Why can percent yield exceed 100 %?

A percent yield over 100 % is a red flag, not a triumph: the isolated product weighs more than stoichiometry allows. The three usual causes are residual solvent (the crystals are still wet), unreacted starting material co-crystallizing with the product, and a weighing error (dirty vial, tared wrong). Dry the sample to constant mass, take an NMR or IR to check purity, and reweigh — a clean recrystallized product almost always drops back under 100 %.

What counts as a 'good' percent yield?

It depends on the reaction. In an undergraduate organic lab, 60–85 % is normal for a one-step synthesis like esterification or a Grignard addition; 40–60 % is common for multi-step procedures where you isolate intermediates. Industrial process chemistry pushes single-step yields above 95 % because a few points matter at ton scale. Percent yield is only meaningful alongside a purity check — 95 % yield of a 60 %-pure product is worse than 70 % yield of a pure one.

Percent yield vs atom economy — what's the difference?

Percent yield measures how much product you actually isolated relative to what the stoichiometry predicts (a lab metric). Atom economy is the fraction of atoms from the reactants that end up in the desired product on paper (a design metric introduced by Barry Trost in 1991). A reaction can have 100 % yield but 20 % atom economy if most reactant mass leaves as by-products; or 100 % atom economy but 50 % yield if the reaction is clean on paper but slow or inefficient in practice. Green chemistry cares about both.

How does percent yield accumulate through a multi-step synthesis?

Multiply the fractional yields, don't add them. Three steps at 80 % each give an overall yield of 0.80 × 0.80 × 0.80 = 51 %, not 240 %. This is why five- or six-step total syntheses often finish below 20 % even when every individual step looks respectable. Report both the per-step yields and the overall yield when writing up a synthesis.

Percent Yield Calculator — With Limiting Reagent

Percent yield compares what you actually isolated from a reaction to what the balanced equation predicted. It’s the single most-reported number in a synthesis writeup and the fastest signal of whether a procedure worked. This calculator handles both flavors of the problem: the direct form where you already know the theoretical mass, and the stoichiometric form where you calculate it from the limiting reagent, molecular weights, and the balanced equation’s mole ratio.

Percent Yield Calculator

Pick a mode. Enter grams (and, in stoich mode, molecular weights and the balanced ratio). Sig figs default to the lowest-precision input.

Formulas are rendered in Show the work above after a successful compute.

The equations

Percent yield itself is a one-liner:

$\% \text{ yield} = \dfrac{m_\text{actual}}{m_\text{theor}} \times 100$

The interesting arithmetic is in the theoretical mass — how many grams of product a given mass of limiting reagent could produce. Convert grams of reagent to moles, apply the mole ratio $r$ from the balanced equation, then convert moles of product back to grams:

$m_\text{theor} = \dfrac{m_\text{lim}}{MW_\text{lim}} \cdot r \cdot MW_\text{prod}$

where $r = \frac{\text{mol product}}{\text{mol limiting reagent}}$ is read directly off the balanced equation. As a helpful auxiliary, the calculator also reports grams lost — the difference $\Delta m = m_\text{theor} - m_\text{actual}$ — which is often the most tangible way to explain “where the mass went” in a lab report.

Worked example — aspirin synthesis (Zumdahl §3.10)

Salicylic acid + acetic anhydride $\to$ aspirin + acetic acid. The reaction is 1:1, so one mole of salicylic acid produces one mole of aspirin. Suppose you start with 6.9 g of salicylic acid (MW 138.12 g/mol) and excess acetic anhydride, and after workup and recrystallization you isolate 7.5 g of dry aspirin (MW 180.16 g/mol).

