Stoichiometry Calculator

Calculate mass, moles, and limiting reactant from a balanced chemical equation.

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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

  1. Balance the equation (coefficients show mole ratios)
  2. Convert grams → moles (divide by molecular weight)
  3. Use mole ratio from balanced equation
  4. Convert moles → grams (multiply by molecular weight)
  5. 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

Frequently Asked Questions

Math of chemical reactions - calculating amounts of reactants/products

Based on balanced equations and mole ratios

1. Balance equation

2. Convert grams → moles

3. Use mole ratio

4. Convert moles → grams

Reactant that runs out first, stopping the reaction

Determines maximum product yield

1. Convert all reactants to moles

2. Divide by coefficient in balanced equation

3. Smallest result = limiting reactant

Coefficients from balanced equation

2H₂ + O₂ → 2H₂O means 2:1:2 ratio

Law of conservation of mass

Coefficients give correct mole ratios for calculations

(Actual yield / Theoretical yield) × 100

Measures reaction efficiency (always <100%)

No. Equations show mole ratios, not gram ratios

Must convert grams → moles → use ratio → moles → grams

Reactants: Starting materials (left side)

Products: What's formed (right side)

No. "Excess" means you have more than needed

Only need to know limiting reactant amount for calculations