Stoichiometry Calculator
Type a chemical equation and the reactant amounts: it checks the balance, finds the limiting reagent and gives the moles and grams of every species, the excess left and the theoretical yield.
Result…
How it's calculated
Moles = mass ÷ molar mass For each reactant: moles ÷ its coefficient; the smallest is the limiting reagent Extent ξ = limiting moles ÷ limiting coefficient Product formed = coefficient × ξ (× molar mass for grams) Excess left = moles − coefficient × ξ
A balanced equation gives the mole ratio between every species. Divide each reactant’s moles by its coefficient: the one with the smallest result runs out first and limits the reaction. That smallest value, the extent of reaction, multiplied by each coefficient gives how many moles react or form, and the molar mass turns moles into grams. The calculator checks that every element balances, and if it doesn’t it can find the smallest whole-number coefficients (by solving the element balance as a set of linear equations). Atomic weights are the CIAAW abridged standard values. Ionic equations with charges are not supported.
Your numbers, step by step
About the Stoichiometry calculator
This stoichiometry calculator works straight from a chemical equation typed as text, such as 2H2 + O2 -> 2H2O. It reads every formula, checks that each element balances on both sides and, if the equation is not balanced, tells you and can balance it with the smallest whole-number coefficients. Give the amounts of the reactants in grams or moles and it finds the limiting reagent, the theoretical yield of every product in moles and grams, and how much of each excess reactant is left over. Chemistry students use it for limiting reagent homework, and teachers and lab staff use it to plan how much of each chemical to weigh out. Molar masses come from the CIAAW abridged standard atomic weights, and the steps show each division by molar mass and coefficient.
How to use it
- Type the reaction into “Chemical equation”: reactants, an arrow (-> or =), then products, each separated by +. Coefficients go in front, as in 2H2O.
- Under “If it is not balanced”, choose “Balance it for me” or “Just tell me”.
- In “Reactant amounts, in order”, give one amount per reactant separated by commas, such as 4 g, 40 g or 0.5 mol, 1 mol. Write excess for a reactant that is in excess.
- Pick what bare numbers mean under “Amounts without a unit are in”.
- Read the limiting reagent, the theoretical yields and the excess left over, and follow the steps for each species.
Worked example
In 2H2 + O2 -> 2H2O with 4 g of hydrogen and 40 g of oxygen: H2 is 2.016 g/mol, so 4 g is 1.984 mol, and ÷ 2 gives 0.992. O2 is 31.998 g/mol, so 40 g is 1.250 mol, and ÷ 1 gives 1.250. Hydrogen has the smaller ratio, so it is the limiting reagent. Water formed = 2 × 0.992 = 1.984 mol, or 35.74 g. Oxygen used is 31.74 g, leaving 8.256 g of oxygen.
Updated
Questions people ask
How do I find the limiting reagent?
Convert each reactant to moles and divide by its coefficient in the balanced equation. The reactant with the smallest result runs out first and limits how much product can form.
What is the formula for theoretical yield in stoichiometry?
Theoretical yield = moles of limiting reagent ÷ its coefficient × the product's coefficient × the product's molar mass. Compare it with what you actually collected in the percent yield calculator.
Why does it say my equation is not balanced?
Some element has a different number of atoms on the two sides. The message lists which ones and, when possible, the balanced version, such as 2H2 + O2 → 2H2O instead of H2 + O2 → H2O.
Why must I use the balanced equation's coefficients rather than the masses?
Coefficients count particles (moles), not grams. A common mistake is comparing the grams of each reactant directly; 4 g of hydrogen is far more moles than 40 g of oxygen.
Can it handle ionic equations or reactions that don't go to completion?
No. Leave out charges, since it balances atoms only. The results assume the reaction goes fully to completion; equilibrium reactions and side reactions give less product.