94 exam-style questions on Counting particles by mass: the mole — topic S1.4 of Structure 1: Models of the particulate nature of matter in the IB Chemistry syllabus — with answers and explanations. Samples below; the full set is free with an account.
Calculate the number of molecules present in 3.0 g of methane, CH4. (M(CH4) = 16.0 g mol⁻¹; NA = 6.02 × 10²³ mol⁻¹.)
Answer: 1.1 × 10²³
n = 3.0 / 16.0 = 0.1875 mol; molecules = 0.1875 × 6.02 × 10²³ ≈ 1.1 × 10²³.
Determine the relative molecular mass of ammonium sulfate, (NH4)2SO4. (Ar: N = 14.0, H = 1.0, S = 32.1, O = 16.0.)
Answer: 132.1
2(14.0 + 4×1.0) + 32.1 + 4×16.0 = 2(18.0) + 32.1 + 64.0 = 36.0 + 96.1 = 132.1.
A hydrocarbon has empirical formula CH2 and a molar mass of 56.0 g mol⁻¹. Determine its molecular formula. (Ar: C = 12.0, H = 1.0.)
Answer: C4H8
Empirical mass = 14.0; 56.0 / 14.0 = 4, so molecular formula = (CH2)4 = C4H8.
In the reaction N2(g) + 3H2(g) → 2NH3(g), 2.0 mol of N2 reacts with 3.0 mol of H2. Determine the maximum amount, in mol, of NH3 that can form.
Answer: 2.0
H2 is limiting: 3.0 mol H2 × (2 NH3 / 3 H2) = 2.0 mol NH3. N2 needed is only 1.0 mol, so N2 is in excess.
Calculate the mass of anhydrous sodium hydroxide required to prepare 500 cm³ of a 0.200 mol dm⁻³ solution. (M(NaOH) = 40.0 g mol⁻¹.)
Answer: 4.00 g
n = c × V = 0.200 × 0.500 = 0.100 mol; mass = 0.100 × 40.0 = 4.00 g.
Which sample contains the greatest number of atoms?
Answer: 1.0 mol of H2
Atoms = mol × atoms per formula unit × NA. Na: 1.0×1 = 1.0; H2O: 0.5×3 = 1.5; CO2: 0.5×3 = 1.5; H2: 1.0×2 = 2.0 mol of atoms — the largest.
That was 6 of 94 Counting particles by mass: the mole questions.
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94 Counting particles by mass: the mole questions — free with an account. Spaced repetition, streaks and full exam simulations included.
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