Worked example — aspirin from 6.9 g salicylic acid

m (salicylic acid, limiting): 6.9 g
MW (salicylic acid): 138.12 g/mol (PubChem CID 338)
MW (aspirin): 180.16 g/mol (PubChem CID 2244)
r (mol aspirin / mol SA): 1
Actual mass aspirin: 7.5 g

Formula
$m_{theor} = \frac{m _{lim}}{M W _{lim}} \cdot r \cdot M W_{prod}$
Substitute
$m_{theor} = \frac{6.9}{138.12} \cdot 1 \cdot 180.16 = 9.00 g$
Note
6.9 g of salicylic acid could produce at most 9.00 g of aspirin; you isolated 7.5 g, so 1.5 g of product was 'lost' to incomplete reaction, transfer losses, or crystals stuck to the filter paper. 83 % is a very reasonable yield for a bench-scale acetylation.
Result
$% yield = \frac{7.5}{9.00} \times 100 = 83.3%$

What a “good” yield looks like

Context	Typical single-step yield
Well-optimized industrial process	95–99 %
Textbook one-step organic synthesis (Fischer esterification, aldol)	60–85 %
Multi-step undergraduate teaching lab, per step	40–70 %
Total synthesis of a complex natural product, overall	often < 5 %

The point of tracking yield is not to punish a low number — it’s to notice when a procedure that used to give 78 % suddenly gives 40 % and figure out what changed (new solvent lot, wet glassware, an off pH).

Common pitfalls

Wrong r. The stoichiometric ratio is the coefficient of the product divided by the coefficient of the limiting reagent in the balanced equation — not their ratio of moles at the end. 4 Al + 3 O₂ → 2 Al₂O₃ gives $r = 2/4 = 0.5$ for Al₂O₃ from Al.
Wrong limiting reagent. If both reagents’ amounts are specified, compute moles of each divided by its coefficient; the smallest ratio is the limiting reagent. Only then apply this calculator.
Wet product. A yield over 100 % almost always means the product still contains solvent or water. Dry to constant mass before reporting.
Hydrates. If you weigh out a hydrated reagent like CuSO₄·5H₂O, use its full hydrate MW (249.68 g/mol), not the anhydrous MW.
Sig figs. The result inherits the precision of your weighings. If your balance reads to 0.01 g and you weigh 6.9 g of starting material, your yield is a 2-sig-fig number no matter how many decimals your calculator shows.

Practice problems

Attempt each on paper, then expand the worked solution to check your arithmetic and sig-figs.

Problem 1 Easy Your reaction is expected to give 12.0 g of product but you isolate 9.6 g of dry, pure product. What is the percent yield?

Answer: 80.0 %

Solve for percent yield (direct)

Actual mass: 9.6 g
Theoretical mass: 12.0 g

Formula
$% yield = \frac{m _{actual}}{m _{theor}} \times 100$
Substitute
$% yield = \frac{9.6}{12.0} \times 100$
Result
$% yield = 80.0%$

Problem 2 Intermediate You burn 5.00 g of magnesium (MW 24.31 g/mol) in excess O₂ to form MgO (MW 40.30 g/mol) in a 1:1 ratio. You isolate 7.50 g of MgO. What is the percent yield?

Answer: 90.5 % (theoretical 8.29 g)

Solve from limiting reagent (1:1 stoichiometry)

m (Mg, limiting): 5.00 g
MW (Mg): 24.31 g/mol
MW (MgO): 40.30 g/mol
r (mol product ÷ mol reagent): 1
Actual mass MgO: 7.50 g

Formula
$m_{theor} = \frac{m _{lim}}{M W _{lim}} \cdot r \cdot M W_{prod}$
Substitute
$m_{theor} = \frac{5.00}{24.31} \cdot 1 \cdot 40.30 = 8.290 g$
Note
1 mol Mg gives 1 mol MgO, so r = 1. The mass ratio 40.30 / 24.31 ≈ 1.66 is why 5.00 g of Mg can produce as much as 8.29 g of MgO — you added oxygen.
Result
$% yield = \frac{7.50}{8.290} \times 100 = 90.5%$

Problem 3 Intermediate You react 4.00 g of aluminum (MW 26.98 g/mol) with excess O₂ to form Al₂O₃ (MW 101.96 g/mol) via 4 Al + 3 O₂ → 2 Al₂O₃. You isolate 6.80 g of Al₂O₃. What is the percent yield?

