The limiting reagent is the cornerstone concept of quantitative chemistry: whichever reactant runs out first caps the entire reaction, no matter how much of the others remains. This calculator identifies it in one step, then calculates the theoretical yield of your chosen product, the excess amounts of every other reactant, and — if you supply your measured result — the percent yield.
How it works
The method follows the standard stoichiometric algorithm taught in every chemistry course:
1. Convert masses to moles. For each reactant the molar mass M is computed by
parsing the molecular formula — including nested brackets such as Ca(OH)2 or Fe2(SO4)3 —
and summing the IUPAC 2021 atomic masses. Then n = mass (g) / M (g/mol).
2. Compute reaction equivalents. Divide each reactant’s moles by its stoichiometric
coefficient: rxn equiv = n / coefficient. This normalises all reactants to the same
scale — it answers the question “how many times could this reactant drive the complete
balanced equation?”
3. Identify the limiting reagent. The reactant with the smallest reaction equivalent is the one that runs out first. Every other reactant is in excess.
4. Calculate theoretical yield. The limiting reagent’s reaction equivalents give the
moles of reaction that can actually proceed: molesRxn = min(rxn equiv). The theoretical
yield of the product is then:
yield (mol) = molesRxn x coefficient_product
yield (g) = yield (mol) x M_product
5. Find excess amounts. For each non-limiting reactant, the moles consumed equal
molesRxn x coefficient_reactant. Subtract from available moles and multiply by the
molar mass to get the unreacted mass in grams.
6. Percent yield (optional). % yield = (actual yield / theoretical yield) x 100.
Enter your measured actual yield and the calculator does this automatically.
Worked example — Haber process
The industrial synthesis of ammonia: N2 + 3 H2 → 2 NH3
Suppose the reactor is loaded with 28 g of N2 and 6 g of H2.
| Reactant | Molar mass | Moles | Coefficient | Rxn equiv |
|---|---|---|---|---|
| N2 | 28.014 g/mol | 0.9995 mol | 1 | 0.9995 |
| H2 | 2.016 g/mol | 2.976 mol | 3 | 0.9921 |
H2 has the smaller reaction equivalent (0.9921), so H2 is the limiting reagent.
- Moles of reaction = 0.9921
- Moles of NH3 produced = 0.9921 x 2 = 1.9841 mol
- Theoretical yield = 1.9841 x 17.031 g/mol = 33.8 g of NH3
- N2 consumed = 0.9921 x 1 = 0.9921 mol; excess N2 = 0.9995 - 0.9921 = 0.0074 mol = 0.21 g
If the actual yield collected is 29 g, the percent yield = (29 / 33.8) x 100 = 85.8%.
Enter the same values in the calculator above — it produces identical results with full step-by-step working displayed under the result cards.
Formula reference
| Quantity | Formula | Units |
|---|---|---|
| Molar mass | M = sum(atomic mass x subscript) | g/mol |
| Moles from mass | n = m / M | mol |
| Reaction equivalents | rxn equiv = n / coefficient | dimensionless |
| Limiting reagent | min(rxn equiv) | — |
| Theoretical yield | molesRxn x coeff_prod x M_prod | g |
| Excess remaining | (n - molesRxn x coeff) x M | g |
| Percent yield | % = (actual / theoretical) x 100 | % |
Tips for accurate results
- Always balance your equation before entering coefficients. An unbalanced equation produces incorrect mole ratios and wrong yield predictions.
- Use the most precise mass measurement available. Even a 0.1 g error propagates through the mole calculation.
- If percent yield exceeds 100%, check whether the product absorbed atmospheric moisture, contained unreacted starting material, or whether the actual yield was measured before drying.
- For reactions that produce multiple products, run the calculator separately for each product by changing the product formula and coefficient.
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