The Mole & Stoichiometry: Chemistry's Counting Unit
Avogadro's number is just chemistry's 'dozen' — and stoichiometry is its grocery-list math. Mole conversions, mass-to-mass calculations, limiting reagent, and percent yield with worked examples.
What is a mole?
A mole is just a counting unit — like a dozen (12) or a gross (144), but vastly larger. Specifically, 1 mole = 6.022 × 10²³ things (atoms, molecules, ions, formula units — whatever you're counting). This is Avogadro's number.
Why such a huge number? Because atoms are incredibly small. The mole is chosen so that 1 mole of carbon-12 atoms weighs exactly 12 grams — bridging the atomic-scale world (amu) to the laboratory-scale world (grams).
So for any element, the atomic mass on the periodic table doubles as its molar mass (in g/mol).
The three fundamental mole bridges
| Converting | Operation | Use |
|---|---|---|
| grams ↔ moles | ÷ or × molar mass | Macroscopic weighing |
| moles ↔ particles | ÷ or × Avogadro's number | Atomic-scale counting |
| moles of gas ↔ volume at STP | ÷ or × 22.4 L/mol | Gas problems |
Example: How many moles in 36.0 g of water?
Molar mass H₂O = 2(1.0) + 16.0 = 18.0 g/mol. So 36.0 g ÷ 18.0 g/mol = 2.00 mol. ✓
Stoichiometry — recipe math
A balanced chemical equation is a recipe. The coefficients are mole ratios. They tell you exactly how much of each ingredient produces how much of each product.
Example: 2 H₂ + O₂ → 2 H₂O. The ratio reads:
- 2 moles H₂ react with 1 mole O₂
- 2 moles H₂ produce 2 moles H₂O
- 1 mole O₂ produces 2 moles H₂O
To bridge between two substances:
moles of A × (coefficient B / coefficient A) = moles of B
Mass-to-mass stoichiometry — the full chain
The classic CBE problem: given grams of one substance, find grams of another. The strategy:
- grams of A → moles of A (divide by molar mass of A)
- moles of A → moles of B (multiply by mole ratio from balanced equation)
- moles of B → grams of B (multiply by molar mass of B)
Example: How many grams of water are produced from 4.0 g of H₂?
4.0 g H₂ ÷ 2.0 g/mol = 2.0 mol H₂
2.0 mol H₂ × (2 mol H₂O / 2 mol H₂) = 2.0 mol H₂O
2.0 mol H₂O × 18.0 g/mol = 36 g H₂O. ✓
Limiting reagent — the bottleneck ingredient
When two reactants are mixed, usually ONE runs out first — the limiting reagent. The other is in EXCESS. Identifying the limiting reagent tells you the maximum product possible.
Strategy: calculate how much of B you'd NEED to react with all of A. If you have MORE B than needed → A is limiting. If LESS → B is limiting.
Example: 3.0 mol H₂ + 1.0 mol O₂ → ? (Equation: 2 H₂ + O₂ → 2 H₂O)
3.0 mol H₂ would need 1.5 mol O₂. Only 1.0 mol O₂ available → O₂ is limiting. Maximum H₂O = 1.0 × (2/1) = 2.0 mol = 36 g.
Percent yield — reality vs theory
The theoretical yield is the maximum possible product from a perfect reaction. The actual yield is what you actually measure. Real reactions never hit 100% — there are side reactions, transfer losses, and incomplete conversions.
Percent yield = (actual ÷ theoretical) × 100%
Typical lab yields: 70–95%. A yield over 100% means the product is wet, contaminated, or contains by-products — never a "winning" result, just bad measurement.
Empirical and molecular formulas
The empirical formula is the simplest whole-number ratio of atoms (CH₂O for glucose). The molecular formula is the actual ratio (C₆H₁₂O₆ for glucose — six times the empirical).
To find an empirical formula from percent composition:
- Assume 100 g — convert each percent to grams.
- Convert grams of each element to moles.
- Divide all by the SMALLEST mole value.
- If results aren't whole numbers, multiply through by a small integer (×2, ×3) to clear fractions.