Moles and Reacting-Mass Calculations

Show your working: that instruction is worth more in this subtopic than anywhere else in 0620. A 4-mark reacting-mass calculation typically carries three method marks and one answer mark, and examiner reports repeat that candidates with one number and no working lose everything when a digit slips. The method is the same fixed four-step routine used across this site: Supplement content, the calculation core of the stoichiometry topic, and the engine reused by titrations, electrolysis and organic yields.

The four-step method (Supplement)

Step 1, the balanced equation: write it down (or copy it from the question) and label what you know and what you want. Step 2, moles of what you know: convert the given mass, gas volume or solution into moles (n = m/M, n = V/24 for gases in dm³ at r.t.p., n = c × V for solutions). Step 3, ratio: read the mole ratio from the coefficients of the balanced equation and scale your moles. Step 4, convert out: turn the new mole figure into the requested unit.

Example with a 2:3 ratio, where most errors occur. What mass of aluminium reacts with excess oxygen to form 20.4 g of aluminium oxide? Step 1, the equation: 4Al + 3O2 → 2Al2O3. Step 2: n(Al2O3) = 20.4 ÷ 102 = 0.20 mol. Step 3: ratio Al : Al2O3 = 4 : 2, so n(Al) = 0.20 × 4/2 = 0.40 mol. Step 4: mass = 0.40 × 27 = 10.8 g. The ratio fraction is “wanted over given”: coefficients of the substance you want divided by the substance you have. The balanced equation must be correct first; formula-writing slips upstream are covered in chemical formulae.

Mass-to-gas and gas-to-mass

The same skeleton handles gases by swapping the conversion at either end. What volume of CO2 forms at r.t.p. when 5.0 g of calcium carbonate decomposes? CaCO3 → CaO + CO2. n(CaCO3) = 5.0 ÷ 100 = 0.050 mol; ratio 1:1; V = 0.050 × 24 = 1.2 dm³. Quote gas volumes in dm³ unless the question demands cm³ (then × 1000).

Limiting reactant (Supplement)

When both reactant amounts are given, neither “excess” declared, check which runs out:

Step	Action
1	Convert both reactants to moles
2	Divide each by its equation coefficient
3	Smaller result = limiting reactant
4	Base every product calculation on it

Example: 4.8 g of magnesium and 4.8 g of oxygen, 2Mg + O2 → 2MgO. n(Mg) = 4.8/24 = 0.20; n(O2) = 4.8/32 = 0.15. Divide by coefficients: 0.20/2 = 0.10 versus 0.15/1 = 0.15. Magnesium is limiting; MgO formed = 0.20 mol = 8.0 g, and oxygen is in excess. The phrase “in excess” in a question is a signal, not filler: it tells you which reactant to ignore. The full drill set, with answer walk-throughs, is in our mole calculations technique guide.

Worked exam question

Zinc reacts with dilute hydrochloric acid: Zn + 2HCl → ZnCl2 + H2. A student adds 3.25 g of zinc to a solution containing 0.080 mol of HCl. (a) Show that the zinc is in excess. (2) (b) Calculate the volume of hydrogen gas produced at r.t.p. (2)

Model answer: (a) n(Zn) = 3.25 ÷ 65 = 0.050 mol (1); 0.050 mol Zn needs 0.10 mol HCl but only 0.080 mol is present. Comparing 0.050/1 with 0.080/2 = 0.040 shows HCl is limiting, so zinc is in excess (1). (b) n(H2) = 0.080 × 1/2 = 0.040 mol (1); V = 0.040 × 24 = 0.96 dm³ (1).

Mark-by-mark: (a) first mark for zinc’s moles, second for a valid numerical comparison. A bare assertion “zinc is extra” scores once at most. (b) the ratio mark requires using HCl (the limiting reagent), halved; candidates who use zinc’s 0.050 mol get 1.2 dm³ and lose both marks unless working shows a consistent method. Final mark needs ×24 with dm³.

The mistakes that cost marks

1. Skipping the ratio. Mass → moles → straight back to mass, ignoring a 2:1 equation, is the single most frequent calculation error. Underline coefficients before you start.
2. Ratio upside-down: multiplying by given/wanted instead of wanted/given. Check direction with common sense: 2HCl per Zn means fewer H2 than HCl.
3. Building product answers on the excess reactant. The limiting reagent governs; that is its definition.
4. Hidden working. Arithmetic done on a calculator and only the answer written surrenders the method marks that exist precisely to protect you.

How examiners want it phrased

Typical student wording	Accepted mark-scheme wording
”I converted it and got 10.8"	"n(Al2O3) = 20.4/102 = 0.20 mol; n(Al) = 0.20 × 4/2 = 0.40 mol; mass = 0.40 × 27 = 10.8 g"
"There’s more zinc than acid"	"Zn would require 0.10 mol HCl; only 0.080 mol is available, so Zn is in excess"
"The acid runs out first"	"HCl is the limiting reactant, so it determines the amount of product"
"About 1 dm³ of gas"	"V = 0.040 mol × 24 dm³/mol = 0.96 dm³”

The Malaysia note

Reacting-mass questions are where Malaysian Extended candidates most visibly split into two groups by October mocks: those with a written routine, and those who recalculate from instinct each time. The routine wins, repeatably. The concept maps directly onto SPM’s “konsep mol” questions, so dual-curriculum students have double the practice material available. If your working currently lives in the calculator instead of on paper, one free trial lesson is enough to install the three-line layout that examiners pay method marks for.

Test yourself

Three fresh calculations. Write the four steps out in full before you click each answer.

Q1 (3 marks). Iron is extracted in the blast furnace: Fe2O3 + 3CO → 2Fe + 3CO2. Calculate the mass of iron produced from 32.0 g of iron(III) oxide. [Mr of Fe2O3 = 160; Ar of Fe = 56]

Show answer

• n(Fe2O3) = 32.0 ÷ 160 = 0.20 mol [1] • Ratio Fe2O3 : Fe = 1 : 2, so n(Fe) = 0.40 mol [1] • Mass of Fe = 0.40 × 56 = 22.4 g [1]

Q2 (3 marks). Magnesium carbonate decomposes on heating: MgCO3 → MgO + CO2. Calculate the volume of carbon dioxide, in dm³ at r.t.p., produced when 4.2 g of magnesium carbonate is fully decomposed. [Mr of MgCO3 = 84]

Show answer

• n(MgCO3) = 4.2 ÷ 84 = 0.050 mol [1] • Ratio 1 : 1, so n(CO2) = 0.050 mol [1] • V = 0.050 × 24 = 1.2 dm³ [1]

Q3 (4 marks, Supplement). Hydrogen reacts with nitrogen: N2 + 3H2 → 2NH3. A mixture of 5.6 g of nitrogen and 0.90 g of hydrogen is reacted. (a) Identify the limiting reactant, showing your working. (2) (b) Calculate the maximum mass of ammonia formed. (2) [Ar: N 14, H 1]

Show answer

• (a) n(N2) = 5.6 ÷ 28 = 0.20 mol; n(H2) = 0.90 ÷ 2 = 0.45 mol [1] • Divide by coefficients: 0.20/1 = 0.20 vs 0.45/3 = 0.15, so hydrogen is the limiting reactant [1] • (b) n(NH3) = 0.45 × 2/3 = 0.30 mol [1] • Mass = 0.30 × 17 = 5.1 g [1]