Stoichiometry calculator
Enter a reaction and the amounts you start with, such as C3H8 44 g, O2 100 g. The calculator balances the equation, finds the limiting reagent and shows how much of each substance is used or formed.
Runs in your browser: nothing you enter is sent anywhere.
How to use it
- Type the reaction. It does not have to be balanced.
- List what you have: formula, number, unit. Use several reactants to find the limiting one.
- Read the amounts used or formed for each substance, and what is left of the excess reactants.
How it works
The equation is balanced exactly (see the equation balancer). Each given mass becomes moles by dividing by the molar mass. Dividing moles by the coefficient gives how many times the equation could run; the smallest value sets the extent. Every substance then uses or forms coefficient × extent moles, and the mass follows from its molar mass.
Worked example
C3H8 + 5 O2 → 3 CO2 + 4 H2O with 44.1 g of propane (about 1 mol) and 100 g of oxygen (3.125 mol). Propane allows 1 / 1 = 1 run; oxygen allows 3.125 / 5 = 0.625. Oxygen is limiting, so 0.625 mol C3H8, 3.125 mol O2, 1.875 mol CO2 and 2.5 mol H2O take part.
Tips
- Check units: mg and kg are accepted as well as g.
- Spell formulas exactly as in the equation, including case.
- Use the limiting reagent, not the larger amount, to predict a yield.
Frequently asked questions
What is the limiting reagent?
The reactant that runs out first. Each given amount in moles is divided by its coefficient; the smallest quotient is how far the reaction can go, and that reactant is limiting. The others are left over in part.
Which atomic masses are used?
The standard atomic weights from the periodic table tool, the same as in the molar mass calculator. Results therefore agree with a molar mass table to the digits shown.
Can I start from a product?
Yes, give that one product alone, such as CO2 66 g, and the tool works back to the reactants needed. Mixing a product with other amounts is not accepted because the reaction extent would be set twice.
Does it give the actual yield?
No. Results are theoretical amounts assuming the reaction goes to completion. Real yields are lower; divide the actual mass by the theoretical mass for the percent yield.