Percentage Yield and Limiting Reagent
A reaction almost never gives you everything the equation promises. Percentage yield is how much you actually got, as a fraction of how much you should have got.
This page covers how to find the limiting reagent, how to work out the theoretical yield from it, the percentage yield formula, and a precipitation practical that lets a class measure a real one. Plus why a yield over 100 % is impossible and why it happens anyway.
Start with the limiting reagent, because everything else depends on it
When two reactants are mixed, one usually runs out first. That one is the limiting reagent, and it decides how much product you can possibly make. The other is in excess and some of it will be left over.
The shortcut that fails
You cannot find the limiting reagent by picking the reactant with fewer moles.
That works when the mole ratio happens to be 1:1 and it fails the moment it is not. Most revision pages teach the shortcut and never mention the exception.
The method that always works
- Convert both masses to moles. n = m ÷ M
- Look at the balanced equation and work out how many moles of the second reactant you would need to use up all of the first
- Compare that with how much you actually have. If you have less than you need, that reactant is limiting
Worked example, where the shortcut gives the wrong answer
Methanol is made industrially from carbon monoxide and hydrogen:
CO(g) + 2H2(g) → CH3OH(g)
56,0 g of carbon monoxide is mixed with 6,00 g of hydrogen.
- n(CO) = 56,0 ÷ 28,0 = 2,00 mol
- n(H2) = 6,00 ÷ 2,00 = 3,00 mol
- To use up 2,00 mol of CO you would need 4,00 mol of H2, because the ratio is 1 : 2
- Only 3,00 mol of H2 is available
- Hydrogen is limiting. Carbon monoxide is in excess.
Notice what the shortcut would have done. Carbon monoxide has fewer moles, 2,00 against 3,00, so the shortcut picks CO. That is the wrong answer.
Theoretical, actual and percentage yield
| Term | What it is |
|---|---|
| Theoretical yield | The maximum mass of product the equation allows, assuming the limiting reagent is completely converted |
| Actual yield | What you actually get out of the flask and onto the balance |
| Percentage yield | The actual yield as a percentage of the theoretical |
percentage yield = (actual yield ÷ theoretical yield) × 100 %
The actual yield is always lower. Reactants are not perfectly pure, side reactions happen, the reaction may not go to completion, and some product is simply left behind on the glassware.
Finishing the methanol example
Hydrogen is limiting at 3,00 mol, and the ratio H2 : CH3OH is 2 : 1.
- n(CH3OH) = 3,00 ÷ 2 = 1,50 mol
- M(CH3OH) = 32,0 g·mol-1
- Theoretical yield = 1,50 × 32,0 = 48,0 g
If the reaction actually produced 40,0 g:
percentage yield = 40,0 ÷ 48,0 × 100 = 83,3 %
A second example, with one reactant in excess
14,0 g of nitrogen reacts with excess hydrogen and produces 12,8 g of ammonia.
N2(g) + 3H2(g) → 2NH3(g)
- n(N2) = 14,0 ÷ 28,0 = 0,500 mol
- The ratio N2 : NH3 is 1 : 2, so n(NH3) = 1,00 mol
- Theoretical yield = 1,00 × 17,0 = 17,0 g
- percentage yield = 12,8 ÷ 17,0 × 100 = 75,3 %
"Excess hydrogen" is doing work in that question. It tells you nitrogen must be the limiting reagent, so you can skip the comparison entirely.
Where this fits in the curriculum
| Subject | Physical Sciences |
|---|---|
| Grade | 11 |
| Term | 2 |
| Topic | Quantitative aspects of chemical change |
| Marks | 40 |
The practical, and it is ours rather than the textbook's
The Grade 11 textbook has no percentage yield practical. Chapter 8 covers the topic as theory and worked examples across two pages, then moves on to empirical formula. There is no experiment anywhere in the chapter.
So the method below is one we have designed, and we would rather say so than let a teacher assume it carries the textbook's authority.
It is worth doing. Percentage yield stays abstract until you have actually lost some. A learner who calculates 83 % from numbers on a page has done arithmetic. A learner who weighs 1,08 g where the theory says 1,25 g, and can point at where the other 0,17 g went, has understood it.
The reaction
Na2CO3(aq) + CaCl2(aq) → CaCO3(s) + 2NaCl(aq)
Sodium carbonate solution plus calcium chloride solution gives a white precipitate of calcium carbonate, which is chalk. Washing soda and a drying salt, and the product is harmless.
Apparatus
| Item | Qty |
|---|---|
| Balance reading to 0,01 g | 1. Essential |
| Filter funnel, glass 75 mm | 1 |
| Filter paper, 90 mm | 1 per group |
| Beakers, 100 ml and 250 ml | 2 each |
| Measuring cylinder, 25 ml | 2 |
| Wash bottle and stirring rod | 1 each |
The balance is not negotiable. Every other practical in Grade 11 can be improvised with borrowed equipment. This one is a mass measured against a mass, and there is no way round it.
