The mole is a counting unit, like "dozen", except the number is 6.022 × 10²³. Every mole problem is a conversion between four quantities — mass, moles, particles and (for gases) volume — and every conversion goes through moles. Learn the triangle and the numericals stop being different problems.
The four conversions
| From | To | Multiply by |
|---|---|---|
| mass (g) | moles | 1 ÷ molar mass |
| moles | mass (g) | molar mass |
| moles | particles | N_A = 6.022 × 10²³ |
| moles | volume at STP (gas) | 22.4 L |
Molar masses you should know: H = 1, C = 12, N = 14, O = 16, Na = 23, Mg = 24, S = 32, Cl = 35.5, Ca = 40, Fe = 56 (g/mol). So H₂O = 18, CO₂ = 44, NaCl = 58.5, H₂SO₄ = 98, CaCO₃ = 100.
Example 1 — grams to moles
How many moles are there in 90 g of water?
Molar mass of H₂O = 18 g/mol. Moles = 90/18 = 5 mol.
Example 2 — moles to molecules
How many molecules are in 5 mol of water?
5 × 6.022 × 10²³ = 3.011 × 10²⁴ molecules. And since each molecule has 3 atoms, that is 9.033 × 10²⁴ atoms.
Example 3 — mass to particles in one go
How many molecules are in 4.4 g of CO₂?
Moles = 4.4/44 = 0.1 mol. Molecules = 0.1 × 6.022 × 10²³ = 6.022 × 10²² molecules.
Example 4 — gas volume
What volume does 8 g of O₂ occupy at STP?
Moles = 8/32 = 0.25 mol. Volume = 0.25 × 22.4 = 5.6 L.
Example 5 — which is heavier?
Which has more mass: 1 mol of CO₂ or 2 mol of H₂O?
CO₂: 1 × 44 = 44 g. H₂O: 2 × 18 = 36 g. 1 mol of CO₂ is heavier. More moles does not mean more mass.
Example 6 — from a reaction
How many grams of CO₂ are produced when 10 g of CaCO₃ decomposes completely? CaCO₃ → CaO + CO₂
Moles of CaCO₃ = 10/100 = 0.1 mol. The equation says 1 mol CaCO₃ gives 1 mol CO₂, so 0.1 mol CO₂ = 0.1 × 44 = 4.4 g.
Example 7 — limiting reagent
4 g of H₂ reacts with 32 g of O₂ to form water. Which reactant limits, and how much water forms? 2H₂ + O₂ → 2H₂O
Moles: H₂ = 4/2 = 2 mol; O₂ = 32/32 = 1 mol. The equation needs 2 mol H₂ per 1 mol O₂ — exactly what we have, so neither is in excess and both are used up. Water formed = 2 mol = 36 g. (Change it to 4 g H₂ and 16 g O₂: then O₂ = 0.5 mol needs only 1 mol H₂, so O₂ is limiting and 1 mol = 18 g of water forms.)
Example 8 — percentage composition
Find the mass percentage of oxygen in H₂SO₄.
Molar mass = 2 + 32 + 64 = 98. Oxygen = 64/98 × 100 = 65.3 %.
Where students go wrong
- Using atomic mass where molecular mass is needed (O = 16 but O₂ = 32).
- Applying 22.4 L to a liquid or a solid. It is for gases at STP only.
- Forgetting to convert to moles before using the equation's ratio. The coefficients compare moles, never grams.
- Rounding N_A to 6 × 10²³ when the answer is expected to three significant figures.
How it is tested
Class 9 asks Examples 1–4. Class 11, NEET and JEE ask Examples 6–8 dressed up: a mixture, a percentage yield, or two steps of reaction. The method never changes — grams → moles → ratio → moles → grams.