Chemistry calculator
Stoichiometry Calculator
This stoichiometry calculator converts grams or moles of one substance into grams and moles of another using the mole ratio from a balanced equation. It also finds the limiting reactant and the theoretical and percent yield. Type a formula like H2O and the molar mass is calculated for you.
Calculate Stoichiometric Amounts
mol B = mol A × (coefficient B ÷ coefficient A)
Coefficients come from the balanced equation. In any "molar mass" box you can type a number in g/mol or a formula such as H2O, Ca(OH)2 or CuSO4·5H2O.
Known substance (A)
Wanted substance (B)
Reactant 1
Reactant 2
Product
Result:
Steps:
How to Use the Stoichiometry Calculator
- Balance your chemical equation first and note the coefficient of each substance. If you need help, use the Chemical Equation Balancer.
- For a simple conversion, stay on Mole ratio (A → B). Enter the known amount in grams or moles, its formula (or molar mass), and both coefficients.
- To compare two reactants, switch to Limiting reactant & yield. Enter both masses, formulas and coefficients, plus the product. Add your actual yield to get the percent yield.
- Press Calculate with Steps to see every conversion, with units, in the order you would write it on paper.
How Stoichiometry Works
Almost every stoichiometry problem follows the same three-step path, often called the mole map:
grams A → ÷ M(A) → moles A → × (b ÷ a) → moles B → × M(B) → grams B
Here M is molar mass in g/mol and a and b are the coefficients of A and B in the balanced equation. The middle step is the only one that uses the equation; the outer steps are unit conversions. Molar mass is the sum of the atomic masses in the formula; for water, 2(1.008) + 15.999 = 18.015 g/mol.
Limiting reactant. When two reactants are given, compute moles ÷ coefficient for each. The smaller value belongs to the limiting reactant, and the product is calculated from it alone.
Percent yield = actual yield ÷ theoretical yield × 100%. Values above 100% usually mean the product is wet or impure.
Worked Examples
Worked example 1: grams to grams (2H₂ + O₂ → 2H₂O)
How much water forms from 4.00 g of H₂? Moles of H₂ = 4.00 g ÷ 2.016 g/mol = 1.984 mol. The ratio H₂O : H₂ is 2 : 2, so moles of H₂O = 1.984 mol. Mass of H₂O = 1.984 mol × 18.015 g/mol = 35.74 g of H₂O.
Worked example 2: burning propane (C₃H₈ + 5O₂ → 3CO₂ + 4H₂O)
How much CO₂ comes from 22.0 g of propane? Moles of C₃H₈ = 22.0 ÷ 44.097 = 0.4989 mol. Moles of CO₂ = 0.4989 × 3/1 = 1.4967 mol. Mass of CO₂ = 1.4967 × 44.009 = 65.87 g of CO₂. The same method shows the reaction needs 0.4989 × 5 = 2.495 mol (79.82 g) of O₂.
Worked example 3: limiting reactant (N₂ + 3H₂ → 2NH₃)
28.0 g of N₂ reacts with 9.0 g of H₂. Moles: N₂ = 28.0 ÷ 28.014 = 0.9995 mol; H₂ = 9.0 ÷ 2.016 = 4.464 mol. Divide by coefficients: N₂ gives 0.9995 ÷ 1 = 0.9995, H₂ gives 4.464 ÷ 3 = 1.488. N₂ has the smaller value, so N₂ is limiting. NH₃ formed = 0.9995 × 2 = 1.999 mol, and 1.999 mol × 17.031 g/mol ≈ 34.04 g. H₂ used = 0.9995 × 3 = 2.999 mol (6.045 g), leaving about 2.96 g of H₂ in excess.
Worked example 4: percent yield (CaCO₃ → CaO + CO₂)
Heating 50.0 g of CaCO₃ produced 25.2 g of CaO. Moles of CaCO₃ = 50.0 ÷ 100.086 = 0.4996 mol, and the ratio is 1 : 1, so theoretical CaO = 0.4996 × 56.077 = 28.01 g. Percent yield = 25.2 ÷ 28.01 × 100% = 89.95%, or 90.0% to three significant figures.
Common Mistakes to Avoid
- Using an unbalanced equation. The coefficients are the ratio; if they are wrong, every later number is wrong.
- Applying the mole ratio to grams. 4 g of H₂ does not make 4 g of H₂O. Convert to moles before using coefficients.
- Wrong molar mass. Remember diatomic elements (H₂, N₂, O₂, Cl₂) and multiply subscripts inside parentheses, as in Ca(OH)₂ = 74.09 g/mol.
- Choosing the limiting reactant by mass. The reactant with less mass is not automatically limiting. Compare moles ÷ coefficient.
- Rounding too early. Keep extra digits until the final answer, then round to the correct significant figures.
Frequently Asked Questions
What is stoichiometry?
Stoichiometry is the part of chemistry that uses a balanced chemical equation to calculate how much of each reactant is needed and how much product forms. It rests on conservation of mass: atoms are rearranged in a reaction, never created or destroyed.
How do you solve a stoichiometry problem step by step?
1) Write and balance the equation. 2) Convert the known quantity to moles (grams ÷ molar mass). 3) Multiply by the mole ratio from the coefficients (wanted ÷ known). 4) Convert the answer to the unit you need (moles × molar mass for grams).
What is a mole ratio?
A mole ratio is the ratio of the coefficients of two substances in a balanced equation. In 2H₂ + O₂ → 2H₂O, the ratio of O₂ to H₂O is 1 : 2, so every mole of oxygen can form two moles of water. Mole ratios compare moles, never grams.
How do you find the limiting reactant?
Convert each reactant to moles, then divide by its coefficient in the balanced equation. The reactant with the smallest result runs out first and is the limiting reactant. It alone determines the theoretical yield; the other reactant is in excess.
What is the difference between theoretical yield and percent yield?
Theoretical yield is the maximum amount of product the limiting reactant can make, calculated with stoichiometry. Percent yield compares what you actually collected with that maximum: percent yield = actual yield ÷ theoretical yield × 100%.
Why must the equation be balanced first?
The mole ratios come directly from the coefficients, and only a balanced equation has the correct coefficients. Using an unbalanced equation such as H₂ + O₂ → H₂O would give the wrong ratio and a wrong answer. You can balance it first with our chemical equation balancer.
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Formula
moles wanted = moles known × coefficient wanted / coefficient known
Coefficients in a balanced chemical equation define the mole ratio between reactants and products.
Worked example
In 2H₂ + O₂ → 2H₂O, 3 mol O₂ can produce 6 mol H₂O when hydrogen is available.
Common mistake
Balance the equation before reading any mole ratio from it.
