Percent Yield Calculator
Percent yield from actual and theoretical mass, or from the limiting reagent.
About
Percent Yield Calculator
Percent yield compares what a reaction actually produced against the most it could have produced. It is the number that separates a reaction that worked from one that technically happened, and it is the first figure written in a lab report.
The formula
percent yield = (actual yield ÷ theoretical yield) × 100
Actual yield is what you weighed after purifying and drying the product. Theoretical yield is what the balanced equation says the limiting reagent could produce if every molecule reacted and nothing was lost.
Recovering 4.6 g where 6.2 g was possible gives 4.6 ÷ 6.2 × 100 = 74.19%.
Working out the theoretical yield
This is the harder half, and it lives in the stoichiometry. Three steps:
- Convert the limiting reagent mass to moles: mass ÷ molar mass
- Multiply by the mole ratio from the balanced equation
- Convert product moles back to mass: moles × molar mass of the product
Take the decomposition of calcium carbonate, CaCO3 → CaO + CO2, starting from 10.0 g:
- Moles of CaCO3: 10.0 ÷ 100.09 = 0.09991 mol
- Mole ratio is 1:1, so 0.09991 mol of CaO
- Mass of CaO: 0.09991 × 56.08 = 5.603 g
Recovering 4.85 g of CaO from that reaction is a yield of 86.6%.
Finding the limiting reagent
The theoretical yield depends on whichever reagent runs out first, and the largest mass is not always the answer. Divide the moles of each reagent by its coefficient in the balanced equation; the smallest result is limiting.
For N2 + 3H2 → 2NH3 with 2 mol of nitrogen and 3 mol of hydrogen: nitrogen gives 2 ÷ 1 = 2, hydrogen gives 3 ÷ 3 = 1. Hydrogen is limiting despite there being more moles of it, and the theoretical yield of ammonia is 2 mol rather than 4.
Everything else present is in excess, and the leftover is what remains when the limiting reagent is used up.
Where the missing mass goes
Yields fall short for reasons that mostly have nothing to do with the chemistry being wrong.
| Cause | What happens |
|---|---|
| Incomplete reaction | Equilibrium stops short of full conversion |
| Side reactions | Reagent consumed making something else |
| Transfer losses | Product left on glassware and filter paper |
| Purification | Recrystallisation and chromatography discard product with the impurities |
| Solubility | Some product stays dissolved in the filtrate |
Transfer and purification losses dominate on a small scale. A reaction run on 100 mg can lose a fifth of the product to the glassware alone, which is why yields improve as batch sizes grow.
Yields multiply along a synthesis
A multi-step route compounds its losses, and the arithmetic is unforgiving.
| Steps at 74% each | Overall yield |
|---|---|
| 1 | 74.2% |
| 2 | 55.0% |
| 3 | 40.9% |
| 4 | 30.3% |
| 5 | 22.5% |
Five respectable steps leave less than a quarter of the starting material. This is why chemists value a shorter route over a higher-yielding long one, and why convergent syntheses, which join two branches near the end, beat linear ones that pass everything through every step.
What a yield above 100% means
It means something is wrong with the measurement, since the reaction cannot create matter. In practice it points to one of three things:
The product is still wet. Residual solvent adds mass and is the most common cause. Dry to constant weight before weighing.
The product is impure. Unreacted starting material or a salt from the workup is being weighed as product.
The theoretical yield is wrong. An unbalanced equation, the wrong limiting reagent or a mistaken molar mass will all produce an impossible figure.
Percent yield, atom economy and conversion
Three related measures answer different questions. Percent yield asks how much of the possible product you actually isolated. Atom economy asks what fraction of the mass of all reactants ends up in the desired product, which is a property of the reaction itself rather than of how well it was run. Conversion asks how much of the starting material was consumed, regardless of what it turned into.
A reaction can have a high yield and poor atom economy, which is common in reactions that use a large protecting group and then discard it. Green chemistry pays attention to the second measure, because it governs how much waste the process generates.
A full worked problem
Esterification of acetic acid with ethanol to make ethyl acetate: CH3COOH + C2H5OH → CH3COOC2H5 + H2O. Starting from 12.0 g of acetic acid with ethanol in excess.
