The mole is just a counting word
A dozen is 12. A mole is 6.022 × 10²³. That is all it is — a number chosen so that one mole of a substance weighs its relative formula mass in grams.
It exists because you cannot count atoms but you can weigh them. The mole is the bridge between the mass you measure on a balance and the particle counts a chemical equation talks about.
Balanced equations are recipes
2H₂ + O₂ → 2H₂O does not say "two grams of hydrogen". It says two moles of hydrogen molecules react with one mole of oxygen molecules. The coefficients are a ratio of particles, never of mass.
Balancing enforces conservation of mass: atoms are rearranged, never created or destroyed. Count each element on both sides and they must match.
The limiting reagent
Reactants are almost never supplied in the exact ratio. Whichever runs out first stops the reaction dead — that is the limiting reagent, and it alone determines how much product you get. The rest is left over in excess.
The method: divide each reactant's moles by its coefficient. The smallest result is limiting. Then scale the product from that number.
Set H₂ = 5 and O₂ = 4. Dividing by coefficients gives 2.5 and 4 — hydrogen is limiting, you get 5 mol of water, and 1.5 mol of oxygen sits unused. Faded molecules are the leftovers.
Percentage yield
The calculation gives the theoretical yield — the perfect-world maximum. Real labs lose product to side reactions, incomplete reactions, and transfers between containers.
Worked example
Burn 4 g of hydrogen in excess oxygen. M(H₂) = 2 g/mol, so n = 4/2 = 2 mol H₂. The ratio 2H₂ : 2H₂O is 1:1, so you get 2 mol of water. M(H₂O) = 18 g/mol, giving 36 g theoretical. If you collect 30 g, the yield is 30/36 × 100 = 83%.
Empirical and molecular formulae
The empirical formula is the simplest whole-number ratio of atoms in a compound; the molecular formula is the actual number of each atom in a molecule. Ethene is C₂H₄ (molecular) but CH₂ (empirical). You find an empirical formula from masses or percentages by converting each element to moles and dividing by the smallest.
To get the molecular formula, work out the empirical formula mass and see how many times it divides into the true relative molecular mass.
Moles in solutions and gases
For solutions, concentration links moles and volume: a 1 mol/dm³ solution has one mole of solute in every cubic decimetre (1000 cm³). For gases, one mole of any gas occupies the same volume at the same conditions — about 24 dm³ at room temperature and pressure. Together with n = m/M, these let you convert freely between mass, moles, solution volume and gas volume.
The number of particles in a mole is the Avogadro constant, 6.02 × 10²³ — so one mole of water contains 6.02 × 10²³ molecules.
Atom economy (A Level)
A reaction can give a high percentage yield yet still waste most of its atoms as unwanted by-products. Atom economy measures how much of the reactant mass ends up in the desired product. A high atom economy means less waste — a central idea of green chemistry, and often more important industrially than yield alone.
A worked example
A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Find its empirical formula. (Ar: C = 12, H = 1, O = 16)
- Assume 100 g: 40.0 g C, 6.7 g H, 53.3 g O.Percentages become masses in a 100 g sample.
- Moles: C = 40.0/12 = 3.33; H = 6.7/1 = 6.7; O = 53.3/16 = 3.33.moles = mass ÷ Ar for each element.
- Divide by the smallest (3.33): C = 1, H = 2, O = 1.This gives the simplest whole-number ratio.
- Empirical formula = CH₂O.The 1 : 2 : 1 ratio of C : H : O.
CH₂O (this is the empirical formula; the molecular formula could be C₂H₄O₂, C₆H₁₂O₆, etc., depending on the relative molecular mass).