Stoichiometry steps: grams to moles to grams

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)

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?

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.
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?

Grams of Na to moles of Na.
$\frac{10.0\ \text{g Na}}{22.99\ \text{g/mol}} = 0.43497\ \text{mol Na}$
Mole ratio from the equation: 2 mol NaCl per 2 mol Na, which is 1 to 1.
$0.43497\ \text{mol Na} \times \frac{2\ \text{mol NaCl}}{2\ \text{mol Na}} = 0.43497\ \text{mol NaCl}$
Moles of NaCl to grams. $M_{\mathrm{NaCl}} = 22.99 + 35.45 = 58.44$ g/mol.
$0.43497\ \text{mol} \times 58.44\ \text{g/mol} = 25.42\ \text{g}$
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?

Molar mass of propane.
$M = 3(12.01) + 8(1.008) = 44.09\ \text{g/mol}$
Grams of propane to moles.
$\frac{22.0\ \text{g}}{44.09\ \text{g/mol}} = 0.49898\ \text{mol C}_3\text{H}_8$
Mole ratio: 5 mol $\mathrm{O_2}$ per 1 mol propane.
$0.49898 \times \frac{5}{1} = 2.4949\ \text{mol O}_2$
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 $\mathrm{KClO_3}$ ?

Molar mass of potassium chlorate.
$M = 39.10 + 35.45 + 3(16.00) = 122.55\ \text{g/mol}$
Grams to moles.
$\frac{12.25\ \text{g}}{122.55\ \text{g/mol}} = 0.099959\ \text{mol KClO}_3$
Mole ratio: 3 mol $\mathrm{O_2}$ per 2 mol $\mathrm{KClO_3}$ .
$0.099959 \times \frac{3}{2} = 0.14994\ \text{mol O}_2$
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.
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

In $\mathrm{N_2 + 3\,H_2 \rightarrow 2\,NH_3}$ , how many moles of ammonia form from 4.5 mol of hydrogen?
1. 1.5 mol
2. 3.0 mol
3. 6.8 mol
4. 9.0 mol
Show answer
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.
Magnesium burns: $\mathrm{2\,Mg + O_2 \rightarrow 2\,MgO}$ . How many grams of MgO form from 6.00 g of Mg?
Show answer
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.
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 $\mathrm{H_2O_2}$ ?
1. 4.00 g
2. 8.00 g
3. 16.0 g
4. 17.0 g
Show answer
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.
In $\mathrm{4\,Fe + 3\,O_2 \rightarrow 2\,Fe_2O_3}$ , what is the mole ratio of $\mathrm{O_2}$ to $\mathrm{Fe_2O_3}$ ?
1. 4 : 3
2. 3 : 2
3. 2 : 3
4. 4 : 2
Show answer
Answer: 3 : 2
Read the coefficients in front of the two formulas: 3 for $\mathrm{O_2}$ and 2 for $\mathrm{Fe_2O_3}$ .
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?
Show answer
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.

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, accessed October 1, 2026
CIAAW, Standard atomic weights, accessed October 1, 2026

Stoichiometry: how much reacts and how much forms

Key ideas

Worked examples

Common mistakes and how to fix them

Frequently asked questions

Sources

Try asking Ducky

Key ideas

Worked examples

Common mistakes and how to fix them

Frequently asked questions

Sources

Related

Try asking Ducky