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Stoichiometry Calculator

Enter a reaction and the amounts of reactants you have, in grams, moles or particles. The calculator balances the equation, finds the limiting reactant, works out how much excess is left over, and gives the theoretical and percentage yield. It can also work backwards from a target amount of product.

Chemistry · ReactionsUpdated Sources cited

It does not need to be balanced. Press Set up after changing it.
2. Amounts

Fill in the reactants you have to find the limiting reactant and yields, or a single product to find the reactants needed. Leave the rest empty.

3. Actual yield (optional)

Result

Fe₂O₃ + 3CO → 2Fe + 3CO₂

Limiting reactant: CO

SpeciesUsed / formedLeft over
Fe₂O₃M = 159.69 g/mol95.02 g0.595 mol4.982 g
COM = 28.01 g/mol50 g1.785 mol0 g
FeM = 55.845 g/mol66.46 g1.19 mol—
CO₂M = 44.009 g/mol78.56 g1.785 mol—
Assumptions behind this result
  • The reaction goes to completion with no side reactions.
  • Avogadro constant N_A = 6.02214076 × 10²³ mol⁻¹ (exact).

Step-by-step solution

  1. Balance the equation

    Mole ratios come from the coefficients of the balanced equation.

    Fe2O3+3CO→2Fe+3CO2\mathrm{Fe_{2}O_{3}} + 3\mathrm{CO} \rightarrow 2\mathrm{Fe} + 3\mathrm{CO_{2}}

  2. Find the molar masses

    Add the standard atomic weights of the atoms in each formula.

    M(Fe2O3)=159.69 g/mol,M(CO)=28.01 g/mol,M(Fe)=55.845 g/mol,M(CO2)=44.009 g/molM(\mathrm{Fe_{2}O_{3}}) = 159.69\ \text{g/mol},\quad M(\mathrm{CO}) = 28.01\ \text{g/mol},\quad M(\mathrm{Fe}) = 55.845\ \text{g/mol},\quad M(\mathrm{CO_{2}}) = 44.009\ \text{g/mol}

  3. Convert Fe₂O₃ to moles

    Divide the mass in grams by the molar mass.

    n=100 g159.69 g/mol=0.62623 moln = \frac{100\ \text{g}}{159.69\ \text{g/mol}} = 0.62623\ \text{mol}

  4. Convert CO to moles

    Divide the mass in grams by the molar mass.

    n=50 g28.01 g/mol=1.7851 moln = \frac{50\ \text{g}}{28.01\ \text{g/mol}} = 1.7851\ \text{mol}

  5. Find the limiting reactant

    Divide each reactant's moles by its coefficient. The smallest result shows which runs out first: CO is the limiting reactant.

    Fe2O3:0.626231=0.62623,CO:1.78513=0.59503\mathrm{Fe_{2}O_{3}}: \frac{0.62623}{1} = 0.62623,\quad \mathrm{CO}: \frac{1.7851}{3} = 0.59503

  6. Theoretical yield of Fe

    Use the mole ratio 2 Fe : 3 CO, then convert to grams.

    n=2×0.59503=1.1901 mol,m=1.1901×55.845=66.458 gn = 2 \times 0.59503 = 1.1901\ \text{mol},\quad m = 1.1901 \times 55.845 = 66.458\ \text{g}

  7. Theoretical yield of CO₂

    Use the mole ratio 3 CO₂ : 3 CO, then convert to grams.

    n=3×0.59503=1.7851 mol,m=1.7851×44.009=78.559 gn = 3 \times 0.59503 = 1.7851\ \text{mol},\quad m = 1.7851 \times 44.009 = 78.559\ \text{g}

  8. Excess Fe₂O₃ left over

    Fe₂O₃ is in excess. Subtract the amount used from the amount available.

    0.62623−0.59503=0.031199 mol=4.9821 g0.62623 - 0.59503 = 0.031199\ \text{mol} = 4.9821\ \text{g}

What the calculator does

  • Balances the equation for you.
  • Converts each amount to moles, from grams, kilograms, milligrams, moles, millimoles or number of particles.
  • Finds the limiting reactant and the amount of each excess reactant left over.
  • Gives the theoretical yield of every product in moles and grams.
  • Calculates the percentage yield from an actual yield.
  • Works backwards: given a target amount of product, it finds the reactants needed.

