IB Chemistry SLTopic 5 — Quantifying Chemical ChangePaper 1 & 2Core skill~12 min read
Reacting Masses
A balanced equation counts particles, but a balance in the lab weighs grams. The mole is the converter between the two, and almost every calculation in this topic is the same short journey through it.
📚 What you need to know
n = m / M, where n is amount in mol, m is mass in g, and M is molar mass in g mol–1.
Rearranged: m = n × M.
The coefficients in the balanced equation give the mole ratio — never the mass ratio.
The route is always mass → moles → moles → mass.
Be precise about the particle: one mole of CaF2 is one mole of formula units, containing one mole of Ca2+ and two moles of F–.
Any mass unit works as long as you are consistent, since the ratios are unaffected.
Moles, mass and molar mass
The central relationship
n = m / M
amount (mol) = mass (g) ÷ molar mass (g mol–1)
Molar mass comes straight from the periodic table in the data booklet: add up the relative atomic masses of every atom in the formula. Use the two-decimal values given there and keep them all the way through, rounding only at the very end.
A mole is just a counting word, like “dozen” — it means 6.02 × 1023 of something. The question is of what. Say “one mole of chlorine” and you could mean Cl atoms or Cl2 molecules, and those differ by a factor of two.
The formula unit is the counting unit. Everything inside it is scaled by its subscript.
The route from mass to mass
Masses cannot be compared directly, because the coefficients in an equation count particles, not grams. Two moles of magnesium and two moles of magnesium oxide are equal in moles but nowhere near equal in mass. So every reacting-mass problem takes the same detour.
Three arrows, and only the middle one involves any chemistry. The outer two are pure arithmetic with the periodic table.
🧩 The method, every time
Write and balance the equation.
Find moles of the substance you know: n = m / M.
Use the coefficients to convert to moles of the substance you want.
Convert back to mass: m = n × M.
Round to a sensible number of significant figures — usually matching the data given.
WORKED EXAMPLE
Calculate the mass of calcium oxide produced when 25.0 g of calcium carbonate is heated. CaCO3(s) → CaO(s) + CO2(g)
Step 1 — molar massesM(CaCO₃) = 40.08 + 12.01 + 3(16.00) = 100.09M(CaO) = 40.08 + 16.00 = 56.08Step 2 — moles of what you known = 25.0 ÷ 100.09 = 0.2498 molStep 3 — ratioCaCO₃ : CaO is 1 : 1, so the moles carry straight across.n(CaO) = 0.2498 molStep 4 — back to mass0.2498 × 56.08 = 14.008mass of CaO = 14.0 gLighter than the carbonate, because 11.0 g left as carbon dioxide gas. Mass is conserved — it just did not all stay in the crucible.
WORKED EXAMPLE
What mass of magnesium is needed to produce 10.0 g of magnesium oxide? 2Mg(s) + O2(g) → 2MgO(s)
Step 1 — start from what you were givenThis time the known substance is the PRODUCT, so the route runs backwards.M(MgO) = 24.31 + 16.00 = 40.31n = 10.0 ÷ 40.31 = 0.2481 molStep 2 — ratio2Mg : 2MgO is 1 : 1.n(Mg) = 0.2481 molStep 3 — mass0.2481 × 24.31 = 6.031mass of Mg = 6.03 gThe route is identical in both directions. Only the starting box changes.
Units do not have to be grams. If you work in tonnes throughout, the moles come out in millions but every ratio still holds, and the answer arrives in tonnes. Just never mix the two in one calculation.
WORKED EXAMPLE
In the Haber process, N2(g) + 3H2(g) ⇌ 2NH3(g). Calculate the maximum mass of ammonia, in tonnes, obtainable from 1.00 tonne of nitrogen.
Step 1 — convert to grams1.00 tonne = 1.00 × 10⁶ gStep 2 — moles of nitrogenM(N₂) = 28.02n = 1.00 × 10⁶ ÷ 28.02 = 3.569 × 10⁴ molStep 3 — ratio 1 : 2n(NH₃) = 2 × 3.569 × 10⁴ = 7.138 × 10⁴ molStep 4 — massM(NH₃) = 17.047.138 × 10⁴ × 17.04 = 1.216 × 10⁶ g1.22 tonnes of ammoniaMore mass out than nitrogen in, because the hydrogen joined it. Nothing was created — you simply were not told the mass of the other reactant.
💡 Exam tip
Balance the equation first. A wrong coefficient poisons everything downstream.
Never apply the coefficients to masses. The ratio is a mole ratio, always.
Keep full precision in your calculator and round once, at the end.
Show the moles line explicitly. It is usually worth a mark even if the final answer slips.
Check the answer is physically sensible — a product mass should be plausible next to the reactant mass you started with.
⚠️ Common mix-up
Multiplying masses by the coefficients instead of converting to moles first.
Using Ar where Mr is needed, especially for diatomic elements: M(O2) is 32.00, not 16.00.
Confusing moles of formula units with moles of ions in a compound.
Rounding at every step and drifting away from the mark scheme’s answer.
Assuming mass is lost when a gas escapes. It is conserved; it simply left the container.
Up next: Avogadro’s Law and Molar Gas Volume — where gases turn out to be far easier than solids, because for them you can skip the weighing entirely.
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