IB Chemistry — Reactivity 2.1: The Amount of Chemical Change (Stoichiometry, Yield, Atom Economy)

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Can you apply the mole ratio directly to masses?

This sub-topic is examined identically at SL and HL. A balanced equation is a recipe written in moles: its coefficients give the mole ratio in which substances react. From that ratio you can predict masses, gas volumes and concentrations — and judge how efficient a reaction is.

👆 Set the amounts and find the limiting reactant & yield · compare atom economies

1. Mole ratio & stoichiometry R2.1.1–2

A balanced equation conserves mass — the same atoms appear on both sides. Its coefficients give the mole ratio. To use it: convert what you know to moles (n = m ÷ M for a mass, n = c × V for a solution, or n = V ÷ 22.7 dm³ mol⁻¹ for a gas at STP), apply the ratio, then convert back. For example, in N₂ + 3H₂ → 2NH₃, one mole of N₂ reacts with three of H₂ to give two of NH₃ — so 0.5 mol N₂ makes 1.0 mol NH₃.

The three "moles" conversionsMass: n = m/M · Solution: n = c × V (c in mol dm⁻³, V in dm³) · Gas at STP (273.15 K, 100 kPa): n = V / 22.7 (V in dm³). Always route through moles — never apply a mole ratio to masses directly.

2. Limiting reactant & theoretical yield R2.1.3

When reactants are not mixed in the exact ratio, one runs out first — the limiting reactant — and it decides how much product forms (the theoretical yield). The other is in excess. To find it, divide the moles of each reactant by its coefficient; the smallest value is limiting. Try it:

mol of reactant 1

mol of reactant 2

3. Percentage yield R2.1.4

Real reactions rarely give every mole the equation promises — some product is lost, side-reactions occur, or the reaction is reversible. The percentage yield compares what you actually got with the theoretical maximum:

percentage yield = (experimental yield ÷ theoretical yield) × 100%

Percentage yield · uses the theoretical yield from the reaction above

Theoretical yield of product (g)

Experimental (actual) yield (g)

4. Atom economy R2.1.5

Percentage yield tells you how much of the theoretical product you captured; atom economy asks a greener question — of all the atoms in the reactants, what fraction end up in the desired product rather than in waste by-products?

atom economy = (M of desired product ÷ total M of all products) × 100%

A reaction can have a high yield but poor atom economy if it makes a lot of waste. Addition reactions (one product) have 100% atom economy; reactions that discard a by-product are lower. Compare:

Learn The Amount of Chemical Change step by step

Free interactive mini-lessons: short screens, worked examples and quick checks. Pick your exam board to open its course map, then tap L on this topic:

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Common mistakes examiners see

Can you apply the mole ratio directly to masses?✗ Yes — just use the coefficients on the grams. ✓ No — the ratio is in moles. Convert mass → moles (n = m/M) first, apply the ratio, then convert back.

How do you find the limiting reactant?✗ Whichever reactant has the smaller mass. ✓ Divide each reactant's moles by its coefficient; the smallest value is limiting — mass alone can mislead.

What is the difference between percentage yield and atom economy?✗ They are two names for the same thing. ✓ Yield = actual ÷ theoretical product (how much you captured). Atom economy = desired product mass ÷ total product mass (how little waste the equation makes).

What molar gas volume does the current IB guide use at STP?✗ 24.0 dm³ mol⁻¹. ✓ 22.7 dm³ mol⁻¹ at STP (273.15 K and 100 kPa) — from the current data booklet.

Does a 100% yield mean 100% atom economy?✗ Yes, they go together. ✓ No — a reaction can give every possible mole of product (high yield) yet still make a bulky by-product, giving low atom economy.

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The Amount of Chemical Change