Stoichiometry calculator: limiting reactant, yield and atom economy

Stoichiometry uses a balanced equation to work out how much of each substance reacts or forms. The method is always the same: change the mass you know into moles, use the mole ratio from the equation, then change back to a mass. This stoichiometry calculator does all of it from any equation, balanced or not: type the amounts you have and it finds the limiting reactant, the theoretical yield of each product, the percentage yield and the atom economy.

It opens on iron oxide reacting with carbon monoxide, example 1 below. Open the full stoichiometry calculator over this page to try your own reaction.

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Stoichiometry Calculator

Find reacting masses, limiting reactants, yield and atom economy

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How to use the stoichiometry calculator

1. Under Reaction, type the equation, for example Fe2O3 + CO -> Fe + CO2, or pick a ready-made one. It doesn't need to be balanced.
2. Under Amounts, give the amount of each reactant you have, in grams or moles. Leave a reactant blank if there is plenty of it.
3. Under Product, choose the product you want. Optionally, type the Actual mass collected for a percentage yield.
4. Read the results: the limiting reactant, the theoretical yield of your product, the percentage yield and the atom economy.
5. Check the table of moles before and after for every substance, the steps underneath, and the before-and-after bar showing that mass is conserved.
6. To practise, press Random problem in the Practice section.

The calculator uses precise relative atomic masses, such as 1.008 for hydrogen and 15.999 for oxygen. If your course rounds them to 1 and 16, your answers may differ slightly in the last figure.

How to do stoichiometry: the method

1. Balance the equation. The numbers in front give the mole ratio.
2. Find the moles of what you know: n = m ÷ M, mass divided by molar mass.
3. Find the limiting reactant if you know more than one amount: divide each reactant's moles by its number in the equation. The smallest runs out first.
4. Use the mole ratio to get the moles of the product.
5. Convert back to a mass: m = n × M. This is the theoretical yield.

Atom economy compares the wanted product with everything that went in, using the balanced equation: the formula mass of the wanted product divided by the total formula mass of the reactants, times 100. In England it is in the AQA GCSE Chemistry specification, as separate-chemistry content.

Worked examples you can check in the calculator

Example 1: limiting reactant and percentage yield

160 g of iron(III) oxide reacts with 56 g of carbon monoxide, and 60 g of iron is collected. These are the calculator's opening values.

CO runs out first. 3 mol of CO make 2 mol of Fe, so 1.999 mol of CO makes 1.999 × 2 ÷ 3 = 1.333 mol of Fe, which is 1.333 × 55.845 = 74.43 g, the theoretical yield. Only 60 g was collected, so the percentage yield is 60 ÷ 74.43 × 100 = 80.6%. About 53.6 g of the iron oxide is left over.

The atom economy for iron is 2 × 55.845 ÷ (159.687 + 3 × 28.01) × 100 = 45.8%. More than half the mass of the reactants ends up as carbon dioxide, which is not the product wanted.

Example 2: grams to grams

4 g of hydrogen burns in 40 g of oxygen. How much water forms? The balanced equation is 2H₂ + O₂ → 2H₂O.

• Hydrogen: 4 ÷ 2.016 = 1.984 mol, and 1.984 ÷ 2 = 0.992.
• Oxygen: 40 ÷ 31.998 = 1.250 mol, and 1.250 ÷ 1 = 1.250.
• Hydrogen gives the smaller number, so it is limiting. 2 mol of H₂ make 2 mol of H₂O, so 1.984 mol of water forms: 1.984 × 18.015 = 35.74 g.

The 8.26 g of oxygen left over plus 35.74 g of water makes 44 g, the same as the 4 g + 40 g that went in. Mass is conserved, and the calculator's before-and-after bar shows it.

Example 3: working in moles

2 mol of nitrogen reacts with 3 mol of hydrogen: N₂ + 3H₂ → 2NH₃. Switch both amounts to moles. Nitrogen gives 2 ÷ 1 = 2 and hydrogen gives 3 ÷ 3 = 1, so hydrogen is limiting. 3 mol of H₂ makes 2 mol of NH₃, which is 2 × 17.031 = 34.06 g, and 1 mol of nitrogen is left over. Having more moles of hydrogen doesn't stop it being the limiting reactant: the equation needs three times as much.

Example 4: one reactant and percentage yield

Heating 50 g of calcium carbonate gives calcium oxide: CaCO₃ → CaO + CO₂. 50 ÷ 100.086 = 0.4996 mol of CaCO₃ gives the same number of moles of CaO, so the theoretical yield is 0.4996 × 56.077 = 28.01 g. If 25 g is collected, the percentage yield is 25 ÷ 28.01 × 100 = 89.2%. With only one reactant there is no limiting reactant to find.

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Why the actual yield is lower

The theoretical yield is the most you could make. Real reactions give less: some product is lost when it is separated or transferred, side reactions make other products, and reversible reactions may not go to completion. A percentage yield above 100% usually means the product is still wet or impure.

Common mistakes

• Not balancing first. The mole ratio comes from the balanced equation. The calculator balances it for you.
• Using the ratio of masses. Equations give ratios of moles, not grams. Always convert to moles first.
• Picking the limiting reactant by the smaller mass or the smaller number of moles. Divide by the numbers in the equation before comparing.
• Using the excess reactant to work out the yield. Only the limiting reactant sets how much product forms.
• Dividing the wrong way for percentage yield. It is actual ÷ theoretical, so it is normally under 100%.

Read next

Chemistry · Moles and quantitiesLimiting reactant calculator: find it in grams or molesFind the limiting reactant from masses or moles, the theoretical yield, and how much of the excess reactant is left, with a start, change and end table.