Stoichiometry calculator

How do you solve a stoichiometry problem?

Balance the equation, turn what you know into moles (grams ÷ molar mass), multiply by the mole ratio from the coefficients, then turn the moles of the substance you want into grams, molecules or liters (OpenStax Chemistry 2e, Section 4.3). Moles of B = moles of A × (coefficient of B ÷ coefficient of A). OpenStax Example 4.8: in 2Al + 3I₂ → 2AlI₃, 0.429 mol of Al needs 0.429 × 3/2 = 0.644 mol of I₂. Example 4.10: to make 16 g of Mg(OH)₂ from MgCl₂ + 2NaOH → Mg(OH)₂ + 2NaCl you need about 22 g of NaOH.

Example: how much oxygen does octane need?

OpenStax Example 4.11: 2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O. 702 g of octane (114.23 g/mol) is 6.145 mol; times 25/2 that is 76.82 mol of O₂, or 2.46 × 10³ g at 32.00 g/mol. This is the pre-filled example: press Calculate to see it, along with the CO₂ and water formed.

How do I find the percent yield?

Choose a product as the substance to find and type the amount you actually collected; the percent yield is actual ÷ theoretical × 100 (OpenStax Section 4.4). OpenStax Example 4.13: CuSO₄ + Zn → Cu + ZnSO₄ with 1.274 g of copper sulfate should give 0.5072 g of copper; collecting 0.392 g is a 77.3% yield. The theoretical amount here assumes the substance you entered is the one that runs out; if you know the amounts of several reactants, use the limiting reactant calculator first.

What does STP mean, and how are liters worked out?

Mark a gas with (g), like O2(g), and the calculator multiplies its moles by the molar volume of an ideal gas at 0 °C (273.15 K). At 1 atm (101.325 kPa), the standard in OpenStax Chemistry 2e, that is 22.414 L/mol (OpenStax rounds it to 22.4 L). IUPAC has used 1 bar (100 kPa) since 1982, which gives 22.711 L/mol; pick it in the gas conditions box. Both numbers come from the exact gas constant and match NIST CODATA 2022. STP is 0 °C, not room temperature: for other conditions use the ideal gas law calculator.

How is this different from the limiting reactant calculator and the equation balancer?

This one answers “I have X grams of A, how much B?”: one known amount, every other substance follows from the mole ratio. If you know the amounts of two or more reactants and want to know which one runs out first, use the limiting reactant calculator. To get only the coefficients, use the chemical equation balancer; for a single substance in moles, grams and molecules, the grams to moles calculator.

Worked example: burning hydrogen

The balanced equation is 2H₂ + O₂ → 2H₂O. Suppose 10.0 g of H₂ reacts completely. With a molar mass of 2.016 g/mol (H = 1.008), 10.0 g is 10.0 ÷ 2.016 = 4.960 mol of H₂. The coefficients give the mole ratios: 2 mol H₂ to 1 mol O₂ to 2 mol H₂O. So the reaction uses 4.960 × 1/2 = 2.480 mol of O₂, which is 2.480 × 31.998 g/mol = 79.4 g (O = 15.999), and forms 4.960 mol of H₂O, or 4.960 × 18.015 g/mol = 89.4 g. Mass is conserved: 89.4 g of reactants make 89.4 g of product. As a gas at 0 °C and 1 atm, the 4.960 mol of H₂ would occupy 4.960 × 22.414 L/mol = 111.2 L. If an experiment collected 70.0 g of water, the percent yield is 70.0 ÷ 89.4 × 100 = 78.3%.

Why the mole ratio, not the mass ratio

Coefficients count particles, not grams. In 2H₂ + O₂ → 2H₂O the ratio of H₂ to O₂ is 2 to 1 by moles but about 1 to 8 by mass (4.0 g against 32.0 g). Always convert to moles first, apply the ratio, and then convert back to the unit you want. The molar mass of each substance comes from adding up its atomic masses, which you can look up on the periodic table; use a few more digits than your final answer needs, and round only at the end.

Common mistakes

• Using an unbalanced equation. Every ratio depends on the coefficients, so balance first; the tool does it for you from the formulas.
• Mixing the formula of an element with its atom: hydrogen and oxygen gases are H₂ and O₂, and their molar masses are about 2 and 32 g/mol, not 1 and 16.
• Applying the molar volume to a liquid or a solid. The 22.4 L figure is for an ideal gas at standard conditions only; water formed here is a liquid and has no gas volume at 0 °C.
• Assuming that a reactant runs out when you gave the amount of only one. If both amounts are known, find the limiting reactant first.
• Reading a percent yield above 100% as a good result; it usually means the product is wet, impure or was weighed wrongly.

Sources: OpenStax Chemistry 2e, Chapter 4 (4.3 Reaction Stoichiometry, 4.4 Reaction Yields); atomic masses rounded from the IUPAC standard atomic weights; molar volume as described in the answers above. Related: limiting reactant calculator, chemical equation balancer, grams to moles calculator and molar mass calculator.