IB DP Chemistry Structure 1.4: Counting particles by mass: the mole – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

This set practises Structure 1.4, “Counting particles by mass: the mole”, in line with the IB Diploma Programme Chemistry guide for first assessment in 2025 (published in February 2023). Statements Structure 1.4.1 to Structure 1.4.6 are all tested. Structure 1.4 has no AHL material, so every question suits SL and HL alike, for the course examined from 2025 onwards.

Work through the study guide and the revision notes before you start. Everything else for the course is on the DP Chemistry course page and in the checklist to print; our short diagnostics show which topic needs work. Mixed questions on the whole component are in the component-level question set.

Use these data throughout: NA = 6.02 × 10²³ mol⁻¹. Ar values: H 1.01, He 4.00, B 10.81, C 12.01, N 14.01, O 16.00, Na 22.99, Mg 24.31, P 30.97, S 32.07, Cl 35.45, K 39.10, Ca 40.08, Mn 54.94, Fe 55.85. Give answers to 3 significant figures unless told otherwise.

Questions

1. A sample contains 0.0350 mol of phosphorus pentachloride, PCl₅.

(a) Calculate the number of PCl₅ molecules in the sample. [1] (b) Calculate the number of chlorine atoms in the sample. [1]

2. Calculate the relative formula mass, Mr, of calcium phosphate, Ca₃(PO₄)₂. [2]

3. Calculate the number of oxygen atoms in 7.50 g of potassium nitrate, KNO₃. [3]

4. Iron(II) sulfate crystals have the formula FeSO₄·7H₂O. Calculate the percentage by mass of water in the crystals. Give your answer to 2 decimal places. [3]

5. An oxide of manganese contains 72.03% manganese by mass. Determine its empirical formula. Show your working. [4]

6. A liquid used in antifreeze contains 38.69% carbon, 9.76% hydrogen and 51.55% oxygen by mass. Its molar mass is 62.08 g mol⁻¹.

(a) Determine the empirical formula of the compound. [3] (b) Deduce its molecular formula. [1]

7. A technician prepares 500.0 cm³ of sodium thiosulfate solution, Na₂S₂O₃(aq), with [Na₂S₂O₃] = 0.0800 mol dm⁻³.

(a) Calculate the mass of Na₂S₂O₃ needed. [3] (b) Express the concentration in g dm⁻³. [1]

8. 20.0 cm³ of magnesium chloride solution, [MgCl₂] = 0.250 mol dm⁻³, is diluted with water to a total volume of 250.0 cm³.

(a) Calculate [MgCl₂] in the diluted solution. [2] (b) State [Cl⁻] in the diluted solution. [1]

9. Ammonia burns in oxygen at high temperature. All substances are gases under the conditions used.

4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)

(a) State Avogadro’s law. [1] (b) 60 cm³ of ammonia is mixed with 100 cm³ of oxygen and reacts completely. Calculate the total volume of gas after the reaction, measured under the same conditions. [4]

10. Deduce which of these samples contains the greatest total number of atoms: 1.00 g of hydrogen gas, H₂; 1.00 g of helium, He; 1.00 g of methane, CH₄. Show your working. [3]

11. A gaseous hydride of boron, X, contains 78.11% boron and 21.89% hydrogen by mass. Its molar mass is 27.68 g mol⁻¹.

(a) Determine the empirical formula of X. [3] (b) Find the molecular formula of X, showing your reasoning. [2] (c) Calculate the number of hydrogen atoms in 0.554 g of X. [2] (d) X burns in oxygen: X(g) + 3O₂(g) → B₂O₃(s) + 3H₂O(l). State the volume of oxygen that reacts with 15 cm³ of X, both measured under the same conditions. [1]

12. A solution of potassium sulfate, K₂SO₄, has a concentration of 8.71 g dm⁻³.

(a) Calculate [K₂SO₄] in mol dm⁻³. [2] (b) State [K⁺] in the solution. [1] (c) Calculate the number of K⁺ ions in 30.0 cm³ of the solution. [2] (d) Calculate the volume, in cm³, of the solution that contains 0.0125 mol of K₂SO₄. [2]

Answers

1. (a) 0.0350 × 6.02 × 10²³ = 2.11 × 10²² molecules [1] (b) 5 Cl per molecule: 5 × 0.0350 × 6.02 × 10²³ = 1.05 × 10²³ chlorine atoms [1] Examiner insight: In part (b) the factor of 5 must appear in your working; an answer equal to part (a) shows you counted molecules, not atoms.

2. Ca: 3 × 40.08 = 120.24; PO₄: 2 × (30.97 + 4 × 16.00) = 189.94 [1] Mr = 120.24 + 189.94 = 310.18 (no units) [1] Examiner insight: Adding “g mol⁻¹” to an Mr is wrong, because Mr is a ratio; keep that unit for molar mass.

3. M(KNO₃) = 39.10 + 14.01 + 3 × 16.00 = 101.11 g mol⁻¹ and n = 7.50 / 101.11 = 0.07418 mol [1] n(O) = 3 × 0.07418 = 0.2225 mol [1] number of O atoms = 0.2225 × 6.02 × 10²³ = 1.34 × 10²³ [1] Examiner insight: Show the moles at each stage; if the final number is wrong, visible working is the only way a marker can see which steps you got right.

4. M(FeSO₄·7H₂O) = 55.85 + 32.07 + 4 × 16.00 + 7 × 18.02 = 278.06 g mol⁻¹ [1] mass of water = 7 × (2 × 1.01 + 16.00) = 126.14 [1] % water = 126.14 / 278.06 × 100 = 45.36% [1] Examiner insight: The question asks for 2 decimal places, so 45.4% does not follow the instruction; always obey a stated precision over the usual 3 s.f.

