IB Chemistry HLTopic 5 — How Much? Quantifying Chemical ChangePaper 1 & 2Core skill~12 min read
Reacting Masses
A balance measures grams. A chemical equation counts particles. Those are two different languages, and moles are the translator. Once you can travel mass → moles → moles → mass, you can answer almost every quantitative question in the course.
📚 What you need to know
n = m / M — moles equal mass in grams divided by molar mass in g mol−1.
The balancing numbers give you the mole ratio, and that ratio is the only bridge between two substances.
You cannot go from mass straight to mass. Moles always sit in the middle.
Molar mass comes from the periodic table: add up the relative atomic masses in the formula.
Be clear what particle you mean: one mole of CaF2 gives one mole of Ca2+ but two moles of F−.
Any mass unit works as long as you are consistent — grams, kilograms or tonnes.
Give answers to the same number of significant figures as the data you were given.
The one equation everything hangs on
Moles from massn = m ÷ M moles = mass (g) ÷ molar mass (g mol−1)
Rearranged, m = n × M. That is the same equation twice, and you will use both directions in almost every question — once at the start to get into moles, once at the end to get back out.
Three arrows, three operations. Learn the shape of this road and you will never be stuck wondering what to do next.
Students lose marks by dividing by the wrong molar mass at the last step. The final multiplication must use the molar mass of the substance you were asked about, not the one you started with. Label every number with what it is.
The mole ratio is just the balancing numbers
This is the step people over-think. If the equation says 2NaHCO3 gives 1Na2CO3, then the ratio is 2 : 1, so you halve the moles. That is all.
This is the reaction inside baking powder. Two moles of the powder release one mole of carbon dioxide — which is what makes a cake rise.
🧩 The method
Write the balanced equation. No equation, no ratio, no marks.
Work out the molar mass of the substance you were given.
Convert to moles:n = m / M. Keep at least four figures at this stage.
Multiply or divide by the mole ratio to get the moles of the substance you want.
Convert back to mass:m = n × M, using the new molar mass.
Round at the very end and add the unit.
Do not round early. If 0.6262 mol becomes 0.63 mol at step three, your final mass can be out by a whole significant figure. Keep the long number in your calculator and round only once, at the end.
Relative masses you will keep needing
Substance
Working
M / g mol−1
CaCO3
40.08 + 12.01 + (3 × 16.00)
100.09
CaO
40.08 + 16.00
56.08
NaHCO3
22.99 + 1.01 + 12.01 + (3 × 16.00)
84.01
Na2CO3
(2 × 22.99) + 12.01 + (3 × 16.00)
105.99
Fe2O3
(2 × 55.85) + (3 × 16.00)
159.70
CO2
12.01 + (2 × 16.00)
44.01
H2O
(2 × 1.01) + 16.00
18.02
Worked examples
WORKED EXAMPLE
Baking powder decomposes on heating: 2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g). What mass of sodium carbonate is left when 8.40 g of NaHCO3 decomposes completely?
Step 1: molar mass of NaHCO₃
22.99 + 1.01 + 12.01 + 48.00 = 84.01 g mol⁻¹Step 2: moles you started withn = 8.40 ÷ 84.01 = 0.1000 molStep 3: use the 2 : 1 ration(Na₂CO₃) = 0.1000 ÷ 2 = 0.0500 molStep 4: back to mass, new molar massm = 0.0500 × 105.99 = 5.2995 g5.30 g of Na₂CO₃ (3 s.f.)Notice the answer is smaller than 8.40 g. Water and carbon dioxide left as gas and vapour, so the solid that remains must weigh less.
WORKED EXAMPLE
Iron is extracted with carbon monoxide: Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g). Calculate the mass of iron produced from 100 g of Fe2O3, and the mass of CO needed.
Step 1: moles of iron(III) oxiden = 100 ÷ 159.70 = 0.6262 molStep 2: two ratios from one equation
Fe is 1 : 2, so n(Fe) = 2 × 0.6262 = 1.2523 mol CO is 1 : 3, so n(CO) = 3 × 0.6262 = 1.8785 molStep 3: massesm(Fe) = 1.2523 × 55.85 = 69.9 g m(CO) = 1.8785 × 28.01 = 52.6 g69.9 g of iron, needing 52.6 g of carbon monoxideOne set of moles, two different ratios. Always go back to the moles of the substance you were given rather than chaining answer onto answer.
WORKED EXAMPLE
Quicklime is made by roasting limestone: CaCO3(s) → CaO(s) + CO2(g). What mass of limestone, in tonnes, is needed to make 1.00 tonne of quicklime?
Step 1: you are working backwards, so start from the product
1.00 tonne = 1.00 × 10⁶ gStep 2: moles of CaOn = 1.00 × 10⁶ ÷ 56.08 = 17832 molStep 3: ratio is 1 : 1n(CaCO₃) = 17832 molStep 4: mass, then back to tonnesm = 17832 × 100.09 = 1.785 × 10⁶ g1.78 tonnes of limestoneYou could also spot that the ratio of masses is 100.09 : 56.08 and scale directly. Same answer, but the mole route works even when the ratio is not 1 : 1.
💡 Exam tip
Label every number. Write “n(CaO) = …” not just “n = …”. It is how you avoid using the wrong molar mass later.
Set out four clear steps. Method marks are given for the moles line and the ratio line even if the arithmetic slips.
Keep four significant figures in the middle of the calculation and round once at the end.
Work in whatever mass unit the question uses, as long as you convert consistently. Tonnes to grams is × 106.
Sanity check the size. If a product should weigh less than the reactant and your answer is bigger, something is wrong.
Watch “completely” and “excess” in the wording. They tell you the reaction went all the way and which reactant limits it.
⚠ Common mix-up
Multiplying masses by the mole ratio. The ratio only ever applies to moles, never to grams.
Using the wrong molar mass at the end — the classic error. New substance, new M.
Inverting the ratio. If 2 moles give 1 mole, you halve. Ask yourself “should this get bigger or smaller?” first.
Rounding at step three. It quietly shifts your final answer.
Forgetting to balance first. An unbalanced equation gives a plausible looking answer that is simply wrong.
Mixing units mid-question — grams in one line, tonnes in the next.
Masses are only half the story. Gases are easier to measure by volume than by weighing, and for gases there is a shortcut that skips the molar mass entirely. Up next: Avogadro’s Law and Molar Gas Volume.
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