How to use it

  1. Type the reaction and press Set up. It does not need to be balanced.
  2. Enter the amount of each reactant you have, with its unit. Leave the others empty.
  3. Optionally, enter the amount of product you actually obtained.
  4. Press Calculate.

The method

Moles from mass:

n=mMn = \frac{m}{M}

Moles from particles: n = N / N_A, where N_A = 6.02214076 × 10²³ mol⁻¹ exactly.

Limiting reactant: divide each reactant’s moles by its coefficient ν. The smallest value is how far the reaction can go:

ξ=min⁡iniνi\xi = \min_i \frac{n_i}{\nu_i}

Theoretical yield of a product with coefficient ν_p is ν_p ξ moles. Percentage yield is

percentage yield=actual yieldtheoretical yield×100%\text{percentage yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\%

Worked example: reducing iron ore

Iron is made by reducing iron(III) oxide with carbon monoxide. Suppose you have 100 g of Fe₂O₃ and 50 g of CO.

  1. Balance: Fe₂O₃ + 3CO → 2Fe + 3CO₂.
  2. Moles: 100 g ÷ 159.69 g/mol = 0.626 mol Fe₂O₃; 50 g ÷ 28.010 g/mol = 1.79 mol CO.
  3. Divide by coefficients: Fe₂O₃: 0.626 ÷ 1; CO: 1.79 ÷ 3 = 0.595. The smaller value belongs to CO, which is limiting.
  4. Theoretical yield of iron: 2 × 0.595 = 1.19 mol, which is 66.5 g.
  5. Excess: 4.98 g of Fe₂O₃ is left unreacted.

Common mistakes

  • Using an unbalanced equation. The mole ratios come from the coefficients.
  • Comparing grams instead of moles to find the limiting reactant.
  • Forgetting to divide by the coefficient. The reactant with fewer moles is not always limiting.
  • Mixing up theoretical and actual yield in the percentage-yield formula.

Practical applications

  • Industrial chemistry. Plants feed reactants in a deliberate ratio, often with one in excess, to use up the more expensive one.
  • Pharmaceuticals. Yields at each step of a synthesis determine the cost of a drug.
  • Environmental chemistry. The mass of CO₂ produced by burning a fuel is a stoichiometry calculation.
  • Laboratory work. Planning how much of each reagent to weigh out.

Assumptions and limitations

The reaction is assumed to go to completion, as written, with no side reactions. Real reactions may reach equilibrium short of completion. Molar masses use IUPAC abridged atomic weights. Reactions given as half-equations with electrons have no masses to calculate and are not accepted.

Frequently asked questions

How do I find the limiting reactant?

Convert each reactant to moles and divide by its coefficient in the balanced equation. The reactant with the smallest result runs out first and limits how much product can form.

Why can't I just compare the masses of the reactants?

Reactions happen between numbers of particles, not masses. 10 g of hydrogen contains far more molecules than 10 g of oxygen, so masses must be converted to moles first.

Why is my percentage yield below 100%?

Real reactions lose product through incomplete reaction, side reactions, and losses during transfer and purification. A yield above 100% usually means the product is still wet or impure.

What if I only know the amount of one reactant?

The calculator then assumes the other reactants are in excess, so that reactant is limiting. It says so in the assumptions.

Can I work out how much reactant I need?

Yes. Leave the reactants empty and enter an amount for one product. The calculator gives the moles and grams of each reactant required.

References

  1. BIPM (2019). The International System of Units (SI Brochure), 9th edition: the mole and the Avogadro constant.
  2. Atkins, P., Jones, L. & Laverman, L. (2016). Chemical Principles (7th ed.), Fundamentals sections L and M. W. H. Freeman.
  3. IUPAC Commission on Isotopic Abundances and Atomic Weights. Abridged standard atomic weights (2024 revision).

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