# Stoichiometry: how much reacts and how much forms

Canonical: https://duckyhelper.com/learn/chemistry/stoichiometry/
Updated: 2026-10-01

Stoichiometry uses a balanced equation to find how much of one substance reacts with, or forms from, another. The coefficients give the mole ratio. Most problems follow one path: grams of A to moles of A (divide by molar mass), moles of A to moles of B (multiply by the mole ratio), then moles of B to grams of B (multiply by molar mass). Always balance the equation first.

## Key ideas

Coefficients in a balanced equation count particles, and so they also count moles. In \(\mathrm{N_2 + 3\,H_2 \rightarrow 2\,NH_3}\), 1 mol of nitrogen reacts with 3 mol of hydrogen to make 2 mol of ammonia. They never count grams.

$$
n_B = n_A \times \frac{b}{a}
$$

- \(n_A\) is the moles of the substance you know (mol).
- \(n_B\) is the moles of the substance you want (mol).
- \(a\) and \(b\) are the coefficients of A and B in the balanced equation. The ratio \(\tfrac{b}{a}\) is the **mole ratio**, with what you want on top.

For a grams-to-grams problem, chain three steps. \(m\) is mass in grams and \(M\) is molar mass in g/mol:

$$
m_B = \frac{m_A}{M_A} \times \frac{b}{a} \times M_B
$$

**The stoichiometry road map**

| Step | From | To | Multiply by |
| --- | --- | --- | --- |
| 1 | grams of A | moles of A | \(\tfrac{1}{M_A}\) (1 mol over grams) |
| 2 | moles of A | moles of B | \(\tfrac{b}{a}\) (mole ratio) |
| 3 | moles of B | grams of B | \(M_B\) (grams over 1 mol) |

> **Note: Atomic masses used on this page**
>
> H 1.008, C 12.01, O 16.00, Na 22.99, Mg 24.31, Cl 35.45, K 39.10 (g/mol).

## Worked examples

**Example 1: moles to moles**

Problem: In \(\mathrm{2\,H_2 + O_2 \rightarrow 2\,H_2O}\), how many moles of water form from 3.0 mol of oxygen?

1. The equation is already balanced. Water and oxygen have coefficients 2 and 1, so the mole ratio is 2 mol water per 1 mol oxygen.
2. Multiply, with what you want on top of the ratio.

   $$
   3.0\ \text{mol O}_2 \times \frac{2\ \text{mol H}_2\text{O}}{1\ \text{mol O}_2} = 6.0\ \text{mol H}_2\text{O}
   $$

Answer: 6.0 mol of water

**Example 2: grams to grams**

Problem: Sodium reacts with chlorine gas: \(\mathrm{2\,Na + Cl_2 \rightarrow 2\,NaCl}\). How many grams of NaCl form from 10.0 g of sodium?

1. Grams of Na to moles of Na.

   $$
   \frac{10.0\ \text{g Na}}{22.99\ \text{g/mol}} = 0.43497\ \text{mol Na}
   $$
2. Mole ratio from the equation: 2 mol NaCl per 2 mol Na, which is 1 to 1.

   $$
   0.43497\ \text{mol Na} \times \frac{2\ \text{mol NaCl}}{2\ \text{mol Na}} = 0.43497\ \text{mol NaCl}
   $$
3. Moles of NaCl to grams. \(M_{\mathrm{NaCl}} = 22.99 + 35.45 = 58.44\) g/mol.

   $$
   0.43497\ \text{mol} \times 58.44\ \text{g/mol} = 25.42\ \text{g}
   $$
4. 10.0 g has 3 significant figures, so the answer gets 3.

Answer: 25.4 g of NaCl

**Example 3: how much reactant do you need?**

Problem: Propane burns: \(\mathrm{C_3H_8 + 5\,O_2 \rightarrow 3\,CO_2 + 4\,H_2O}\). How many grams of oxygen are needed to burn 22.0 g of propane completely?

1. Molar mass of propane.

   $$
   M = 3(12.01) + 8(1.008) = 44.09\ \text{g/mol}
   $$
2. Grams of propane to moles.

   $$
   \frac{22.0\ \text{g}}{44.09\ \text{g/mol}} = 0.49898\ \text{mol C}_3\text{H}_8
   $$
3. Mole ratio: 5 mol \(\mathrm{O_2}\) per 1 mol propane.

   $$
   0.49898 \times \frac{5}{1} = 2.4949\ \text{mol O}_2
   $$
4. Moles of \(\mathrm{O_2}\) to grams, with \(M_{\mathrm{O_2}} = 32.00\) g/mol.

   $$
   2.4949\ \text{mol} \times 32.00\ \text{g/mol} = 79.84\ \text{g}
   $$

Answer: 79.8 g of oxygen

**Example 4: a decomposition with 4 significant figures**

Problem: Heating potassium chlorate releases oxygen: \(\mathrm{2\,KClO_3 \rightarrow 2\,KCl + 3\,O_2}\). How many grams of \(\mathrm{O_2}\) come from 12.25 g of \(\mathrm{KClO_3}\)?

