: “A student calculates a percentage yield of 107%. Suggest why.”
Start from the principle. A yield above 100% is impossible. You cannot make more product than the atoms you started with allow. Mass is conserved. So a figure above 100% is not a brilliant result — it is a message that something is wrong, and your job is to say what.
- The product was not dry. This is the answer the examiner wants nine times out of ten. Water left in the crystals is extra mass on the balance that was never your product. The fix is to dry it and reweigh until the mass stops changing.
- The product was impure. Unreacted starting material or a side product got weighed along with the thing you wanted.
- An arithmetic error. Usually dividing theoretical by actual instead of the other way round, or using the reactant mass as the theoretical yield.
- The balance was not zeroed, or the mass of the container was not subtracted.
A student makes copper sulfate crystals. The theoretical yield is 14.0 g. They weigh their crystals straight after filtering and get 15.0 g, a yield of 107%. Explain what has gone wrong and what they should do.
What went wrong: the crystals were still wet. The 15.0 g includes water from the filtering, which is not copper sulfate, so the measured mass is too high.
What to do: dry the crystals — in a warm oven, a desiccator, or between filter papers — then weigh them again. Repeat until two weighings agree. That is called drying to constant mass, and using that phrase is usually worth the mark on its own.
Working backwards, the way industry does it
In a factory nobody asks “what yield did I get”. They ask “I need this much product, so how much do I have to start with”. That is the same formula, rearranged.
Or, more simply: divide the amount you need by the yield as a decimal.
A plant needs to produce 500 kg of a drug. The process runs reliably at 82% yield. What theoretical yield must the batch be designed for?
82% as a decimal is 0.82
500 ÷ 0.82 = 609.75…
theoretical yield needed = 610 kg
So they must scale the reactants up for a 610 kg batch to end up with 500 kg. Note that you divide by 0.82. A very common error is to multiply, giving 410 kg, which is less than the amount needed — an answer that cannot possibly be right, and a free check that you have divided the correct way.
This is also why yield matters commercially. At 82% yield, every tonne of raw material buys 820 kg of saleable product and 180 kg of waste to dispose of. Lifting the yield by five points is worth real money, which is why so much industrial chemistry is about catalysts, temperature and pressure rather than new reactions.
Percentage yield against atom economy
Both are percentages. Both sound like efficiency. They measure completely different things, and questions deliberately put them side by side.
Percentage yield
How much of the possible product you actually managed to collect. It is about your practical skill and the conditions.
Measured after the experiment.
A tidier chemist gets a higher one.
Atom economy
How much of the reactant mass ends up in the product you wanted, rather than in a by-product. It is about the reaction you chose.
Calculated before you do anything.
Only a different equation changes it.
The sharpest way to hold them apart: you could do an experiment perfectly, lose not one crystal, and still have a terrible atom economy, because half the mass you started with was always going to leave as a by-product. And you could pick a reaction with 100% atom economy and still get a 40% yield because you spilled it.
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Working out atom economy
Remember to multiply each Mr by its big number in the equation.
Some textbooks write the bottom as “total Mr of all the products” instead. Both give exactly the same answer, because the mass of the reactants equals the mass of the products in a balanced equation. Use whichever your specification uses.
Copper can be made from copper(II) oxide in two ways. Calculate the atom economy of each and say which is greener.
Route A: CuO + H2 → Cu + H2O
Route B: 2CuO + C → 2Cu + CO2
(Mr: Cu = 63.5, CuO = 79.5, H2 = 2.0, C = 12.0)
Route A — reactants are CuO and H₂
wanted : Cu = 63.5
all reactants : 79.5 + 2.0 = 81.5
63.5 ÷ 81.5 × 100 = 77.9
Route A atom economy = 77.9%
Route B — note the 2s
wanted : 2 × 63.5 = 127.0
all reactants : (2 × 79.5) + 12.0 = 171.0
127.0 ÷ 171.0 × 100 = 74.3
Route B atom economy = 74.3%
Route A is greener on atom economy, at 77.9% against 74.3%, because water is a lighter by-product than carbon dioxide. A full answer would add that route A also avoids releasing CO2, while noting that hydrogen is more expensive and harder to store than carbon. Atom economy is one factor in “greener”, not the whole of it.
