Chemistry

How to Do Stoichiometry: Step-by-Step Worked Examples

Convert grams to moles, use mole ratios, find the limiting reactant, and calculate percent yield with fully worked chemistry problems

By AI Picture Answer Education Team · Published September 25, 2026 · Updated September 25, 2026

What is stoichiometry?

Stoichiometry is the math of chemical reactions. It uses a balanced chemical equation to predict how much product a reaction makes, or how much of each reactant you need. The key insight is that the coefficients in a balanced equation describe moles, not grams. So almost every stoichiometry problem follows the same path: convert what you are given into moles, use the mole ratio, then convert back to whatever unit the question asks for.

You can check each problem below with the stoichiometry calculator, which walks through the gram-to-mole-to-gram conversions.

The stoichiometry roadmap:

grams of A → moles of A → moles of B → grams of B

Divide by the molar mass of A, multiply by the mole ratio (B/A), then multiply by the molar mass of B.

How to solve a stoichiometry problem

Balance the equation. Mole ratios are only correct if the equation is balanced. If you need help, use the chemical equation balancer.
Convert the given quantity to moles. For a mass, divide by the molar mass (g/mol).
Apply the mole ratio. Multiply by (coefficient of wanted substance) / (coefficient of given substance).
Convert moles to the requested unit. For grams, multiply by the molar mass of the wanted substance.
Check units and significant figures, and make sure the answer is reasonable.

Molar masses used in this guide

We use atomic masses H = 1.008, C = 12.011, N = 14.007, and O = 15.999 g/mol. That gives H₂ = 2.016, O₂ = 31.998, H₂O = 18.015, N₂ = 28.014, NH₃ = 17.031, C₃H₈ = 44.097, and CO₂ = 44.009 g/mol. A mole calculator can handle the gram-to-mole step for any formula.

Worked examples

Example 1: Grams to grams

How many grams of water form when 4.0 g of hydrogen gas reacts completely with oxygen?

2H₂ + O₂ → 2H₂O

Step 1: The equation is balanced: 4 H and 2 O on each side.

Step 2: Convert grams of H₂ to moles: 4.0 g ÷ 2.016 g/mol = 1.984 mol H₂.

Step 3: Apply the mole ratio. The coefficients show 2 mol H₂ make 2 mol H₂O, so 1.984 mol H₂ × (2 mol H₂O / 2 mol H₂) = 1.984 mol H₂O.

Step 4: Convert moles of H₂O to grams: 1.984 mol × 18.015 g/mol = 35.7 g H₂O.

Answer: 35.7 g of water (36 g when rounded to the two significant figures in 4.0 g).

Check with conservation of mass: The O₂ needed is 1.984 ÷ 2 = 0.992 mol, or 0.992 × 31.998 = 31.7 g. Reactant mass 4.0 + 31.7 = 35.7 g, which equals the product mass.

Example 2: Finding the limiting reactant

28.0 g of N₂ reacts with 9.0 g of H₂. How many grams of ammonia can form, and how much of the excess reactant is left over?

N₂ + 3H₂ → 2NH₃

Step 1: Convert both reactants to moles. N₂: 28.0 ÷ 28.014 = 0.9995 mol. H₂: 9.0 ÷ 2.016 = 4.464 mol.

Step 2: Compare what is needed. 0.9995 mol N₂ needs 0.9995 × 3 = 2.999 mol H₂. You have 4.464 mol H₂, which is more than enough.

Step 3: N₂ runs out first, so N₂ is the limiting reactant. Base the product on N₂.

Step 4: Mole ratio: 0.9995 mol N₂ × (2 mol NH₃ / 1 mol N₂) = 1.999 mol NH₃.

Step 5: Convert to grams: 1.999 mol × 17.031 g/mol = 34.0 g NH₃.

Step 6: Leftover H₂: 4.464 − 2.999 = 1.465 mol, and 1.465 × 2.016 = 2.95 g, or about 3.0 g of H₂.

