Stoichiometry Calculator
Calculate mass, moles, and limiting reactant from a balanced chemical equation.
Last updated: August 11, 2026
Stoichiometry: Recipe Math for Chemistry
Stoichiometry = using balanced equations to calculate amounts. Like a recipe: 2 eggs + 1 cup flour = 12 cookies. Chemistry: 2H₂ + O₂ → 2H₂O means 2 moles hydrogen + 1 mole oxygen = 2 moles water. Or in grams: 4g H₂ + 32g O₂ = 36g H₂O. Mass is always conserved.
Three key calculations: (1) Mole-to-mole ratios from coefficients, (2) Converting grams ↔ moles using molecular weight, (3) Finding limiting reactant (ingredient that runs out first). Master these and you can predict any reaction yield.
Basic Stoichiometry Steps
- Balance the equation (coefficients show mole ratios)
- Convert grams → moles (divide by molecular weight)
- Use mole ratio from balanced equation
- Convert moles → grams (multiply by molecular weight)
- Check work (mass in = mass out)
Example Problem: Complete Walkthrough
Problem: How many grams of water form when 10g hydrogen reacts with excess oxygen?
Equation: 2H₂ + O₂ → 2H₂O
Step-by-Step Solution
- 1. Given: 10g H₂
- 2. Convert to moles: 10g ÷ 2.016 g/mol = 4.96 mol H₂
- 3. Mole ratio: 2 H₂ : 2 H₂O = 1:1
- 4. Moles H₂O: 4.96 mol (same as H₂)
- 5. Convert to grams: 4.96 × 18.015 = 89.4g H₂O
Quick Formula
Grams A × (MW B / MW A) × (mole ratio B/A)
10g H₂ × (18.015/2.016) × (2/2)
= 10 × 8.936 × 1
= 89.4g H₂O
Limiting Reactant Concept
What is a Limiting Reactant?
The reactant that runs out first, stopping the reaction.
Example: Making sandwiches with 10 bread slices + 4 cheese slices. Cheese limits you to 4 sandwiches (2 bread per sandwich). Bread is in excess.
Problem: 5g H₂ reacts with 20g O₂. Which is limiting?
Equation: 2H₂ + O₂ → 2H₂O
| Reactant | Given Mass | Moles | Needed Ratio | Result |
|---|---|---|---|---|
| H₂ | 5g | 5 ÷ 2.016 = 2.48 mol | Need 2:1 ratio | Limiting |
| O₂ | 20g | 20 ÷ 32 = 0.625 mol | Have 2.48÷2 = 1.24 mol needed | Excess |
Common Stoichiometry Problems
| Problem Type | Given | Find | Method |
|---|---|---|---|
| Mass-to-Mass | Grams A | Grams B | g → mol → ratio → mol → g |
| Mass-to-Moles | Grams A | Moles B | g → mol → ratio |
| Moles-to-Mass | Moles A | Grams B | ratio → mol → g |
| Limiting Reactant | Grams A & B | Which limits | Compare mole ratios |
| Percent Yield | Actual & Theory | % Efficiency | (Actual/Theoretical) × 100 |
Mole Ratio Examples
| Equation | Mole Ratios |
|---|---|
| N₂ + 3H₂ → 2NH₃ | 1 N₂ : 3 H₂ : 2 NH₃ |
| 2Na + Cl₂ → 2NaCl | 2 Na : 1 Cl₂ : 2 NaCl |
| CH₄ + 2O₂ → CO₂ + 2H₂O | 1 CH₄ : 2 O₂ : 1 CO₂ : 2 H₂O |
| C₃H₈ + 5O₂ → 3CO₂ + 4H₂O | 1 C₃H₈ : 5 O₂ : 3 CO₂ : 4 H₂O |
Quick Tips & Common Mistakes
Do This
- Always balance equation first
- Use dimensional analysis (track units)
- Check if mass is conserved
- Identify limiting reactant in multi-reactant problems
- Round final answer appropriately (sig figs)
Avoid This
- Using unbalanced equations (wrong ratios)
- Confusing coefficients with subscripts
- Forgetting to convert grams ↔ moles
- Assuming excess reactant is limiting
- Mixing up reactants and products