5. % O = 100 − 72.03 = 27.97% [1] Mn: 72.03 / 54.94 = 1.311 mol; O: 27.97 / 16.00 = 1.748 mol [1] ratio Mn : O = 1 : 1.333 [1] × 3 gives 3 : 4, so empirical formula Mn₃O₄ [1] Examiner insight: Rounding 1 : 1.333 to 1 : 1 gives MnO, which is a real oxide but the wrong answer; recognise .33 as a signal to multiply by 3.

6. (a) C: 38.69 / 12.01 = 3.221; H: 9.76 / 1.01 = 9.663; O: 51.55 / 16.00 = 3.222 [1] divide by smallest: C 1.00, H 3.00, O 1.00 [1] empirical formula CH₃O [1] (b) empirical formula mass = 12.01 + 3 × 1.01 + 16.00 = 31.04; 62.08 / 31.04 = 2, so C₂H₆O₂ [1] Examiner insight: For the molecular formula, show the division of molar mass by empirical formula mass; stating C₂H₆O₂ with no reason risks losing credit if the multiple is not evident.

7. (a) n = C × V = 0.0800 × 0.5000 = 0.0400 mol [1] M(Na₂S₂O₃) = 2 × 22.99 + 2 × 32.07 + 3 × 16.00 = 158.12 g mol⁻¹ [1] m = 0.0400 × 158.12 = 6.32 g [1] (b) 0.0800 × 158.12 = 12.6 g dm⁻³ [1] Examiner insight: Using V = 500.0 instead of 0.5000 dm³ makes the mass a thousand times too large; check that a lab mass looks sensible before moving on.

8. (a) n(MgCl₂) = 0.250 × 0.0200 = 5.00 × 10⁻³ mol [1] [MgCl₂] = 5.00 × 10⁻³ / 0.2500 = 0.0200 mol dm⁻³ [1] (b) two Cl⁻ per formula unit: [Cl⁻] = 0.0400 mol dm⁻³ [1] Examiner insight: Divide by the total final volume (250.0 cm³), not by the volume of water added (230.0 cm³).

9. (a) At the same temperature and pressure, equal volumes of all gases contain the same number of molecules. [1] (b) 60 cm³ NH₃ needs 60 × 5/4 = 75 cm³ O₂, so O₂ is in excess by 100 − 75 = 25 cm³ [1] NO formed = 60 cm³ (ratio 4 : 4) [1] H₂O formed = 60 × 6/4 = 90 cm³ (all gases here) [1] total = 25 + 60 + 90 = 175 cm³ [1] Examiner insight: Read the conditions: here water is a gas, so it counts; in many combustion questions it is cooled to a liquid and does not.

10. H₂: n = 1.00 / 2.02 = 0.495 mol, atoms = 2 × 0.495 = 0.990 mol [1] He: n = 1.00 / 4.00 = 0.250 mol of atoms; CH₄: n = 1.00 / 16.05 = 0.0623 mol, atoms = 5 × 0.0623 = 0.312 mol [1] Hydrogen gas has the most atoms (5.96 × 10²³) [1] Examiner insight: Comparing moles of atoms is enough; you need not multiply each by NA, but you must multiply moles of molecules by the atoms per molecule.

11. (a) B: 78.11 / 10.81 = 7.226 mol; H: 21.89 / 1.01 = 21.67 mol [1] ratio B : H = 1 : 3.00 [1] empirical formula BH₃ [1] (b) empirical formula mass = 10.81 + 3 × 1.01 = 13.84; 27.68 / 13.84 = 2 [1] molecular formula B₂H₆ [1] (c) n(B₂H₆) = 0.554 / 27.68 = 0.02001 mol, so n(H) = 6 × 0.02001 = 0.1201 mol [1] number of H atoms = 0.1201 × 6.02 × 10²³ = 7.23 × 10²² [1] (d) ratio X : O₂ = 1 : 3, so 45 cm³ of O₂ [1] Examiner insight: Part (c) needs the molecular formula from (b); if (b) is wrong, set out (c) clearly so your method can still be followed from your own value.

12. (a) M(K₂SO₄) = 2 × 39.10 + 32.07 + 4 × 16.00 = 174.27 g mol⁻¹ [1] [K₂SO₄] = 8.71 / 174.27 = 0.0500 mol dm⁻³ [1] (b) [K⁺] = 2 × 0.0500 = 0.100 mol dm⁻³ [1] (c) n(K⁺) = 0.100 × 0.0300 = 3.00 × 10⁻³ mol [1] number = 3.00 × 10⁻³ × 6.02 × 10²³ = 1.81 × 10²¹ K⁺ ions [1] (d) V = n / C = 0.0125 / 0.0500 = 0.250 dm³ [1] = 250 cm³ [1] Examiner insight: Part (d) asks for cm³; giving 0.250 with no unit, or with dm³, does not answer the question set.

Where marks are usually lost

• Leaving out the atoms-per-formula-unit factor when the question names atoms or ions.
• Giving Mr a unit, or giving molar mass without one.
• Rounding the mole ratio in an empirical formula too early, so 1.333 becomes 1.3 and then 1.
• Forgetting the oxygen percentage when only the metal’s percentage is given.
• Using cm³ directly in n = CV.
• Dividing by the volume of water added instead of the final volume after dilution.
• Writing the concentration of an ion equal to the concentration of the salt.
• Forgetting the excess reactant gas, or counting a liquid product, in a final gas volume.
• Not following a stated precision, such as “2 decimal places”.

Next steps

• Revision notes for Structure 1.4
• Study guide for Structure 1.4
• IB DP Chemistry course hub
• Printable IB DP Chemistry checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

Source document: Diploma Programme Chemistry guide (International Baccalaureate Organization), for first assessment in 2025, published in February 2023. This page follows its Structure 1.4 statements.