What this calculator does
Find out how many grams of each substance a reaction uses or makes, starting from one amount you know. Stoichiometry is the bookkeeping of chemistry. A mole is a chemist's counting unit, a fixed (huge) number of particles. A balanced equation says how many moles of each substance react with, or form from, every other one. Molar mass (the weight of one mole) turns those moles into grams you can weigh. Put the two together, and one measured amount tells you the amount of everything else.
Type an equation, balanced or not, and the calculator balances it with exact arithmetic. Then it follows the standard route. Grams of the substance you know become moles. The equation's ratio turns those into moles of the substance you want. And those moles become grams.
A second mode compares two reactant amounts. A reactant is a starting substance. The calculator finds the limiting reactant, the one that runs out first. It also reports how much of the other reactant is left over. A table shows every substance for the same amount of reaction.
Where it's used
Why this math matters outside the classroom:
- Chemistry homework. Students find that 100 g of propane burns to about 299 g of carbon dioxide by converting through 2.27 mol of propane.
- Lab preparation. A lab technician works out that 10 g of iron needs about 4.30 g of oxygen to rust completely to iron(III) oxide.
- Limiting reactants. With 100 g of propane but only 200 g of oxygen, oxygen runs out first and limits the carbon dioxide to about 165 g.
- Industrial batches. Process engineers scale a recipe by mole ratios so that no expensive reactant is left over at the end of a batch.
- Fuel emissions. Environmental science classes estimate the carbon dioxide released per kilogram of a fuel from its combustion equation.
Grade level and standards: Grades 10–11; Common Core State Standards for Mathematics HSN.Q.A.1. © Copyright 2010. National Governors Association Center for Best Practices and Council of Chief State School Officers. All rights reserved.
How to use it
- Type the equation with plus signs between substances and an arrow between the sides, for example C3H8 + O2 -> CO2 + H2O. The numbers in front of each substance (its coefficients) are worked out for you.
- Count the substances from the left, starting substances first, then products (what forms). Choose the one you know by its number. Enter its amount in grams, milligrams, kilograms or moles.
- For a limiting-reactant problem, choose "Two reactant amounts" and enter the second reactant and its amount.
- Choose the substance to find and read its mass and moles. Then check the table for all the others.
What the inputs mean
- Chemical equation (balanced or not)
- Capital letters matter in element symbols: Co is cobalt, but CO is carbon monoxide. Brackets, hydrate dots and state labels such as (g) or (aq) are fine. Charged particles (ions) aren't supported.
- Known substance
- Substance 1 is the first reactant written. In C3H8 + O2 -> CO2 + H2O, substance 2 is O2 and substance 4 is H2O. The table under the result lists the numbering.
- Second reactant amount
- Only used in limiting-reactant mode. Both known substances must be reactants.
- Find the amount of
- Choose a product to see how much forms, or a reactant to see how much is consumed.
Formula and method
moles known = mass known ÷ molar mass known moles wanted = moles known × (coefficient wanted ÷ coefficient known) mass wanted = moles wanted × molar mass wanted limiting reactant: the smaller of (moles A ÷ coefficient A) and (moles B ÷ coefficient B) excess left = moles of excess − coefficient of excess × (moles limiting ÷ coefficient limiting)
In a balanced equation, the numbers in front (coefficients) compare moles, never grams. So every conversion goes through moles. Divide a substance's moles by its coefficient to get the extent of reaction. That's how many "rounds" of the equation that amount can supply. Multiply the extent by another coefficient to get that substance's moles.
When you give two reactants, each one supports a different number of rounds. The reaction stops when the first one is used up. So the smaller number wins, and that reactant is limiting. Molar masses come from IUPAC standard atomic weights, the official values from chemistry's international body.
Worked example
How much carbon dioxide forms when 100 g of propane burns completely? Equation: C3H8 + O2 -> CO2 + H2O.
- Balanced: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O.
- Molar mass of C₃H₈ = 3 × 12.011 + 8 × 1.008 = 44.097 g/mol, so 100 g ÷ 44.097 = 2.2677 mol.
- Mole ratio 3 CO₂ : 1 C₃H₈ gives 2.2677 × 3 = 6.8032 mol of CO₂.
- Molar mass of CO₂ = 44.009 g/mol, so 6.8032 × 44.009 = 299.40 g.
- With only 200 g of oxygen instead: 200 ÷ 31.998 = 6.2504 mol, ÷ 5 = 1.2501, which is less than propane's 2.2677, so oxygen limits. CO₂ = 1.2501 × 3 × 44.009 = 165.04 g, and 44.88 g of propane is left unburned.
Result: Burning 100 g of propane in plenty of oxygen makes about 299.40 g of carbon dioxide. With only 200 g of oxygen, it makes 165.04 g.
Understanding the result
The answer is the most you could get, if the reaction ran to the end perfectly. Real reactions recover less. The percent yield measures how much less.
Notice that 100 g of propane gives nearly 300 g of carbon dioxide. No mass is created. The oxygen taken from the air adds the difference. You can see the starting and final masses balance in the table.
Assumptions and limitations
- The reaction goes to completion exactly as written, with no side reactions.
- Every substance is pure, and the known amounts are exact.
- Molar masses use IUPAC standard atomic weights.
- Gas volumes and solution strengths aren't converted directly. Change them to moles first with the ideal gas law or molarity (moles per liter).
- Equations with charged particles (ions) or electrons aren't supported.
- Some equations can balance in more than one separate way. These are rejected, since there's no single mole ratio.
Frequently asked questions
Why can't I go straight from grams to grams?
The numbers in front of each substance count particles, and different substances have particles of different weights. One mole of propane weighs 44 g, but one mole of oxygen weighs 32 g. So a 1 : 5 mole ratio isn't a 1 : 5 mass ratio. Going through moles takes care of this.
Do I need to balance the equation first?
No. The calculator balances it with the smallest whole numbers that work and shows the result as the first step. If you type numbers in front of substances, they're replaced by the correct ones.
How do I know which reactant is limiting?
Divide each reactant's moles by its number in front in the balanced equation. The smaller result belongs to the limiting reactant. Comparing plain masses or plain moles can give the wrong answer, because those numbers differ.
What does the extent of reaction mean in the table?
Every row uses the same number of "rounds" of the balanced equation. So the table shows the moles and grams of each substance that react or form together. The grams you start with add up to the grams you end with.
How this calculator is checked
Method: Mole-ratio stoichiometry from the balanced equation (exact rational null-space balancing) with molar masses from IUPAC standard atomic weights; limiting reactant by smallest moles-to-coefficient ratio. The calculation is covered by automated tests with known reference values and edge cases. Last updated 2026-10-18. Found a problem? Tell us.
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