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Topic 3 · 3.3

The mole and calculations

Core and Extended · mostly Supplement · Papers 2, 4, 5 and 6

Core candidates need only the first objective (concentration units). Everything else on this page is EXTENDED (Supplement) content — and it appears on almost every Paper 4.

🎯What you need to be able to do

  • State that concentration can be measured in g/dm³ or mol/dm³.
  • State that one mole contains \( 6.02 \times 10^{23} \) particles (the Avogadro constant) EXTENDED.
  • Use amount (mol) \( = \) mass (g) \( \div \) molar mass (g/mol), and the molar gas volume of 24 dm³ at r.t.p. EXTENDED.
  • Calculate reacting masses, limiting reactants, gas volumes, and concentrations in g/dm³ and mol/dm³ EXTENDED.
  • Use titration results to find moles, concentration or volume EXTENDED.
  • Calculate empirical and molecular formulae, percentage yield, percentage composition by mass and percentage purity EXTENDED.

📚The chemistry

The mole

A mole is the amount of substance containing \( 6.02 \times 10^{23} \) particles (atoms, molecules or ions) — the Avogadro constant. The mass of one mole is the molar mass: the Ar or Mr in grams. One mole of CO2 (Mr 44) weighs 44 g.

\( n = \dfrac{m}{M} \) (mol, g, g/mol)
particles \( = n \times 6.02 \times 10^{23} \)
gas volume at r.t.p. \( = n \times 24 \) dm³
\( n = c \times V \) (mol/dm³ × dm³)
A map with moles at the centre and four quantities around it: mass in grams (divide by molar mass to get moles, multiply to go back), number of particles (divide or multiply by 6.02 times 10 to the 23), gas volume at room temperature and pressure (divide or multiply by 24 cubic decimetres), and a solution's volume in cubic decimetres times concentration.
Every calculation goes through moles. Convert what you know into moles, use the equation’s ratio, then convert out.

Volumes are often given in cm³: divide by 1000 to get dm³. A concentration in g/dm³ is the mol/dm³ value times the molar mass.

Calculations from equations

  1. Write the balanced equation.
  2. Convert the known quantity to moles.
  3. Use the mole ratio from the equation.
  4. Convert the answer to the unit asked for (mass, gas volume, concentration…).

If amounts of two reactants are given, one will run out first: the limiting reactant. Compare the moles you have with the ratio in the equation; the product is calculated from the limiting reactant, and the other is in excess.

Titrations use the same steps: moles of the solution whose concentration is known (\( c \times V \)), then the ratio, then the unknown concentration (\( n \div V \)).

Formulae and percentages

  • Empirical formula: divide each mass (or percentage) by Ar, then divide by the smallest to get the whole-number ratio.
  • Molecular formula: divide Mr by the empirical formula mass and multiply up.
  • Percentage yield \( = \dfrac{\text{actual yield}}{\text{theoretical yield}} \times 100 \).
  • Percentage composition of an element \( = \dfrac{\text{total } A_\mathrm{r} \text{ of that element}}{M_\mathrm{r}} \times 100 \).
  • Percentage purity \( = \dfrac{\text{mass of pure substance}}{\text{mass of impure sample}} \times 100 \).
A bar comparison for magnesium reacting with hydrochloric acid, Mg + 2HCl. Available: 0.10 mol magnesium and 0.10 mol HCl. Needed to use all the magnesium: 0.20 mol HCl, so HCl runs out first. 0.10 mol HCl reacts with only 0.05 mol magnesium, leaving 0.05 mol magnesium in excess and making 0.05 mol hydrogen.
Practice question 9: the equation needs 2 HCl per Mg, so 0.10 mol HCl is used up by only 0.05 mol Mg. HCl is limiting.

✏️Worked example EXTENDED

Ar: H 1, C 12, O 16, Na 23, S 32, Ca 40. (a) Calculate the number of moles and the number of molecules in 8.8 g of carbon dioxide. [2] (b) \( \mathrm{CaCO_3(s) + 2HCl(aq) \rightarrow CaCl_2(aq) + H_2O(l) + CO_2(g)} \). Calculate the volume of CO2, in cm³ at r.t.p., made from 5.0 g of calcium carbonate. [3] (c) 25.0 cm³ of aqueous sodium hydroxide is neutralised by 20.0 cm³ of 0.100 mol/dm³ sulfuric acid: \( \mathrm{H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O} \). Calculate the concentration of the sodium hydroxide. [3] (d) A hydrocarbon is 85.7% carbon and 14.3% hydrogen by mass, and its Mr is 56. Find its empirical and molecular formulae. [3]

(a) \( n = \tfrac{8.8}{44} = 0.20 \) mol; molecules \( = 0.20 \times 6.02 \times 10^{23} = 1.2 \times 10^{23} \).

(b) \( n(\mathrm{CaCO_3}) = \tfrac{5.0}{100} = 0.050 \) mol. Ratio 1 : 1, so 0.050 mol CO2. Volume \( = 0.050 \times 24 = 1.2 \) dm³ \( = 1200 \) cm³.

