Stoichiometry uses the coefficients in a balanced chemical equation to calculate how much reactant is needed or how much product can form. If a chemistry problem asks how many grams of product result from a given mass of reactant, stoichiometry is the method you need.
The short answer: Balance the equation, convert your starting quantity to moles, use the mole ratio from the balanced equation, and convert to the requested unit. Keep units in every step so they can cancel correctly.
In this IntoChemistry guide, you will learn the method, work through four example types, and practice limiting reactants and percent yield.
What Is Stoichiometry?
A balanced chemical equation describes the relative numbers of particles—and therefore moles—that react and form. Consider ammonia synthesis:
N₂ + 3 H₂ → 2 NH₃
The coefficients mean that 1 mol N₂ reacts with 3 mol H₂ to form 2 mol NH₃ (assuming a complete reaction and the other reactant is available). They do not mean that 1 gram of nitrogen reacts with 3 grams of hydrogen. Converting masses to moles is essential because different substances have different molar masses.
Stoichiometry is used when planning syntheses, checking chemical reaction yields, and interpreting quantitative laboratory results.
What You Need to Know First
Moles and molar mass
A mole contains approximately 6.022 × 10²³ specified entities. Molar mass, in g/mol, converts between a substance’s mass and amount in moles:
- Moles = mass (g) ÷ molar mass (g/mol).
- Mass (g) = moles × molar mass (g/mol).
Use atomic masses consistently. In this article: H = 1.01, N = 14.01, O = 16.00 g/mol. Thus H₂ = 2.02, N₂ = 28.02, O₂ = 32.00, NH₃ = 17.04, and H₂O = 18.02 g/mol.
Dimensional analysis and conversion factors
A conversion factor is a fraction representing equal quantities. For ammonia synthesis, 2 mol NH₃ / 1 mol N₂ converts moles of nitrogen to moles of ammonia. Arrange it so unwanted units cancel; never select a factor merely because its numbers look familiar.
Balancing equations comes first
Check the number of each element’s atoms on both sides of the reaction. Change coefficients, never chemical subscripts, to balance an equation.
How to Solve Stoichiometry Problems in Five Steps
- Write and balance the chemical equation.
- Identify the given and the goal. Underline the starting quantity and the unit requested for the final answer.
- Convert the starting amount to moles if it is expressed as a mass or another non-mole unit.
- Multiply by the balanced-equation mole ratio between the given and requested substances.
- Convert to the requested unit, then check units, reasonable magnitude, and significant figures.
For mole-to-mole problems, you can skip the mass-to-moles and moles-to-mass steps.
Four Types of Stoichiometry Problems
All four examples use the same balanced reaction, N₂ + 3 H₂ → 2 NH₃, so you can focus on how the conversion setup changes.
1. Mole-to-mole: nitrogen to ammonia
Problem: If 0.750 mol N₂ reacts with excess H₂, how many moles of NH₃ can form?
0.750 mol N₂ × (2 mol NH₃ / 1 mol N₂) = 1.50 mol NH₃.
Notice that the N₂ units cancel. You do not need molar mass because the question begins and ends in moles.
2. Mole-to-gram: nitrogen to ammonia
Problem: How many grams of NH₃ can form from 0.600 mol N₂ with excess H₂?
0.600 mol N₂ × (2 mol NH₃ / 1 mol N₂) × (17.04 g NH₃ / 1 mol NH₃) = 20.448 g = 20.4 g NH₃ (3 significant figures).
3. Gram-to-mole: hydrogen to ammonia
Problem: How many moles of NH₃ can form from 18.0 g H₂ when N₂ is in excess?
18.0 g H₂ × (1 mol H₂ / 2.02 g H₂) × (2 mol NH₃ / 3 mol H₂) = 5.9406… mol = 5.94 mol NH₃.
Mass cannot be used directly in the mole ratio: first convert grams of H₂ to moles of H₂.
4. Gram-to-gram: nitrogen to ammonia
Problem: How many grams of NH₃ can form from 10.0 g N₂ with excess H₂?
10.0 g N₂ × (1 mol N₂ / 28.02 g N₂) × (2 mol NH₃ / 1 mol N₂) × (17.04 g NH₃ / 1 mol NH₃) = 12.16… g = 12.2 g NH₃.
Remember the conversion pathway: grams of given → moles of given → moles of wanted → grams of wanted. This is the IntoChemistry Stoichiometry Roadmap.
How to Solve Limiting Reactant Problems
When amounts of two reactants are supplied, do not assume either is in excess. The limiting reactant is used up first, and it determines the maximum theoretical amount of product. The other reactant is present in excess.
Worked example: Which reactant limits water production?
Balanced equation: 2 H₂ + O₂ → 2 H₂O.
Given: 4.00 g H₂ and 24.0 g O₂.
- Available H₂: 4.00 g ÷ 2.02 g/mol = 1.980… mol H₂.