Answer: 34.0 g of NH₃, with about 3.0 g of H₂ left over.

Check: H₂ used = 9.0 − 2.95 = 6.05 g. Total reacted = 28.0 + 6.05 = 34.05 g, which matches the 34.0 g of ammonia within rounding.

Example 3: Theoretical and percent yield

A student burns 22.0 g of propane and collects 58.0 g of carbon dioxide. What is the percent yield?

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Step 1: Check the balance: 3 C, 8 H, and 10 O on each side.

Step 2: Convert propane to moles: 22.0 g ÷ 44.097 g/mol = 0.4989 mol C₃H₈.

Step 3: Mole ratio: 0.4989 mol × (3 mol CO₂ / 1 mol C₃H₈) = 1.4967 mol CO₂.

Step 4: Theoretical yield: 1.4967 mol × 44.009 g/mol = 65.87 g, or 65.9 g CO₂.

Step 5: Percent yield = (actual ÷ theoretical) × 100 = (58.0 ÷ 65.87) × 100 = 88.1%.

Answer: Theoretical yield 65.9 g CO₂; percent yield 88.1%.

Check: A percent yield below 100% is expected, since some product is usually lost. A result above 100% would signal a calculation or measurement error.

Example 4: Moles to moles

How many moles of O₂ are needed to burn 0.50 mol of propane?

Step 1: The quantity is already in moles, so skip the molar mass step.

Step 2: Mole ratio from the balanced equation: 0.50 mol C₃H₈ × (5 mol O₂ / 1 mol C₃H₈) = 2.5 mol O₂.

Answer: 2.5 mol O₂

Common mistakes to avoid

Watch out for these:

Using an unbalanced equation. Every ratio comes from the coefficients, so an unbalanced equation makes every later step wrong.
Using the mole ratio on grams. 2H₂ + O₂ → 2H₂O does not mean 2 g of H₂ makes 2 g of water. Convert to moles first.
Using atomic mass for a diatomic gas. Oxygen gas is O₂, so its molar mass is 31.998 g/mol, not 15.999 g/mol.
Choosing the limiting reactant by mass. The reactant with fewer grams is not always limiting. Compare moles adjusted by the coefficients.
Rounding too early. Keep extra digits through the calculation and round only the final answer.
Flipping the mole ratio. Put the substance you want on top and the substance you have on the bottom so the units cancel.

Practice problems

Try these on your own:

How many moles of NH₃ form from 6.0 mol of H₂ with excess N₂?
How many grams of O₂ react with 8.0 g of H₂ in 2H₂ + O₂ → 2H₂O?
A reaction has a theoretical yield of 40.0 g and an actual yield of 32.0 g. What is the percent yield?

Click to see answers

6.0 mol H₂ × (2 mol NH₃ / 3 mol H₂) = 4.0 mol NH₃
8.0 ÷ 2.016 = 3.968 mol H₂; × (1/2) = 1.984 mol O₂; × 31.998 = 63.5 g O₂
(32.0 ÷ 40.0) × 100 = 80.0%

Frequently asked questions

What are the basic steps of stoichiometry?

Balance the equation, convert the given amount to moles, multiply by the mole ratio from the coefficients, and convert the result to the unit you need.

How do I find the limiting reactant?

Convert every reactant to moles, then use the mole ratio to see which one would run out first. That reactant limits how much product can form.

Why can't I use grams directly in the mole ratio?

The coefficients in a balanced equation count particles (moles), and different substances have different molar masses. Grams must be converted to moles first.

Need help with stoichiometry?

Stoichiometry builds on careful unit conversion, the same skill you use when you work step by step through synthetic division or check each case of an absolute value equation: write every step, and check the result. For chemistry homework, confirm your answers with the stoichiometry calculator and the molarity calculator for solution problems.

Working from a lab sheet or textbook? Snap a photo of your homework problem and AI Picture Answer will lay out the conversions step by step.

Snap a photo of your homework problem

Upload a picture of any chemistry problem and get a step-by-step solution in seconds

Try it free