The problems are on the mole calculations worksheet. Try them on your own first, then compare with the answers below.
Quick answer: The mole is the chemist's counting unit: one mole contains 6.02 × 10²³ particles, the Avogadro constant. These worked answers correct the 20 problems of the Chemistry 9-10 worksheet on mole calculations, from mole and Avogadro to baking soda: each solution shows the formula, the substitution and the final answer with its unit.
Also, compare your working with these answers, not only your final numbers. Then, check that you used the molar mass of the whole particle, followed the units through each step and gave answers to a sensible number of significant figures.
Level: Chemistry grades 9-10 (GCSE · IGCSE · O Level · Class 9-10) · Curriculum: HS-PS1-7, GCSE C3
Problem 1 Mole and Avogadro
1. Therefore, we get 6.02 × 10²³ per mole.2. Also, we get 6.02 × 10²³ atoms.
3. 2 × 6.02 × 10²³ = 1.204 × 10²⁴ ≈ 1.20 × 10²⁴ atoms.
6.02 × 10²³ per mol; 1.20 × 10²⁴ atoms in 2 mol.
Problem 2 Mass of one mole of atoms
1. The molar mass in g/mol equals the relative atomic mass in grams.C 12 g, Mg 24 g, S 32 g, Cu 63.5 g.
Problem 3 Relative formula masses
1. Moreover, we get 12 + 2 × 16 = 44.2. Thus, we get 23 + 16 + 1 = 40.
3. 40 + 12 + 3 × 16 = 100.
4. Therefore, we get 2 × 1 + 32 + 4 × 16 = 98.
44, 40, 100 and 98.
Problem 4 Moles of water
1. Also, we get H₂O: 2 × 1 + 16 = 18 g/mol.2. n = 36 ÷ 18 = 2.0 mol.
2.0 mol
Problem 5 Mass from moles
1. Moreover, we get m = n × M.2. Thus, we get m = 0.50 × 58.5 = 29.25 g, which is 29.3 g to 3 significant figures.
29.3 g
Problem 6 Reading the balances
1. H₂O: 18 g/mol. CO₂: 44 g/mol. NaCl: 58.5 g/mol.2. Each sample is 18.0 ÷ 18 = 1.00, 44.0 ÷ 44 = 1.00 and 58.5 ÷ 58.5 = 1.00 mol.
3. Each contains 6.02 × 10²³ particles (molecules or formula units).
1.00 mol each, 6.02 × 10²³ particles each.
Problem 7 Atoms from moles
1. Therefore, we get N = n × 6.02 × 10²³ = 0.20 × 6.02 × 10²³ = 1.204 × 10²³.1.20 × 10²³ atoms
Problem 8 Brackets in formulae
1. Also, we get 40 + 2 × 17 = 74.2. 2 × 27 + 3 × (32 + 64) = 54 + 288 = 342.
3. Moreover, we get 2 × (14 + 4) + 12 + 48 = 36 + 60 = 96.
4. Thus, we get 24 + 2 × (14 + 48) = 24 + 124 = 148.
74, 342, 96 and 148.
Problem 9 Moles from mass
1. M = 44. n = 4.4 ÷ 44 = 0.10 mol.2. Therefore, we get M = 98. n = 49 ÷ 98 = 0.50 mol.
3. Also, we get M = 27. n = 2.7 ÷ 27 = 0.10 mol.
0.10 mol, 0.50 mol, 0.10 mol
Problem 10 Mass from moles, harder
1. M = 100. m = 0.25 × 100 = 25 g.2. Moreover, we get M = 18. m = 3.0 × 18 = 54 g.
3. Thus, we get M = 56. m = 0.10 × 56 = 5.6 g.
25 g, 54 g, 5.6 g
Problem 11 Molecules in a raindrop
1. n = 0.050 ÷ 18 = 0.002778 mol, which is 2.78 × 10⁻³ mol.2. Therefore, we get N = 0.002778 × 6.02 × 10²³ = 1.67 × 10²¹.
