Stoichiometry Calculator
Calculate amounts of reactants and products
How Stoichiometry Works
Stoichiometry is the calculation of reactants and products in chemical reactions using the balanced equation and mole ratios. It allows you to determine how much of a product is formed from a given amount of reactant, or how much reactant is needed.
Enter a chemical equation, specify the known substance and amount, and choose the target substance. The calculator will balance the equation, apply mole ratios, and show each conversion step.
The Stoichiometry Process
Stoichiometric calculations follow a consistent set of steps. First, the chemical equation must be balanced so the mole ratios between reactants and products are known. Then, the given quantity is converted to moles (if not already in moles), the mole ratio from the balanced equation is applied to find moles of the target substance, and finally the result is converted to the desired unit.
For example, to find how many grams of water are produced from 4 grams of hydrogen gas: 2H2 + O2 → 2H2O. Convert 4 g H2 to moles (4 ÷ 2.016 = 1.98 mol), apply the 2:2 mole ratio to get 1.98 mol H2O, then convert to grams (1.98 × 18.015 = 35.7 g H2O).
Supported Units
- Grams (g): The most common laboratory unit for mass.
- Moles (mol): The SI unit for amount of substance, central to all stoichiometric conversions.
- Kilograms (kg) and Milligrams (mg): For industrial-scale or micro-scale calculations.
- Liters (L): For gas volumes at standard temperature and pressure (STP), where 1 mole of an ideal gas occupies 22.414 L.
Step-by-Step Solutions
This calculator shows every conversion step in detail, including the balanced equation, molar masses used, mole ratio applied, and unit conversions performed. This makes it an effective learning tool for understanding the stoichiometric method, not just getting an answer.
Worked example: ammonia synthesis with a limiting reagent
Reacting 28.0 g of nitrogen with 9.00 g of hydrogen in the Haber process, N2 plus 3H2 gives 2NH3. The question is how much ammonia forms.
- Confirm the equation is balanced. N2 plus 3H2 gives 2NH3 has two nitrogen atoms and six hydrogen atoms on each side, so it is.
- Convert both reactants to moles. Nitrogen is 28.0 divided by 28.014, which is 1.00 mol. Hydrogen is 9.00 divided by 2.016, which is 4.46 mol.
- Test which reactant runs out first. Every mole of nitrogen needs three moles of hydrogen, so 1.00 mol of nitrogen needs 3.00 mol of hydrogen. There are 4.46 mol available, so hydrogen is in excess and nitrogen is the limiting reagent.
- Apply the mole ratio from the limiting reagent. Two moles of ammonia form per mole of nitrogen, giving 2.00 mol of ammonia.
- Convert to mass. 2.00 mol times the molar mass of ammonia, 17.031 g/mol, gives 34.1 g.
Answer. The reaction yields 34.1 g of ammonia, and about 1.46 mol of hydrogen is left over unreacted.
Common mistakes
These are the errors that come up most often, and each one changes the answer rather than merely looking untidy.
Calculating from the wrong reactant
If you run the calculation from the reactant present in excess you will get a yield that cannot physically occur. Always test which reactant limits the reaction before applying a mole ratio, even when one reactant looks obviously abundant.
Using coefficients as if they were masses
The coefficients in a balanced equation are ratios of moles, never of grams. Two moles of hydrogen and two moles of water appear in the same equation but weigh 4.03 g and 36.03 g respectively.
Inverting the mole ratio
Going from reactant to product, the product coefficient goes on top. Writing the ratio upside down produces an answer wrong by the square of the ratio, which is easy to miss when the numbers are close together.
Skipping the balancing step
An unbalanced equation gives mole ratios that violate conservation of mass. Balance first, every time, even if the equation looks familiar.
Confusing theoretical yield with what you will actually get
Stoichiometry gives the theoretical maximum. Real reactions lose material to side reactions, incomplete conversion, and transfer losses. A percent yield below 100 is normal, not necessarily an error.
Frequently asked questions
What is a limiting reagent?
The limiting reagent is the reactant that is completely consumed first and therefore caps how much product can form. Every other reactant is in excess and some of it remains when the reaction stops. Identify it by converting each reactant to moles and dividing by its coefficient in the balanced equation; the smallest result is limiting.
How do I calculate percent yield?
Divide the mass you actually obtained by the theoretical yield from stoichiometry, then multiply by 100. If theory predicts 34.1 g and you isolate 28.0 g, the percent yield is about 82 percent.
Can I use litres instead of grams for gases?
Yes, provided the gas is at standard temperature and pressure, where one mole of an ideal gas occupies 22.414 litres. Away from those conditions you should use the ideal gas law to convert volume to moles instead.
Why does the calculator show every step?
Because the method matters more than the number. Stoichiometry problems in an exam are marked on the working, and seeing the mole conversion, the ratio, and the final unit conversion laid out separately makes it much easier to find where your own attempt diverged.