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IGCSE Moles and Stoichiometry Made Simple: The Complete Notes

These IGCSE moles and stoichiometry notes exist because most students meet the mole concept as a pile of separate formulae: one for mass, one for gas volume, one for concentration, another for empirical formula, and try to memorise all of them at once. That’s usually where the confusion starts. Under exam pressure, five or six unconnected formulae are hard to recall accurately, and it’s easy to reach for the wrong one.

The mole concept in Cambridge IGCSE Chemistry (0620) is actually one idea, applied in different situations. Once that idea is clear, stoichiometry stops being a memory test and becomes a repeatable process. These IGCSE moles and stoichiometry notes build it in order, the way it’s meant to be understood: as a ladder, where each step depends on the one below it.

Step 1: Ar and Mr, turning a formula into a mass

Relative atomic mass (Ar) comes straight from the periodic table: it’s the larger of the two numbers next to each element’s symbol, not the smaller one (the smaller number is the atomic number, and mixing the two up is one of the most common early errors in this topic).

Relative formula mass (Mr) is just the Ar values added together according to the formula. For sodium carbonate, Na₂CO₃:

Mr = (2 × 23) + 12 + (3 × 16) = 106

This step feels like simple arithmetic, but it’s actually the bridge between a chemical formula and a real, measurable mass in grams. Get it wrong here, and every calculation built on top of it is wrong too.

Step 2: what a mole actually is

A mole is a counting unit: the same idea as “a dozen,” except the number is enormous, because atoms are enormously small. One mole of any substance contains 6.02 × 10²³ particles (the Avogadro constant), and the mass of one mole, in grams, equals its Ar or Mr.

This is the conceptual leap worth pausing on. Students who treat “mole” as an abstract exam word rather than an actual quantity of substance tend to struggle with everything that follows, and it’s rarely because the maths is hard; usually it’s because they’re manipulating a symbol they don’t really picture.

Step 3: the three routes into moles

That’s the core of these IGCSE moles and stoichiometry notes: every stoichiometry question, underneath all the different wording, is about getting to moles and then getting back out again. There are exactly three ways in:

From mass: moles = mass (g) ÷ molar mass (g/mol)
From gas volume at r.t.p.: moles = volume (dm³) ÷ 24
From solution concentration: moles = concentration (mol/dm³) × volume (dm³)

Each of these also runs in reverse: mass = moles × Mr, volume = moles × 24, concentration = moles ÷ volume. Learning them as three relationships with an inverse each, rather than six separate formulae, cuts the memory load in half.

One detail that trips students up specifically in the gas and concentration routes: volumes are often given in cm³ but the formulae need dm³. Divide by 1000 to convert cm³ to dm³ before using either formula, and do it as the first step rather than an afterthought at the end.

Step 4: the balanced equation as a ratio machine

This is the step that turns “moles” into “stoichiometry.” The coefficients in a balanced equation give the mole ratio between reactants and products. A student who’s solid on Steps 1–3 but skips this one will correctly find the moles of the substance they’re given, then quietly assume a 1:1 ratio for the substance they actually need, which is the single most common source of lost marks in this whole topic.

The habit that fixes it: read the ratio off the equation explicitly, in writing, before calculating anything. Not mentally. Written down, every time.

Step 5: applying the ladder

This is where these IGCSE moles and stoichiometry notes turn into exam technique. Once Steps 1–4 are solid, everything else in stoichiometry is the same four-move sequence (equation, moles of what you know, ratio, moles/mass/volume of what you need) applied to a specific situation:

Limiting reactant: calculate moles of each reactant given, compare each to what the equation requires, and identify which one runs out first. The reactant left over doesn’t affect how much product forms.
Percentage yield: (actual yield ÷ theoretical yield) × 100. The theoretical yield has to come from a full stoichiometric calculation based on the limiting reactant — never a guess.
Percentage composition or purity: mass of the pure component ÷ total mass, ×100.
Empirical and molecular formula: convert masses (or percentages) to moles, divide every value by the smallest number of moles to get a ratio, then convert that ratio to whole numbers.
Titrations: use the volume and concentration of the solution you know to find its moles, apply the equation’s ratio, then find the moles, concentration, or volume of the second solution.

Seen this way, a student isn’t learning eight unrelated formulae; they’re applying one process to eight different contexts, which is the whole point of building these IGCSE moles and stoichiometry notes as a ladder rather than a list.

Multi-step questions are just two familiar steps, joined

The questions that feel hardest, mass of one reactant to mass of a completely different product, aren’t a new type of maths, just a longer one. There’s no single formula that jumps straight from one to the other: mass → moles (Step 3), moles → moles (Step 4, the ratio bridge), moles → mass (Step 3 again, for the new substance). Once a student sees the bridge in the middle, “hard” multi-step problems stop feeling unfamiliar.

Complete formula reference

The table below is the compressed version of these IGCSE moles and stoichiometry notes: every formula in one place, plus the exact spot students lose marks.

Find this	Formula	Watch for
Relative formula mass, Mr	Sum of (Ar × number of each atom)	Use Ar, not atomic number
Moles from mass	moles = mass (g) ÷ Mr	Mass always in grams
Mass from moles	mass = moles × Mr
Moles from gas volume	moles = volume (dm³) ÷ 24	r.t.p. only; convert cm³ → dm³ by ÷1000
Moles from concentration	moles = concentration (mol/dm³) × volume (dm³)	Convert cm³ → dm³ by ÷1000
Mole ratio between substances	Read from the balanced equation’s coefficients	Never assume 1:1
Percentage yield	(actual yield ÷ theoretical yield) × 100	Theoretical yield needs a full calculation
Empirical formula ratio	moles of each element ÷ smallest moles value	Never round — multiply to clear fractions

The takeaway from these IGCSE moles and stoichiometry notes

Stoichiometry looks like a long list of formulae from the outside. From the inside, it’s one idea, moles as a counting unit, applied through three ways in, one ratio step, and a handful of applied skills built on top. Students who learn it as a connected ladder, rather than a list to memorise, tend to find that the “harder” topics later in the syllabus (electrolysis calculations, energetics, titrations) get noticeably easier too, because they’re reusing the same toolkit.

If your child understands each piece on its own but struggles to connect them under exam conditions, that’s a completely normal stage, and it’s exactly what we work through in Nowa Link’s IGCSE moles and stoichiometry coaching: building the connections between the steps until the whole ladder feels like one process instead of eight, rather than re-explaining the mole concept from scratch.

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