Stoichiometry Calculator
Enter a known substance and amount, the mole ratio coefficients, and the target substance to find how much product is formed.
What stoichiometry computes
A balanced chemical equation encodes mole ratios. The coefficients tell you how the reactants and products are related quantitatively, and stoichiometry is the procedure for translating “how much of A do I have?” into “how much of B will I get?”
The general path is always the same three-step conversion:
grams of A → moles of A → moles of B → grams of B
- Mass to moles using the molar mass of A (n = m/M).
- Moles of A to moles of B through the mole ratio from the balanced equation.
- Moles of B back to mass (or volume, for a gas at STP) using the molar mass of B.
The middle step is where the balanced equation does its work; the outer steps are unit conversions. The reason you cannot shortcut directly from grams to grams is that the ratio is in moles, and one mole of one substance does not weigh the same as one mole of another.
Inputs
- A balanced equation (e.g.,
2H2 + O2 -> 2H2O). - The known substance and its amount, in grams, moles, or liters of gas at STP.
- The target substance.
The calculator parses the equation, applies the three-step path, and shows each conversion explicitly.
Worked examples
Mass to mass. How many grams of water form when 4.0 g H₂ reacts with excess O₂? Equation: 2H₂ + O₂ → 2H₂O. Moles of H₂ = 4.0 / 2.016 = 1.984. H₂:H₂O = 2:2 = 1:1, so moles of H₂O = 1.984. Mass = 1.984 × 18.015 = 35.7 g.
Mass to gas volume at STP. How many liters of CO₂ at STP from 12.0 g of carbon? C + O₂ → CO₂. Moles of C = 12.0 / 12.011 = 0.999. C:CO₂ = 1:1. Volume at STP = 0.999 × 22.414 = 22.4 L.
Moles to mass. 0.500 mol NaOH + HCl → NaCl + H₂O. NaOH:NaCl = 1:1, so 0.500 mol NaCl. Mass = 0.500 × 58.44 = 29.2 g.
Methane combustion. Grams of O₂ needed to burn 32.0 g CH₄? CH₄ + 2O₂ → CO₂ + 2H₂O. Moles of CH₄ = 32.0 / 16.04 = 1.995. CH₄:O₂ = 1:2, so moles of O₂ = 3.990. Mass = 3.990 × 32.00 = 127.7 g.
Where stoichiometry lives
- Predicting reaction yields before running a synthesis.
- Scaling a literature procedure up or down by an arbitrary factor.
- Calculating raw material requirements for an industrial run.
- Finding emission masses from combustion equations.
- Drug synthesis, where the molar ratios determine reagent quantities and excess calculations.
Pair this with the Limiting Reagent Calculator when you have specific amounts of two or more reactants and need to find which one runs out first.