- Available O₂: 24.0 g ÷ 32.00 g/mol = 0.750 mol O₂.
- To use 0.750 mol O₂, the reaction needs 1.50 mol H₂. There is more than enough H₂, so O₂ is limiting.
- Water produced: 0.750 mol O₂ × (2 mol H₂O / 1 mol O₂) × (18.02 g/mol) = 27.03 g ≈ 27.0 g H₂O.
You can verify the limiting reactant another way: calculate the theoretical product yield independently from each reactant. The smaller product yield identifies the limiting reactant.
How to Calculate Percent Yield
Theoretical yield is the maximum product predicted from stoichiometry; actual yield is the amount collected experimentally.
Percent yield = (actual yield ÷ theoretical yield) × 100%.
If the water example above produces an actual yield of 23.5 g, percent yield = (23.5 g ÷ 27.03 g) × 100% = 86.9%. Carry unrounded values in intermediate calculations and round your final answer appropriately.
In a laboratory, a yield below 100% might reflect incomplete reaction, transfers, or product losses. A reported yield above 100% often warrants checking for moisture or impurities rather than assuming unusually successful chemistry.
Five Common Stoichiometry Mistakes
- Using an unbalanced equation. Every mole ratio becomes unreliable if the equation is wrong.
- Comparing coefficients directly to grams. Coefficients describe mole relationships, not mass relationships.
- Skipping a molar-mass conversion. Unless the given quantity is already in moles, convert before using the reaction ratio.
- Inverting a conversion factor. Write the units explicitly and confirm that they cancel.
- Rounding too early. Keep guard digits until the final answer; then use appropriate significant figures and units.
Stoichiometry Practice Problems
Try these without looking at the answer key. Use the molar masses printed on the downloadable worksheet. Assume complete reaction and that any reactant whose amount is not given is in excess.
Beginner: mole-to-mole
- For 2 H₂ + O₂ → 2 H₂O, calculate moles of water formed from 1.25 mol O₂.
- For N₂ + 3 H₂ → 2 NH₃, calculate moles of NH₃ formed from 0.600 mol N₂.
- For 2 KClO₃ → 2 KCl + 3 O₂, calculate moles of O₂ produced from 0.800 mol KClO₃.
- For CaCO₃ → CaO + CO₂, calculate moles of CO₂ produced from 2.40 mol CaCO₃.
Intermediate: mass conversions
- For N₂ + 3 H₂ → 2 NH₃, calculate grams of NH₃ formed from 3.00 mol H₂.
- For CaCO₃ → CaO + CO₂, calculate moles of CO₂ produced from 50.0 g CaCO₃.
- For CH₄ + 2 O₂ → CO₂ + 2 H₂O, calculate grams of CO₂ produced from 16.0 g CH₄.
- For 4 Al + 3 O₂ → 2 Al₂O₃, calculate grams of Al₂O₃ produced from 5.40 g Al.
Advanced: limiting reactants and percent yield
- For N₂ + 3 H₂ → 2 NH₃, determine the limiting reactant and theoretical grams of NH₃ from 14.0 g N₂ and 2.00 g H₂.
- For 2 Mg + O₂ → 2 MgO, determine the limiting reactant and theoretical grams of MgO from 12.0 g Mg and 10.0 g O₂.
- If problem 10 yields 16.5 g MgO experimentally, calculate percent yield.
- For 2 KClO₃ → 2 KCl + 3 O₂, calculate theoretical grams of O₂ from 24.5 g KClO₃ and percent yield if 8.40 g O₂ is collected.
Download the 12-question stoichiometry worksheet (PDF) and the separate worked answer key (PDF). Try the questions before checking your answers.
Frequently Asked Questions
What is the easiest way to do stoichiometry?
Write the conversion as a single chain of factors and cancel units. For most beginning students, following a consistent sequence is more reliable than memorizing separate formulas.
Why do I have to convert to moles?
Balanced equation coefficients count particles in relative amounts. Moles let you compare those amounts directly, regardless of the substances’ masses.
Can I do stoichiometry without dimensional analysis?
Yes, you can divide a calculation into several separate equations. Dimensional analysis is especially helpful for catching unit mistakes.
How do I choose the correct mole ratio?
Read both coefficients from the balanced equation. Put the substance you want on top and the starting substance on the bottom, so the starting moles cancel.
How should I study for a stoichiometry exam?
Practice balancing equations, molar masses, simple mole ratios, and gram-to-gram calculations before moving to limiting reactants and yields. Explain the reason for each conversion out loud.
Final Checklist and Next Steps
Before submitting an answer, check: balanced equation → correct given and wanted → mole conversion → mole ratio → target unit → sensible rounding.
Explore the IntoChemistry General Chemistry Resources hub for related topics. Use the Stoichiometry Learning Center to find this guide and its practice materials.