Learn Index · Matter and Chemistry Grade 11–12

The mole and stoichiometry

The mole counts particles, and amounts of substances in a reaction are related by the coefficients in the balanced equation.

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Counting atoms by weighing

Atoms are far too small to count individually. Instead, chemists count them in huge groups, just as bakers count eggs by the dozen. The chemist’s counting unit is the mole (mol).

One mole contains 6.022 × 10²³ particles. This is Avogadro’s number, named after Amedeo Avogadro. If you had a mole of pennies, you could give every person on Earth hundreds of billions of dollars.

Since 2019, the mole has been defined as exactly 6.022 140 76 × 10²³ particles.

Molar mass

The molar mass is the mass of one mole of a substance, in g/mol. It equals the atomic or formula mass from the periodic table.

Substance Calculation Molar mass
Carbon (C) 12.01 12.01 g/mol
Water (H₂O) 2(1.01) + 16.00 18.02 g/mol
Carbon dioxide (CO₂) 12.01 + 2(16.00) 44.01 g/mol
Sodium chloride (NaCl) 22.99 + 35.45 58.44 g/mol
Glucose (C₆H₁₂O₆) 6(12.01) + 12(1.01) + 6(16.00) 180.18 g/mol

Conversions

The mole is the bridge between mass, particles and volume of gas.

To convert… Use
grams → moles divide by molar mass
moles → grams multiply by molar mass
moles → particles multiply by 6.022 × 10²³
particles → moles divide by 6.022 × 10²³
moles of gas → litres at STP (0 °C, 1 atm) multiply by 22.4 L/mol

Example: How many moles are in 90.0 g of water?

90.0 g ÷ 18.02 g/mol = 4.99 mol

How many molecules is that? 4.99 × 6.022 × 10²³ = 3.01 × 10²⁴ molecules

Stoichiometry

Stoichiometry uses a balanced equation to calculate amounts of reactants and products. The coefficients give the mole ratio.

2H₂ + O₂ → 2H₂O

The mole ratio of H₂ : O₂ : H₂O is 2 : 1 : 2.

The three-step method

grams A → moles A → moles B → grams B

Example: How many grams of water form when 8.0 g of H₂ reacts completely?

  1. grams to moles: 8.0 g ÷ 2.02 g/mol = 3.96 mol H₂
  2. mole ratio: 3.96 mol H₂ × (2 mol H₂O / 2 mol H₂) = 3.96 mol H₂O
  3. moles to grams: 3.96 mol × 18.02 g/mol = 71 g H₂O

Limiting reactants

In real reactions, reactants are rarely in the perfect ratio. The limiting reactant runs out first and decides how much product forms. The other is the excess reactant.

Analogy: If you have 10 bread slices and 3 cheese slices, you can make only 3 cheese sandwiches. Cheese is limiting.

Example: 4.0 mol H₂ reacts with 3.0 mol O₂.

  • 4.0 mol H₂ needs 2.0 mol O₂. We have 3.0 mol, so O₂ is in excess.
  • H₂ is limiting. It produces 4.0 mol H₂O.
  • 1.0 mol O₂ is left over.

Tip: calculate the product from each reactant. The smaller answer tells you the limiting reactant.

Percent yield

The theoretical yield is the maximum product predicted by stoichiometry. The actual yield is what you really get.

percent yield = actual yield ÷ theoretical yield × 100%

If you expected 71 g of water but collected 64 g: 64 ÷ 71 × 100% = 90%

Yields are less than 100% because of incomplete reactions, side reactions, and product lost during transfer or purification.

Real-world stoichiometry

  • Airbags: sodium azide decomposes (2NaN₃ → 2Na + 3N₂) to make exactly the right volume of nitrogen gas to fill the bag in about 30 milliseconds.
  • Medicine: drug makers calculate precise amounts to avoid waste of expensive reagents.
  • Fertilizer: the Haber process (N₂ + 3H₂ → 2NH₃) makes ammonia for fertilizers that help feed billions of people.

Common mistakes

  • Using grams in the mole ratio. Ratios apply to moles only.
  • Forgetting to balance the equation first.
  • Assuming the reactant with the smaller mass is limiting. Compare moles, adjusted by the ratio.

Try it yourself

  1. Find the molar mass of calcium carbonate (CaCO₃).
  2. How many moles are in 50.0 g of CaCO₃?
  3. For N₂ + 3H₂ → 2NH₃, how many grams of NH₃ form from 14.0 g of N₂?
  4. If 10.0 mol H₂ reacts with 3.0 mol N₂, which is limiting?

(Answers: 100.09 g/mol; 0.500 mol; 0.500 mol N₂ → 1.00 mol NH₃ → 17.0 g; N₂ is limiting, since 3.0 mol N₂ needs only 9.0 mol H₂)

Quick recap

  • A mole is 6.022 × 10²³ particles.
  • Molar mass converts between grams and moles.
  • Stoichiometry uses mole ratios from balanced equations.
  • The limiting reactant determines the maximum product.
  • Percent yield compares actual and theoretical yields.