Revise How much? The amount of chemical change for IB Chemistry (R2.1)

A balanced equation is a recipe written in moles. Its coefficients give the mole ratio — the ratio in which substances react and form — and this ratio is the heart of every stoichiometry calculation. The one rule that never bends is that the ratio applies to MOLES, never directly to masses. So the reliable route is always the same three moves: convert the known mass to moles with $n = m/M$, cross the mole ratio to the substance you want, then convert back to a mass with $m = nM$.

$n = \frac{m}{M} \qquad\text{and}\qquad m = nM$

When a substance is a gas, you often measure its volume rather than its mass. Two ideas make this easy. First, the molar gas volume $V_m$ is the volume occupied by one mole of any ideal gas at a stated temperature and pressure; at STP (0 °C and 100 kPa) it is $22.7$ dm³ mol⁻¹, so $n = V/V_m$. Second, Avogadro's law states that equal volumes of gases at the same temperature and pressure contain equal numbers of moles. The powerful consequence is that for reacting GASES the volume ratio equals the mole ratio directly — you can work in volumes without ever converting to moles.

$n = \frac{V}{V_m} \qquad (V_m = 22.7\ \text{dm}^3\,\text{mol}^{-1}\ \text{at STP})$

For example, in the complete combustion of methane, CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), 100 cm³ of methane reacts with exactly 200 cm³ of oxygen and produces 100 cm³ of carbon dioxide (all measured at the same temperature and pressure), simply because the gas coefficients are 1 : 2 : 1. The liquid water is not counted — Avogadro's law is about gases only.

For reactions in solution, amount is measured through concentration and volume rather than mass. Molar concentration is the amount of solute per unit volume of solution, $c = n/V$, in mol dm⁻³. Rearranged as $n = c \times V$, it is the door into every titration calculation. A titration reacts a measured volume of one solution with a solution of known concentration until the reaction is exactly complete (the equivalence point, usually shown by an indicator). Because you know the concentration and volume on one side, you can find the moles there, cross the mole ratio, and then find the unknown concentration on the other side.

$c = \frac{n}{V} \quad\Longleftrightarrow\quad n = c \times V \qquad (c\ \text{in mol dm}^{-3},\ V\ \text{in dm}^3)$

In real reactions the reactants are rarely mixed in the exact stoichiometric ratio. The reactant that runs out first is the limiting reactant, and it alone controls how much product forms; whatever is left over is the excess reactant and plays no further part. The reliable test is to convert each reactant to moles and divide by its coefficient in the balanced equation — the smallest value is limiting. Do not simply pick the reactant with the smallest mass or even the smallest number of moles; the coefficients decide.

The theoretical yield is the maximum product predicted by stoichiometry from the limiting reactant, assuming the reaction goes to completion with no losses. In a real lab the actual yield you collect is usually lower, so the percentage yield measures how well the reaction ran: $\%\,\text{yield} = \dfrac{\text{actual}}{\text{theoretical}} \times 100$, comparing like with like (both in moles, or both as mass of the same substance). Atom economy is a different idea — it is a design measure of how much of the reactant mass ends up in the useful product: $\%\,\text{atom economy} = \dfrac{M(\text{desired product})}{M(\text{all products})} \times 100$. A reaction can have a high yield but poor atom economy if it also produces a lot of unwanted by-product. Yield describes one particular run; atom economy is fixed by the equation itself.

This is the showcase for a calculation topic. In Paper 2 the marks are analytic: each is tied to a specific line of working (a method mark, M, or an answer mark, A) and — crucially — the method marks and error-carried-forward (ECF) mean a wrong number early on does not cost you every mark that follows. That protection only exists if your method is written down. Study how each of these three marks is earned by a specific line.

Every quantitative topic ahead is a variation on these moves. Back-titrations and redox titrations reuse $n = c \times V$ across more than one mole ratio; gas-law problems replace the molar gas volume with the ideal gas equation when conditions are non-standard; thermochemistry and rates multiply moles by an energy change or a rate constant. Master the three-move habit — convert to moles, cross the mole ratio, convert back — plus the limiting-reactant test, and the rest of the quantitative course becomes familiar ground.