The 100% case worth memorising
Addition reactions always have an atom economy of 100%, because everything on the left joins into one product and nothing is left over. Making ethanol by adding steam to ethene is the standard example:
C₂H₄ + H₂O → C₂H₅OH
There is only one product, so everything counts
wanted : C₂H₅OH = 46.0
all reactants : 28.0 + 18.0 = 46.0
46.0 ÷ 46.0 × 100 = 100
atom economy = 100%
If an equation has only one product, you can write 100% without doing any arithmetic. Spotting that in an exam saves two minutes. It does not mean the yield will be 100% — the ethene and steam reaction runs at about 5% conversion per pass through the reactor, which is exactly why the two measures are different.
The four percentages students mix up
Under time pressure these blur together, because all four are a small number over a big one times 100. Read the question for which pair of numbers it is asking you to compare.
| What it is called | Top of the fraction | Bottom of the fraction | It tells you |
|---|---|---|---|
| Percentage yield | actual yield | theoretical yield | how much of the possible product you got |
| Atom economy | Mr of the wanted product | total Mr of all reactants | how much reactant mass is not wasted as by-product |
| Percentage purity | mass of the pure substance | mass of the whole sample | how much of your sample is the real thing |
| Percentage error | the uncertainty on the instrument | the reading you took | how precise your measurement was |
Two quick tells. If the question gives you two masses of the same substance, it is yield or purity. If it gives you an equation and no masses at all, it has to be atom economy, because that is the only one you can do without weighing anything.
Practice questions
Seven questions, getting harder. The answers are hidden — work each one on paper first, then tap to check. Reading a solution feels like learning and is not.
Use these where you need them: H = 1.0, C = 12.0, O = 16.0, S = 32.1, Ca = 40.0, Zn = 65.4.
1. A student making aspirin expects a theoretical yield of 5.8 g and collects 4.6 g. Calculate the percentage yield.
4.6 ÷ 5.8 = 0.7931…
× 100 = 79.31…
percentage yield = 79.3%
2. An ester synthesis has a theoretical yield of 16.0 g. The student isolates 11.2 g. Calculate the percentage yield.
11.2 ÷ 16.0 = 0.70
percentage yield = 70.0%
3. 25.0 g of calcium carbonate is heated: CaCO3 → CaO + CO2. 11.2 g of calcium oxide is collected. Calculate the percentage yield.
Mᵣ : CaCO₃ = 100.0, CaO = 56.0
moles CaCO₃ = 25.0 ÷ 100.0 = 0.250
Ratio is 1 : 1
moles CaO = 0.250
theoretical = 0.250 × 56.0 = 14.0 g
11.2 ÷ 14.0 × 100 = 80.0
percentage yield = 80.0%
Note you cannot compare 11.2 with 25.0 — different substances. That would give 44.8%, which is the common wrong answer.
4. 4.0 g of hydrogen reacts with 16.0 g of oxygen: 2H2 + O2 → 2H2O. 15.3 g of water is collected. Calculate the percentage yield.
Two masses given, so find the limiting reactant
H₂ : 4.0 ÷ 2.0 = 2.00 → ÷ 2 = 1.00
O₂ : 16.0 ÷ 32.0 = 0.50 → ÷ 1 = 0.50 ← LIMITING
Ratio O₂ : H₂O is 1 : 2
moles H₂O = 0.50 × 2 = 1.00
theoretical = 1.00 × 18.0 = 18.0 g
15.3 ÷ 18.0 × 100 = 85.0
percentage yield = 85.0%
Hydrogen looks generous at 2.00 mol against 0.50 mol, but after dividing by the equation numbers oxygen is the one that runs out.
5. A student weighs their crystals immediately after filtering and calculates a percentage yield of 112%. Explain why, and say what they should do.
A yield above 100% is impossible, so the measured mass must be too high. The crystals were still wet — the water left from filtering was weighed along with the product.
They should dry the crystals, in a warm oven or a desiccator, and reweigh. Repeat until two successive weighings agree, which is called drying to constant mass.