2.78 × 10⁻³ mol; 1.67 × 10²¹ molecules
Problem 12 Atoms in methane
1. Also, we get N = 0.40 × 6.02 × 10²³ = 2.408 × 10²³ ≈ 2.41 × 10²³ molecules.2. 1 carbon and 4 hydrogen atoms = 5 atoms.
3. Moreover, we get 0.40 × 5 × 6.02 × 10²³ = 1.204 × 10²⁴ ≈ 1.20 × 10²⁴ atoms.
1.20 × 10²⁴ atoms
Problem 13 Mass against moles
1. Thus, we get 12 g.2. Gradient = 20 g ÷ 0.50 mol = 40 g/mol, which is the molar mass.
3. Therefore, we get M(NaOH) = 23 + 16 + 1 = 40 g/mol, which matches.
12 g; gradient 40 g/mol.
Problem 14 A tablet of paracetamol
1. Also, we get 500 mg = 0.500 g.2. 8 × 12 + 9 × 1 + 14 + 2 × 16 = 96 + 9 + 14 + 32 = 151 g/mol.
3. Moreover, we get n = 0.500 ÷ 151 = 0.00331 mol = 3.31 × 10⁻³ mol.
3.31 × 10⁻³ mol
Problem 15 Spot the errors
1. Wrong: oxygen gas is O₂, so M = 32 g/mol.2. Wrong: there are two OH groups, so Mr = 40 + 2 × 17 = 74.
3. Wrong: 0.50 mol is 0.50 × 6.02 × 10²³ = 3.01 × 10²³ particles.
4. Wrong: n = m ÷ M.
32 g/mol; 74; 3.01 × 10²³; n = m ÷ M
Problem 16 Magnesium or aluminium
1. Thus, we get Mg: 10 ÷ 24 = 0.417 mol. Al: 10 ÷ 27 = 0.370 mol.2. Mg: 0.4167 × 6.02 × 10²³ = 2.51 × 10²³. Al: 0.3704 × 6.02 × 10²³ = 2.23 × 10²³.
3. Magnesium has more atoms: 0.4167 ÷ 0.3704 = 1.125, so 12.5 % more.
Magnesium has 12.5 % more atoms.
Problem 17 Iron in haematite
1. Therefore, we get 112 ÷ 160 × 100 = 70.0 %.2. Also, we get 0.700 × 2.50 kg = 1.75 kg.
3. n = 160 ÷ 160 = 1.00 mol.
70.0 % iron; 1.75 kg; 1.00 mol
Problem 18 Ions in a solution
1. Each formula unit gives 2 Al³⁺ and 3 SO₄²⁻.2. Moreover, we get Al³⁺: 2 × 0.20 = 0.40 mol. SO₄²⁻: 3 × 0.20 = 0.60 mol.
3. Total = 1.00 mol of ions = 6.02 × 10²³ ions.
0.40 mol, 0.60 mol; 6.02 × 10²³ ions
Problem 19 Mass of one particle
1. Thus, we get 18 ÷ 6.02 × 10²³ = 2.99 × 10⁻²³ g.2. 12 ÷ 6.02 × 10²³ = 1.99 × 10⁻²³ g.
3. Therefore, we get n = 1.00 ÷ 12 = 0.0833 mol, so N = 0.0833 × 6.02 × 10²³ = 5.02 × 10²² atoms.
2.99 × 10⁻²³ g; 1.99 × 10⁻²³ g; 5.02 × 10²² atoms
Problem 20 Baking soda
1. Also, we get 23 + 1 + 12 + 3 × 16 = 84 g/mol.2. n = 8.4 ÷ 84 = 0.10 mol.
3. One Na⁺ per formula unit: 0.10 × 6.02 × 10²³ = 6.02 × 10²².
4. Three O per formula unit: 0.30 mol of O atoms = 0.30 × 6.02 × 10²³ = 1.81 × 10²³.
84 g/mol; 0.10 mol; 6.02 × 10²² Na⁺; 1.81 × 10²³ O atoms