Answer: 89.9 % (theoretical 7.56 g)

Solve from limiting reagent (r = 1/2)

m (Al, limiting): 4.00 g
MW (Al): 26.98 g/mol
MW (Al₂O₃): 101.96 g/mol
r = mol Al₂O₃ / mol Al: 2/4 = 0.5
Actual mass Al₂O₃: 6.80 g

Formula
$m_{theor} = \frac{m _{lim}}{M W _{lim}} \cdot r \cdot M W_{prod}$
Substitute
$m_{theor} = \frac{4.00}{26.98} \cdot 0.5 \cdot 101.96 = 7.561 g$
Note
The balanced equation gives 2 mol Al₂O₃ per 4 mol Al, so r = 0.5. Read r straight off the balanced equation — a wrong r is the single most common source of a wrong theoretical yield.
Result
$% yield = \frac{6.80}{7.561} \times 100 = 89.9%$

Problem 4 Hard You expect a theoretical yield of 5.00 g. After 'drying', you weigh 5.35 g of product. What is the reported percent yield, and what should you do?

Answer: 107 % — the product is still wet or impure; re-dry and reweigh

> 100 % yield — investigate before reporting

Actual mass: 5.35 g
Theoretical mass: 5.00 g

Formula
$% yield = \frac{m _{actual}}{m _{theor}} \times 100$
Substitute
$% yield = \frac{5.35}{5.00} \times 100$
Note
Mass cannot appear from nowhere. Common causes: residual water or solvent, un-reacted starting material that co-crystallized, or a weighing error. Dry to constant mass in a vacuum oven or over P₂O₅, then reweigh; the number almost always falls under 100 %.
Result
$% yield = 107%$

Frequently asked questions

Why can percent yield exceed 100 %?: A percent yield over 100 % is a red flag, not a triumph: the isolated product weighs more than stoichiometry allows. The three usual causes are residual solvent (the crystals are still wet), unreacted starting material co-crystallizing with the product, and a weighing error (dirty vial, tared wrong). Dry the sample to constant mass, take an NMR or IR to check purity, and reweigh — a clean recrystallized product almost always drops back under 100 %.
What counts as a 'good' percent yield?: It depends on the reaction. In an undergraduate organic lab, 60–85 % is normal for a one-step synthesis like esterification or a Grignard addition; 40–60 % is common for multi-step procedures where you isolate intermediates. Industrial process chemistry pushes single-step yields above 95 % because a few points matter at ton scale. Percent yield is only meaningful alongside a purity check — 95 % yield of a 60 %-pure product is worse than 70 % yield of a pure one.
Percent yield vs atom economy — what's the difference?: Percent yield measures how much product you actually isolated relative to what the stoichiometry predicts (a lab metric). Atom economy is the fraction of atoms from the reactants that end up in the desired product on paper (a design metric introduced by Barry Trost in 1991). A reaction can have 100 % yield but 20 % atom economy if most reactant mass leaves as by-products; or 100 % atom economy but 50 % yield if the reaction is clean on paper but slow or inefficient in practice. Green chemistry cares about both.
How does percent yield accumulate through a multi-step synthesis?: Multiply the fractional yields, don't add them. Three steps at 80 % each give an overall yield of 0.80 × 0.80 × 0.80 = 51 %, not 240 %. This is why five- or six-step total syntheses often finish below 20 % even when every individual step looks respectable. Report both the per-step yields and the overall yield when writing up a synthesis.

Sources

Zumdahl & Zumdahl, Chemistry, 10th ed. (2018), §3.10 — Percent Yield
Harris, Quantitative Chemical Analysis, 10th ed. (2020), §7
PubChem CID 2244 — Aspirin (MW 180.158 g/mol)
PubChem CID 338 — Salicylic acid (MW 138.12 g/mol)
Trost, B. M. (1991) 'The atom economy — a search for synthetic efficiency', Science 254, 1471–1477

Last updated: 2026-09-18.