6. A company needs 240 kg of product from a process that runs at 75% yield. What theoretical yield must the batch be designed for?
Divide by the yield as a decimal, do not multiply
240 ÷ 0.75 = 320
theoretical yield needed = 320 kg
Check: 75% of 320 is 240. Correct. Multiplying would give 180 kg, which is less than the amount needed and therefore impossible.
7. Hydrogen is made by reacting zinc with sulfuric acid: Zn + H2SO4 → ZnSO4 + H2. Calculate the atom economy for making hydrogen, and comment.
Mᵣ H₂SO₄ = 2.0 + 32.1 + 64.0 = 98.1
wanted : H₂ = 2.0
all reactants : 65.4 + 98.1 = 163.5
2.0 ÷ 163.5 × 100 = 1.22
atom economy = 1.2%
Dreadful. Over 98% of the mass you started with ends up as zinc sulfate, not hydrogen. The reaction is fine for making a test tube of gas in a lab and hopeless for making hydrogen industrially, which is why nobody does it that way. Notice you did not need a single mass measurement to work this out.
Where the marks get lost
- Comparing the product mass with the reactant mass. They are different substances. The theoretical yield has to be calculated, not borrowed from the question.
- Dividing the wrong way round. Actual goes on top. If your answer is over 100% and the product was dry, you have flipped it.
- Ignoring the limiting reactant when two masses are given, or picking the one you have more moles of without dividing by the equation numbers first.
- Forgetting the mole ratio. If the equation says 1 : 2, the moles of product are double, not equal.
- Mixing grams and moles in the same fraction. Both numbers must be the same kind of thing.
- Using an unbalanced equation. Everything downstream is then wrong, and the working earns nothing.
- Answering “some was lost” and stopping. Name the reason and tie it to the method described in the question.
- Multiplying instead of dividing when working backwards from a target mass.
- Confusing atom economy with percentage yield. If the question gives no masses, it cannot be asking for yield.
Before you write the final answer, ask: is it between 0 and 100? If not, either you divided the wrong way round or your product was wet — and only one of those is something you can fix with a calculator.
The whole thing on one card
Photograph this before the paper.
Yield and atom economy — quick reference
- Actual on top, theoretical underneath, both in the same units, then × 100.
- The theoretical yield is calculated, never the reactant mass from the question.
- Two masses given? Find the limiting reactant. Divide each one’s moles by its number in the equation; smallest wins.
- “In excess” means skip that step — that reactant cannot be limiting.
- Over 100% means the product was wet, impure, or you divided upside down.
- Dry to constant mass is the phrase examiners want for fixing a wet product.
- Atom economy needs no masses at all, only the equation.
- One product only → atom economy is 100%. All addition reactions.
- Yield is about your lab work. Atom economy is about your choice of reaction.
- Final check: is the answer between 0 and 100?
This topic sits on top of moles, so if the mole steps are what slowed you down rather than the yield itself, that is the thing to fix first. Our IB Chemistry revision notes work through moles and reacting masses before they get here.
Frequently asked questions
What is the formula for the yield of a reaction?
How do you calculate percentage yield?
What is the theoretical yield?
What is the actual yield?
Can percentage yield be more than 100%?
Why is percentage yield never 100%?
What is a good percentage yield?
What is the limiting reactant?
What does “in excess” mean in a yield question?
What is the difference between percentage yield and atom economy?
How do you calculate atom economy?
Which reactions have 100% atom economy?
How do you find the mass you need to start with for a given yield?
Do you use mass or moles in the percentage yield formula?
Why does a multi-step synthesis have such a low overall yield?
Is percentage yield the same as percentage purity?
Losing marks on the mole steps rather than the yield?
That is where almost everyone loses them. The division at the end is never the problem — it is getting to the theoretical yield. Our tutors teach IB and A-level chemistry and will work through your own past papers rather than a generic worksheet.
Book a free 30-minute meetingRevising quantitative chemistry? Start with our IB revision notes, or read how to calculate relative atomic mass, which is the step before this one. Writing a chemistry investigation? See our guide to IB internal assessments, or read about IB Diploma tutoring.