The numbers, fixed so the arithmetic stays clean
| Moles | |
|---|---|
| 25,0 cm³ of 0,50 mol·dm-3 Na2CO3 | 0,0125 mol |
| 25,0 cm³ of 0,75 mol·dm-3 CaCl2 | 0,01875 mol |
The ratio is 1 : 1, so sodium carbonate is limiting and the calcium chloride is deliberately in excess.
n(CaCO3) = 0,0125 mol, and M(CaCO3) = 100 g·mol-1, so:
Theoretical yield = 1,25 g
Method
- Weigh a filter paper and write its mass and your group name on it in pencil.
- Measure 25,0 cm³ of the sodium carbonate into a beaker and 25,0 cm³ of the calcium chloride into another.
- Pour one into the other and stir. A white precipitate forms at once.
- Fold the paper into the funnel, stand it in a 250 ml beaker, and pour the mixture through a little at a time.
- Rinse the reaction beaker with water from the wash bottle and pour that through too. Twice. Precipitate left in the beaker is yield you have thrown away.
- Wash the precipitate on the paper with a little more water.
- Leave it to dry, labelled. Oven at 100 °C for an hour, or a warm dry place overnight.
- Next lesson: weigh the dry paper with the precipitate, subtract the paper, and calculate.
This practical takes two lessons. The precipitate has to be bone dry before it is weighed and that does not happen in a period.
What you should see
| Theoretical yield | 1,25 g |
| Typical actual yield | 1,0 to 1,2 g |
| Typical percentage yield | 80 to 95 % |
That range is the point. High enough that the experiment feels like a success, low enough that the missing mass has to be explained.
Why a yield over 100 % is impossible, and why you will still see one
The theoretical yield is the maximum the atoms you started with can make. More than 100 % would mean creating matter.
And yet almost every class produces at least one group reporting 104 %.
Their precipitate is wet. The extra mass is water.
The fix is to dry it longer and reweigh, then dry and reweigh again. When the mass stops changing, it is dry. That is called drying to constant mass and it is the correct way to do it.
Do not simply tell a group their answer is wrong. Ask them what could possibly make a product weigh more than the equation allows. They will get to water on their own, and they will never again weigh a damp solid.
Where the missing mass actually goes
| Loss | How much |
|---|---|
| Precipitate left in the reaction beaker | The biggest one, and the reason for the two rinses |
| Fine particles through the filter paper | Look at the filtrate. Cloudy means yield went with it |
| Precipitate on the stirring rod and funnel | Small but real |
| A little dissolved in the wash water | Calcium carbonate is barely soluble, but not perfectly insoluble |
| Not fully dry | This one goes the other way and pushes the yield too high |
Get the class to list these before you tell them. It is the best discussion in the practical and it is worth eight of the forty marks.
If it does not work
| What you see | What caused it |
|---|---|
| Percentage yield over 100 % | The precipitate is not dry. By far the commonest result |
| The filtrate is cloudy | Fine precipitate passing through. Pour more slowly, and note it as a loss |
| Filtering takes forever | Too much poured at once, or the paper has torn |
| Yield under 60 % | The reaction beaker was not rinsed, or a lot went through the paper |
| No precipitate at all | Solutions made up wrong, or water used instead of one of them. Check the labels |
| Yields vary widely between groups | Normal. Plot them on the board. The spread is a result in itself |
Why industry cares, which is the answer to "when will I use this"
A plant with enough raw material for 10 000 tonnes a year, designed for 95 % yield but running at 85 %, produces 8 500 tonnes instead of 9 500.
That is 1 000 tonnes a year of raw material paid for and never sold.
Percentage yield is a costing question before it is a chemistry question, and that framing reaches learners who are not otherwise interested in stoichiometry.
How the 40 marks are made up
| Section | Marks |
|---|---|
| The calculation, from the equation | 14 |
| The practical, results and percentage yield | 14 |
| Evaluation, where the mass went | 8 |
| Conclusion | 4 |
This split is ours, because the textbook prints neither a practical nor a mark allocation for this topic.
The calculation and evaluation sections together are 22 marks and need no apparatus at all, which is deliberate. A school without a balance can still set most of this.
The three marks most often dropped: forgetting to subtract the mass of the filter paper, not converting cm³ to dm³, and answering the evaluation with "human error" instead of naming a physical place the product went.
Free worksheet and marking memo
Both free, no sign up, straight to the PDF.
- Learner worksheet, 40 marks, with the limiting reagent and percentage yield calculations, the practical results table and the evaluation section set out
- Marking memorandum, with every calculation worked line by line, sample practical results, and a note on the six places learners most often drop marks
Related practicals
- Titration, Grade 12. The other quantitative practical, and it needs the same care with concentrations
- Endothermic and exothermic reactions, Grade 11. Same chemistry bench
- Intermolecular forces, Grade 11. Prescribed, Term 2
Buy this experiment
We are putting together a Percentage Yield Kit with the funnels, filter paper, beakers, measuring cylinders and the two chemicals in one box, plus a printed teacher guide and the marking memo. Coming shortly.
The one thing to check before you order anything is your balance. This practical needs one reading to 0,01 g. A balance reading to 1 g cannot resolve the difference between 1,08 g and 1,25 g, which is the entire result.
Our laboratory balances start well below what most schools expect to pay.