# What does stoichiometry predict?

Balances a chemical equation and turns a known mass or number of moles of one substance into the moles and grams of every other substance in the reaction.

- Page: https://www.acalculator.org/chemistry/stoichiometry-calculator
- JSON spec: https://www.acalculator.org/chemistry/stoichiometry-calculator.json
- Version: 898132e0da14

## Default answer

Example with the default inputs (Chemical equation C3H8 + O2 = CO2 + H2O, Known substance C3H8, Known amount 44, Amount in Grams, Find CO2): 44 g (0.9978 mol) of C3H8 gives 131.737 g (2.9934 mol) of CO2.

## Inputs

| Key | Label | Description |
| --- | --- | --- |
| eq | Chemical equation | The reaction, balanced or not: formulas joined by +, with = or → between the reactants and the products. |
| k | Known substance | The formula of the substance whose amount you know, as it appears in the equation. |
| n | Known amount | How much of the known substance, in the unit chosen below. |
| u | Amount in | Grams or moles. |
| t | Find | The formula of the substance to find, as it appears in the equation. |

## Outputs

| Key | Label | Description |
| --- | --- | --- |
| targetMass | Mass to find | Moles to find × its molar mass. |
| targetMoles | Moles to find | Known moles × (its coefficient ÷ the known coefficient). |
| knownMoles | Known moles | The known amount in moles: grams ÷ molar mass. |
| knownMass | Known mass | The known amount in grams: moles × molar mass. |
| balanced | Balanced equation | The equation with the smallest whole-number coefficients. |
| ratio | Mole ratio | Moles to find per mole of the known substance. |
| all | Every substance | The moles and grams of each substance for this amount of reaction. |

## Method

Balance the equation; known moles n = grams ÷ molar mass (or as typed); moles of X = n × (coefficient of X ÷ coefficient of the known); grams of X = moles of X × molar mass of X.

## Assumptions

- The reaction goes to completion as written, and the known substance sets the amount (no limiting reactant check).
- Molar masses use the CIAAW 2024 abridged standard atomic weights; electrons count as having no mass.

## Worked examples

1. eq = Al + I2 = AlI3, k = Al, n = 0.429, u = mol, t = I2 gives targetMoles = 0.6435, balanced = 2Al + 3I₂ → 2AlI₃. Source: OpenStax, Chemistry 2e, §4.3 Reaction Stoichiometry, https://openstax.org/books/chemistry-2e/pages/4-3-reaction-stoichiometry (Example 4.8: 0.429 mol Al × 3 mol I₂ ÷ 2 mol Al = 0.644 mol I₂).
2. eq = MgCl2 + NaOH = Mg(OH)2 + NaCl, k = Mg(OH)2, n = 16, u = g, t = NaOH gives targetMass = 21.946604, knownMoles = 0.274353. Source: OpenStax, Chemistry 2e, §4.3 Reaction Stoichiometry, https://openstax.org/books/chemistry-2e/pages/4-3-reaction-stoichiometry (Example 4.10: about 22 g NaOH); CIAAW, Abridged Standard Atomic Weights (2024), https://ciaaw.org/abridged-atomic-weights.htm.
3. eq = C8H18 + O2 = CO2 + H2O, k = C8H18, n = 702, u = g, t = O2 gives targetMass = 2,458.001698, balanced = 2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O. Source: OpenStax, Chemistry 2e, §4.3 Reaction Stoichiometry, https://openstax.org/books/chemistry-2e/pages/4-3-reaction-stoichiometry (Example 4.11: 2.46 × 10³ g O₂); CIAAW, Abridged Standard Atomic Weights (2024), https://ciaaw.org/abridged-atomic-weights.htm.
4. eq = C3H8 + O2 = CO2 + H2O, k = C3H8, n = 44, u = g, t = CO2 gives targetMass = 131.736581, targetMoles = 2.993401. Source: OpenStax, Chemistry 2e, §4.3 Reaction Stoichiometry, https://openstax.org/books/chemistry-2e/pages/4-3-reaction-stoichiometry (Example 4.9: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O); CIAAW, Abridged Standard Atomic Weights (2024), https://ciaaw.org/abridged-atomic-weights.htm.

## FAQ

### How do I solve a stoichiometry problem?

Balance the equation, change the known mass to moles with its molar mass, multiply by the mole ratio from the coefficients, and change back to grams. For 16 g of Mg(OH)₂ in MgCl₂ + 2NaOH → Mg(OH)₂ + 2NaCl: 16 ÷ 58.319 = 0.2744 mol, × 2 = 0.5487 mol NaOH, × 39.997 = 21.95 g.

### What is a mole ratio?

The ratio of two coefficients in the balanced equation. In 2Al + 3I₂ → 2AlI₃, 3 mol of I₂ react for every 2 mol of Al, so 0.429 mol of Al needs 0.429 × 3 ÷ 2 = 0.6435 mol of I₂.

### Does the equation need to be balanced first?

No. Type it balanced or not; the calculator works out the smallest whole-number coefficients itself and shows the balanced equation.

### How do I find the limiting reactant?

Run the calculator once for each reactant you have, finding the same product. The reactant that gives the least product is limiting, and that smaller amount is the theoretical yield. This page assumes the known substance sets the amount.

### Which atomic weights does the calculator use?

The abridged standard atomic weights of CIAAW (2024), the same table as the molar mass calculator: H 1.0080, C 12.011, O 15.999 and so on. Textbooks that round to H = 1.01 or O = 16.00 get slightly different answers.

### How many grams of CO₂ come from burning propane?

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O. 44 g of propane is 44 ÷ 44.097 = 0.9978 mol, which makes 3 × 0.9978 = 2.9934 mol of CO₂, or 131.74 g.

## Sources

- OpenStax, Chemistry 2e, §4.3 Reaction Stoichiometry, Examples 4.8 (aluminium and iodine), 4.9 (propane), 4.10 (magnesium hydroxide) and 4.11 (octane). https://openstax.org/books/chemistry-2e/pages/4-3-reaction-stoichiometry (retrieved 2026-10-01)
- CIAAW (Commission on Isotopic Abundances and Atomic Weights), Abridged Standard Atomic Weights, 2024. https://ciaaw.org/abridged-atomic-weights.htm (retrieved 2026-10-01)