Stoichiometry answers questions like 'how much water forms from 4 g of hydrogen?'. Enter the known mass, both molar masses and the coefficients from the balanced equation to get the mass produced or needed.
Balance the equation first.
Enter the mass and molar mass of A, and both coefficients.
Enter B's molar mass and read its mass.
Route: g A ÷ M_A × (b ÷ a) × M_B = g B
Stoichiometry is the arithmetic of chemical reactions. It uses a balanced equation to tell you how much of one substance reacts with, or produces, a given amount of another. This calculator takes the mass of a known substance A, its molar mass, the coefficients of A and B from the balanced equation and the molar mass of B, and returns the moles of each and the mass of B. An optional percentage yield gives the mass you can realistically expect.
The grams-moles-ratio-moles-grams route is taught in Class 11 and underlies a large share of chemistry numericals in board exams, JEE and NEET. Industry relies on it as well, for example in estimating ammonia output from a fertiliser plant's nitrogen feed or cement clinker output from limestone.
1. Balance the chemical equation first. Coefficients must give equal numbers of each atom on both sides.
2. Convert the known mass of A into moles: n_A = mass_A ÷ M_A.
3. Use the mole ratio from the equation: n_B = n_A × (coefficient of B ÷ coefficient of A).
4. Convert moles of B into grams: mass_B = n_B × M_B. This is the theoretical yield.
5. If a percentage yield is given, multiply: actual mass = theoretical mass × yield ÷ 100.
6. If more than one reactant is given, repeat for each and take the smallest product mass, since that reactant runs out first.
A balanced equation counts particles, not grams. In N₂ + 3H₂ = 2NH₃, one molecule of nitrogen reacts with three of hydrogen to give two of ammonia. Since a mole is a fixed number of particles, the same ratio holds in moles. But molecules have different masses, so the ratio does not hold in grams: 28 g of nitrogen needs about 6 g of hydrogen, not 84 g. Converting to moles, applying the ratio, then converting back is the only reliable route.
The mass the calculator finds before applying yield is the theoretical yield, the most you could get if every molecule of A reacted as written. Real reactions fall short because of side reactions, reversible equilibria, incomplete mixing and losses while filtering or transferring. Percentage yield = actual ÷ theoretical × 100. Industrial processes such as the Haber process deliberately run at moderate conversion per pass and recycle unreacted gas, so the per-pass yield can be far below 100% while the overall plant efficiency stays high.
When two reactants are measured out, one usually runs out first; it is the limiting reagent and fixes how much product forms. The other is in excess, and some of it remains. To find the limiting one, calculate the product each reactant could give on its own, or compare the mole ratio available with the ratio in the equation. The calculator works from one known substance at a time, so run it once for each reactant and keep the smaller answer.
A chemistry teacher in Nagpur asks how much ammonia (17.031 g/mol) can form from 56 g of nitrogen (28.014 g/mol) in N₂ + 3H₂ = 2NH₃ with plenty of hydrogen, if the yield is 80%.
Moles of A: 56 ÷ 28.014 = 1.999 mol
Mole ratio from the balanced equation: mol B = 1.999 × 2 ÷ 1 = 3.998001 mol
Mass of B: 3.998001 × 17.031 = 68.09 g
Actual yield: 68.09 × 80% = 54.472 g
Answer: Mass of B 54.472 g; Moles of A 1.999 mol; Moles of B 3.998001 mol
Using an unbalanced equation, which gives a wrong mole ratio.
Applying the mole ratio directly to grams instead of moles.
Inverting the ratio, multiplying by coefficient of A over B instead of B over A.
Ignoring the limiting reagent when both reactant amounts are given.
Including coefficients in the molar mass, for example treating 2H₂O as 36 g/mol.
Class 11 and 12 mole concept and stoichiometry problems for boards, JEE and NEET.
Planning quantities of reagents for lab preparations.
Estimating product output and raw material needs in fertiliser, cement and pharmaceutical plants.
Calculating how much CO₂ is released from burning a given mass of fuel.
Checking percentage yield in organic chemistry practicals.
What is the limiting reagent?
The reactant that runs out first. Run the calculation for each reactant; the smallest product mass is the answer.
Where do I get molar masses?
Use the Molar Mass Calculator.