Revise Counting particles by mass: the mole for IB Chemistry (S1.4)

A mole is a counting word for a number, in the same way that a 'dozen' means twelve. The number a mole stands for is Avogadro's constant, symbol $N_A$ (older texts and the data booklet also use $L$), and its value is $6.02 \times 10^{23}$ mol⁻¹. One mole of any entity contains that many of them: one mole of carbon atoms is $6.02 \times 10^{23}$ atoms, and one mole of water molecules is $6.02 \times 10^{23}$ molecules. Because the number is fixed, always say what the entities are — 'one mole of oxygen' is ambiguous until you state whether you mean O atoms or O₂ molecules.

The mole is useful because it is anchored to mass. The molar mass ($M$) of a substance is the mass of one mole of it, with units g mol⁻¹, and you calculate it by adding the relative atomic masses ($A_r$) of every atom in the formula. For an element the molar mass is just its $A_r$ written with units; for a compound you sum across the whole formula. For example, $M(\text{CaCO}_3) = 40.08 + 12.01 + 3(16.00) = 100.09$ g mol⁻¹. Molar mass is numerically the same as the relative molecular mass $M_r$, but $M_r$ is a unitless ratio while $M$ carries units and is what you use in calculations.

$n = \frac{m}{M}$

where $n$ is the amount in moles (mol), $m$ is the mass in grams (g), and $M$ is the molar mass (g mol⁻¹). This single relationship, and its rearrangements $m = nM$ and $M = m/n$, appears in almost every calculation in this topic. Notice the direction: you DIVIDE the mass by the molar mass. Multiplying is the single most common slip on the whole topic.

An empirical formula is the simplest whole-number ratio of atoms in a compound; a molecular formula is the actual number of atoms in one molecule. Water is both H₂O, but glucose has empirical formula CH₂O and molecular formula C₆H₁₂O₆ — the molecular formula here is six times the empirical unit. Experimental data (combustion results, or percentage composition) always deliver the empirical formula first, because they reveal only the ratio of masses present.

A balanced equation is a recipe written in moles: its coefficients give the mole ratio in which substances react and form. This ratio is the heart of every stoichiometry calculation, but it applies to MOLES, never directly to masses. So the reliable route is always the same three moves: convert what you are given into moles, cross the mole ratio to the substance you want, then convert back to the quantity the question asks for.

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, $c = n/V$, with units mol dm⁻³ (equivalently mol/L, sometimes written M). The recurring trap is volume units: 1 dm³ = 1000 cm³, so a volume given in cm³ must be divided by 1000 before it goes into the formula. In the magnesium example above, converting 100 cm³ to 0.100 dm³ is exactly the step that makes the numbers come out right.

$c = \frac{n}{V} \quad\Longleftrightarrow\quad n = c \times V$

Real reactions rarely give as much product as the equation predicts. The percentage yield compares what you actually obtained with the theoretical maximum from stoichiometry: $\%\,\text{yield} = \dfrac{\text{actual}}{\text{theoretical}} \times 100$, comparing like with like (both in moles, or both as mass of the same substance). Atom economy is different — 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 produces a lot of unwanted by-product; both matter in green chemistry.

This is the showcase for a calculation topic. In Paper 2, the marks are analytic: each is tied to a specific piece 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 is only true if your method is written down. Study how each mark below is earned by a specific line.

Every quantitative topic ahead is built on these conversions. Titration calculations are concentration-and-volume moles crossed through a mole ratio; the ideal gas law swaps mass-to-moles for pressure–volume–temperature; thermochemistry multiplies moles by an energy change per mole. Master the three-move habit — convert to moles, cross the mole ratio, convert back — and the rest of the course becomes variations on a method you already own.