(c) \( n(\mathrm{H_2SO_4}) = 0.100 \times \tfrac{20.0}{1000} = 0.00200 \) mol. Ratio 1 : 2, so \( n(\mathrm{NaOH}) = 0.00400 \) mol. Concentration \( = \dfrac{0.00400}{0.0250} = 0.160 \) mol/dm³.

(d) C: \( \tfrac{85.7}{12} = 7.14 \); H: \( \tfrac{14.3}{1} = 14.3 \). Divide by 7.14: C 1, H 2, so the empirical formula is CH2 (mass 14). \( 56 \div 14 = 4 \): molecular formula C4H8.

Check it. Units ride along: in (c), mol \( \div \) dm³ gives mol/dm³. In (d), Mr of C4H8 is \( 48 + 8 = 56 \) ✓.
Ignoring the ratio, or the cm³. The two classic losses on titration questions are forgetting that one H2SO4 reacts with two NaOH (giving 0.0800 instead of 0.160) and using 20.0 instead of 0.0200 dm³.

📝Practise

All Supplement unless stated. Ar: H 1, C 12, N 14, O 16, Na 23, Mg 24, Cl 35.5, K 39, Ca 40. Molar gas volume 24 dm³ at r.t.p.

1. (Multiple choice.) How many moles are in 4.0 g of sodium hydroxide, NaOH? A: 0.010. B: 0.10. C: 1.0. D: 10.
B. Mr \( = 23 + 16 + 1 = 40 \); \( 4.0 \div 40 = 0.10 \).
2. (Theory.) Calculate the volume of 3.2 g of oxygen gas, O2, at r.t.p. [2]
\( 3.2 \div 32 = 0.10 \) mol; \( 0.10 \times 24 = 2.4 \) dm³.
3. (Theory.) 5.85 g of sodium chloride is dissolved to make 250 cm³ of solution. Calculate the concentration in mol/dm³ and in g/dm³. [3]
\( 5.85 \div 58.5 = 0.100 \) mol in 0.250 dm³: 0.400 mol/dm³. In g/dm³: \( 5.85 \div 0.250 = 23.4 \) g/dm³.
4. (Theory.) 22.50 cm³ of 0.200 mol/dm³ hydrochloric acid neutralises 25.0 cm³ of aqueous potassium hydroxide. Calculate the concentration of the potassium hydroxide. [3]
\( n(\mathrm{HCl}) = 0.200 \times 0.02250 = 0.00450 \) mol; ratio 1 : 1; \( c = 0.00450 \div 0.0250 = 0.180 \) mol/dm³.
5. (Theory.) A compound contains 0.72 g of carbon, 0.12 g of hydrogen and 0.96 g of oxygen. Its Mr is 180. Find its empirical and molecular formulae. [3]
C \( 0.72 \div 12 = 0.06 \); H \( 0.12 \div 1 = 0.12 \); O \( 0.96 \div 16 = 0.06 \). Ratio 1 : 2 : 1, empirical formula CH2O (mass 30). \( 180 \div 30 = 6 \): C6H12O6.
6. (Theory.) Calculate the percentage by mass of nitrogen in ammonium nitrate, NH4NO3. [2]
Mr \( = 80 \); nitrogen \( 2 \times 14 = 28 \); \( \tfrac{28}{80} \times 100 = 35\% \).
7. (Theory.) Heating 10.0 g of calcium carbonate produced 5.0 g of calcium oxide. Calculate the percentage yield. [3]
Theoretical: \( 10.0 \div 100 = 0.100 \) mol CaCO3 gives 0.100 mol CaO \( = 5.6 \) g. Yield \( = \tfrac{5.0}{5.6} \times 100 = 89\% \) (89.3%).
8. (Theory.) A 5.0 g sample of impure calcium carbonate reacts with excess acid to give 1.056 dm³ of carbon dioxide at r.t.p. Calculate the percentage purity of the sample. [3]
\( n(\mathrm{CO_2}) = 1.056 \div 24 = 0.0440 \) mol \( = n(\mathrm{CaCO_3}) \); mass \( = 0.0440 \times 100 = 4.40 \) g. Purity \( = \tfrac{4.40}{5.0} \times 100 = 88\% \).
9. (Theory.) 2.4 g of magnesium is added to 50.0 cm³ of 2.00 mol/dm³ hydrochloric acid: \( \mathrm{Mg + 2HCl \rightarrow MgCl_2 + H_2} \). Identify the limiting reactant and calculate the volume of hydrogen made at r.t.p. [4]
Mg: \( 2.4 \div 24 = 0.100 \) mol. HCl: \( 2.00 \times 0.0500 = 0.100 \) mol. 0.100 mol Mg would need 0.200 mol HCl, so HCl is limiting. \( n(\mathrm{H_2}) = 0.100 \div 2 = 0.0500 \) mol; volume \( = 0.0500 \times 24 = 1.2 \) dm³.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • Royal Society of Chemistry — moles and titration calculation practice sheets
  • Your calculator — use the standard-form key (×10x) for Avogadro calculations