The problems are on the mole concept worksheet. Try them on your own first, then compare with the answers below.
Quick answer: A mole is 6.022 × 10²³ particles, and its mass in grams is the molar mass M, so n = m/M links mass to amount. These worked answers correct the 20 problems of the AP Chemistry worksheet on mole concept, from molar mass to combustion analysis of an organic acid: each solution shows the formula, the substitution and the final answer with its unit.
Each answer shows the conversion, the substitution and the final value in bold. Compare your method as well as your result, since free-response scoring awards marks for each correct step.
Level: AP Chemistry · Curriculum: AP Chem Unit 1
Problem 1 Molar mass
1. M = 40.08 + 2(16.00 + 1.008) = 40.08 + 34.016 = 74.10 g/mol.2. Each formula unit holds 2 OH⁻, so 2 mol of hydroxide ions per mole of compound.
Problem 2 Water in moles
1. n = m/M = 54.0/18.015 = 3.00 mol.2. N = nNA = 3.00 × 6.022 × 10²³ = 1.81 × 10²⁴ molecules.
Problem 3 Mass from moles
1. m = nM = 0.250 × 40.00 = 10.0 g.2. One Na⁺ per formula unit: N = 0.250 × 6.022 × 10²³ = 1.51 × 10²³ ions.
Problem 4 Iron nails
1. N = 0.500 × 6.022 × 10²³ = 3.01 × 10²³ atoms.2. m = nM = 0.500 × 55.85 = 27.9 g.
Problem 5 Reading a mass spectrum
1. There are two peaks, so two isotopes.2. Average = 0.692 × 62.93 + 0.308 × 64.93 = 43.55 + 20.00 = 63.55 u.
3. The value matches copper.
Problem 6 Percent composition of carbon dioxide
1. %C = 12.011/44.009 × 100 = 27.29 %.2. %O = 100 − 27.29 = 72.71 %.
Problem 7 Pure or mixed?
1. Distilled water: pure substance. Brass: mixture (alloy of copper and zinc). Seawater: mixture. Gold bar: pure substance (if 24 carat). Air: mixture. Pure: water, gold; mixtures: brass, seawater, air.2. Evaporate a measured sample: it leaves a solid residue, and its boiling temperature is not fixed. A pure liquid leaves no residue.
Problem 8 Formula of an iron oxide
1. n(Fe) = 5.58/55.845 = 0.0999 mol; n(O) = 2.40/15.999 = 0.150 mol.2. Ratio Fe : O = 0.0999 : 0.150 = 1 : 1.5 = 2 : 3. Fe₂O₃.
3. Iron(III) oxide.
Problem 9 From empirical to molecular formula
1. 12.011 + 2(1.008) = 14.03 g/mol.2. 70.1/14.03 = 5.0, so the formula is (CH₂)₅ = C₅H₁₀.
Problem 10 Nitrogen in a fertiliser
1. M = 2(14.007 + 4 × 1.008) + 32.06 + 4(15.999) = 36.078 + 32.06 + 63.996 = 132.13 g/mol.2. %N = 28.014/132.13 × 100 = 21.2 %.
3. m(N) = 0.2120 × 250 = 53.0 g.
Problem 11 Chlorine isotopes
1. ³⁵Cl (34.97 u) 75.8 % and ³⁷Cl (36.97 u) 24.2 %.2. Average = 0.758 × 34.969 + 0.242 × 36.966 = 26.51 + 8.95 = 35.45 u.
3. No atom has this mass: it is a weighted average of the two isotopes, so the spectrum shows only the real atomic masses.
Problem 12 Water of crystallisation
1. 3.50 − 1.710 = 1.790 g.2. n(H₂O) = 1.790/18.015 = 0.0994 mol; n(MgSO₄) = 1.710/120.37 = 0.01421 mol.
3. Ratio = 0.0994/0.01421 = 7.0, so x = 7.
Problem 13 Combustion of a hydrocarbon
1. n(C) = n(CO₂) = 0.880/44.009 = 0.0200 mol.2. n(H) = 2 × 0.360/18.015 = 0.0400 mol.
3. C : H = 0.0200 : 0.0400 = 1 : 2. Empirical formula CH₂.
Problem 14 Salt and sand
1. 8.00 − 5.20 = 2.80 g.2. 2.80/8.00 × 100 = 35.0 %.
3. n = 2.80/58.44 = 0.0479 mol.
Problem 15 A gas mixture
1. m(N₂) = 0.300 × 28.014 = 8.40 g; m(O₂) = 0.100 × 31.998 = 3.20 g. Total = 11.60 g.2. %N₂ = 8.40/11.60 × 100 = 72.4 %.
3. x(N₂) = 0.300/0.400 = 0.750.
Problem 16 Isotope abundances from an average
1. 68.926x + 70.925(1 − x) = 69.72.2. 70.925 − 1.999x = 69.72, so x = 1.205/1.999 = 0.603.
3. Gallium-69: 60.3 %; gallium-71: 39.7 %.
Problem 17 Mass spectrum of a diatomic molecule
1. The molecule can contain ⁷⁹Br⁷⁹Br, ⁷⁹Br⁸¹Br or ⁸¹Br⁸¹Br, with masses 158, 160, 162.2. P(158) = 0.507² = 0.257; P(160) = 2 × 0.507 × 0.493 = 0.500; P(162) = 0.493² = 0.243. Ratio about 1 : 2 : 1.
3. The figure shows 25.7 %, 50.0 % and 24.3 %, which agrees.
Problem 18 Vitamin C
1. C = 40.9/12.011 = 3.41 mol; H = 4.58/1.008 = 4.54 mol; O = 54.5/15.999 = 3.41 mol.2. Dividing by 3.41: 1 : 1.33 : 1. Multiply by 3: 3 : 4 : 3. C₃H₄O₃ (88.06 g/mol).
3. 176/88.06 = 2.0, so the molecular formula is C₆H₈O₆.
Problem 19 Nitrogen in a mixed fertiliser
1. 28.014/80.04 = 0.3500.2. m(NH₄NO₃) = 2.80/0.3500 = 8.00 g.
3. m(KCl) = 2.00 g, so 20.0 %; n = 2.00/74.55 = 0.0268 mol.
Problem 20 Combustion analysis of an organic acid
1. m(C) = 0.7328 × 12.011/44.009 = 0.2000 g; m(H) = 0.3000 × 2.016/18.015 = 0.0336 g.2. m(O) = 0.5000 − 0.2000 − 0.0336 = 0.2664 g.
3. Moles: C 0.01665, H 0.0333, O 0.01665, ratio 1 : 2 : 1. Empirical formula CH₂O (30.03 g/mol); 60.05/30.03 = 2, so the molecular formula is C₂H₄O₂.