1. Molar mass of potassium chlorate.

   $$
   M = 39.10 + 35.45 + 3(16.00) = 122.55\ \text{g/mol}
   $$
2. Grams to moles.

   $$
   \frac{12.25\ \text{g}}{122.55\ \text{g/mol}} = 0.099959\ \text{mol KClO}_3
   $$
3. Mole ratio: 3 mol \(\mathrm{O_2}\) per 2 mol \(\mathrm{KClO_3}\).

   $$
   0.099959 \times \frac{3}{2} = 0.14994\ \text{mol O}_2
   $$
4. Moles to grams. 12.25 g has 4 significant figures, so keep 4.

   $$
   0.14994\ \text{mol} \times 32.00\ \text{g/mol} = 4.798\ \text{g}
   $$

Answer: 4.798 g of oxygen

## Common mistakes and how to fix them

- **Skipping the balance step.** An unbalanced equation gives the wrong mole ratio. Fix: count atoms on both sides before anything else. See [how to balance chemical equations](https://duckyhelper.com/learn/chemistry/balancing-chemical-equations/).
- **Using the mole ratio on grams.** The coefficients compare moles, not grams. Fix: always convert to moles before you use the ratio.
- **Flipping the ratio.** Fix: write the units on every number. The unit you want goes on top, and the unit you have goes on the bottom so it cancels.
- **Using the wrong molar mass at the end.** In step 3 you need the molar mass of B, the substance you are solving for. Fix: label each molar mass with its formula.
- **Too many significant figures.** Fix: the answer gets the same number of significant figures as the least precise measurement given in the problem.

**Practice problems**

1. In \(\mathrm{N_2 + 3\,H_2 \rightarrow 2\,NH_3}\), how many moles of ammonia form from 4.5 mol of hydrogen?
   A. 1.5 mol
   B. 3.0 mol
   C. 6.8 mol
   D. 9.0 mol

   Answer: 3.0 mol. The ratio is 2 mol \(\mathrm{NH_3}\) per 3 mol \(\mathrm{H_2}\): \(4.5 \times \tfrac{2}{3} = 3.0\) mol. 6.8 mol comes from flipping the ratio.

2. Magnesium burns: \(\mathrm{2\,Mg + O_2 \rightarrow 2\,MgO}\). How many grams of MgO form from 6.00 g of Mg?

   Answer: 9.95 g. \(6.00 \div 24.31 = 0.24681\) mol Mg. The ratio is 1 to 1, so 0.24681 mol MgO. \(M_{\mathrm{MgO}} = 24.31 + 16.00 = 40.31\) g/mol, and \(0.24681 \times 40.31 = 9.949\), which is 9.95 g.

3. Hydrogen peroxide breaks down: \(\mathrm{2\,H_2O_2 \rightarrow 2\,H_2O + O_2}\). How many grams of oxygen gas form from 17.0 g of \(\mathrm{H_2O_2}\)?
   A. 4.00 g
   B. 8.00 g
   C. 16.0 g
   D. 17.0 g

   Answer: 8.00 g. \(M_{\mathrm{H_2O_2}} = 34.02\) g/mol, so \(17.0 \div 34.02 = 0.4997\) mol. The ratio is 1 mol \(\mathrm{O_2}\) per 2 mol \(\mathrm{H_2O_2}\): 0.2499 mol. Then \(0.2499 \times 32.00 = 8.00\) g.

4. In \(\mathrm{4\,Fe + 3\,O_2 \rightarrow 2\,Fe_2O_3}\), what is the mole ratio of \(\mathrm{O_2}\) to \(\mathrm{Fe_2O_3}\)?
   A. 4 : 3
   B. 3 : 2
   C. 2 : 3
   D. 4 : 2

   Answer: 3 : 2. Read the coefficients in front of the two formulas: 3 for \(\mathrm{O_2}\) and 2 for \(\mathrm{Fe_2O_3}\).

5. Methane burns: \(\mathrm{CH_4 + 2\,O_2 \rightarrow CO_2 + 2\,H_2O}\). How many grams of water form from 8.0 g of methane?

   Answer: 18 g. \(M_{\mathrm{CH_4}} = 16.04\) g/mol, so \(8.0 \div 16.04 = 0.4988\) mol. Times 2 for the ratio gives 0.9975 mol water, and \(0.9975 \times 18.02 = 17.97\) g. 8.0 g has 2 significant figures, so 18 g.

## Frequently asked questions

### What does stoichiometry mean?

It comes from the Greek words for element and measure. In practice it means using a balanced equation to calculate amounts: how much of a reactant you need, how much product you can make, or how much is left over. Every stoichiometry problem uses the mole ratio from the coefficients.

### Do I always have to go through moles?

Yes, whenever the amounts are given in grams. The coefficients only compare particles, which means moles. The one shortcut: if both amounts are already in moles, you only need the mole ratio step.

### How is this different from a limiting reactant problem?

In a basic stoichiometry problem you are told one amount and assume the other reactants are in excess. In a limiting reactant problem you get amounts for two reactants and first have to find which one runs out. See the [limiting reactant guide](https://duckyhelper.com/learn/chemistry/limiting-reactant/).

### Can I use dimensional analysis instead of the formula?

Yes, and many teachers prefer it. Write the starting amount, then multiply by fractions whose units cancel: grams over molar mass, the mole ratio, then molar mass over moles. It is the same math as the formula on this page, just written as one long line.

## Sources

- [OpenStax Chemistry 2e, 4.3 Reaction Stoichiometry](https://openstax.org/books/chemistry-2e/pages/4-3-reaction-stoichiometry), accessed 2026-10-01
- [CIAAW, Standard atomic weights](https://www.ciaaw.org/atomic-weights.htm), accessed 2026-10-01

## Related

- [The mole and molar mass](https://duckyhelper.com/learn/chemistry/mole-and-molar-mass/)
- [How to find the limiting reactant](https://duckyhelper.com/learn/chemistry/limiting-reactant/)
- [How to calculate percent yield](https://duckyhelper.com/learn/chemistry/percent-yield/)
- [Chemistry study guides](https://duckyhelper.com/learn/chemistry/)

## Try asking Ducky

- "Check my setup for problem 2. Is my mole ratio upside down?"
- "I got 79.84 g but the answer key says 79.8. Why does that count as different?"
- "Give me one more grams-to-grams problem like this one and watch me solve it."

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