- Moles of acetic acid: 12.0 ÷ 60.05 = 0.19983 mol
- Mole ratio is 1:1, so 0.19983 mol of ester is possible
- Theoretical yield: 0.19983 × 88.11 = 17.61 g
- Recovering 11.2 g gives 11.2 ÷ 17.61 × 100 = 63.6%
The modest yield is expected here, since esterification reaches an equilibrium rather than going to completion. Driving it further means removing water as it forms or using a large excess of one reagent, both of which push the equilibrium toward the product.
Improving a yield
The lever depends on why the yield is low, which is why diagnosing before adjusting saves time.
An equilibrium-limited reaction responds to removing a product or adding excess of the cheaper reagent. A slow reaction responds to a catalyst, more heat or longer time. Losses in the workup respond to better technique: rinsing the flask into the filter, choosing a solvent the product is less soluble in when cold, and drying to constant weight rather than guessing.
Where side reactions are consuming material, more heat usually makes things worse rather than better, since the unwanted pathway often has the higher activation energy.
Reporting it properly
A yield figure means little without the mass it came from and the purity of what was weighed. Standard practice is to state the isolated mass, the moles, the percentage and how the identity and purity were confirmed.
Isolated yield, meaning material actually purified and weighed, is the honest figure. A yield read from a spectrum or a chromatogram of the crude mixture measures how much formed rather than how much survived purification, and the two can differ substantially. Reporting which one is being quoted is part of the result.
Yield on scale-up
A reaction that gives 85% on 100 mg rarely gives 85% on 100 g without work. Heat transfer, mixing and the time taken to add a reagent all change with the size of the vessel, and each can shift the balance between the wanted reaction and its competitors.
The direction of the change is not fixed. Transfer losses fall as a share of the batch, which helps. Against that, a large vessel heats and cools slowly, so a reaction held at temperature for longer may develop more side product. Process chemists treat a scale-up as a fresh optimisation rather than as the same reaction in a bigger flask.
Common mistakes
Using the wrong limiting reagent. Divide moles by the coefficient, and take the smallest.
Forgetting to balance the equation. The mole ratio comes from the coefficients, and an unbalanced equation gives the wrong ratio.
Weighing a wet product. Dry to constant weight, or the yield is measuring solvent.
Comparing yields across different scales. A 60% yield on 50 mg and a 60% yield on 50 g are not the same achievement.
Common questions
Frequently asked questions
It depends heavily on the reaction. Simple precipitations and acid-base reactions routinely clear 90%. Multi-step organic synthesis often runs at 50% to 80% per step, and anything above 90% in that context is notable. The useful comparison is against published yields for the same transformation.
Convert the limiting reagent to moles, multiply by the mole ratio from the balanced equation, then multiply by the molar mass of the product. For 10 g of calcium carbonate decomposing to calcium oxide in a 1:1 ratio, that gives 5.60 g.
The reactant that runs out first and therefore caps the product. Find it by dividing the moles of each reactant by its coefficient in the balanced equation; the smallest result is limiting. The reagent present in the greatest mass is often not the limiting one.
Not genuinely, since the reaction cannot create mass. A figure above 100% means the product is wet, contains impurities, or that the theoretical yield was calculated wrongly. Drying to constant weight resolves most cases.
Common causes are incomplete reaction, side reactions consuming the reagent, product left on glassware, and losses during filtering or recrystallisation. On small scales, transfer losses alone can account for a fifth of the material.
They multiply. Three steps at 80% give 0.8 cubed, or 51.2% overall. This is why a shorter route often beats a longer one with better individual yields, and why convergent syntheses are preferred where possible.
Percent yield measures how much of the possible product you isolated, so it reflects how the reaction was run. Atom economy measures what fraction of the reactant mass ends up in the product, which is fixed by the reaction equation itself and matters for waste.
The percentage should not, since both the actual and theoretical figures scale together. In practice small scales give lower percentages because fixed losses, such as material left on glassware, are a larger share of a smaller batch.