The problems are on the chemical analysis worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Chemical analysis identifies what a sample contains. These worked answers correct the 20 problems of the Chemistry 9-10 worksheet on chemical analysis, from pure or impure to plan to separate four powders: each solution shows the formula, the substitution and the final answer with its unit.
Each answer gives the rule, then the working, then the result in bold. Also, read the working for the method: exam markers reward correct chemistry and correct equations as well as final numbers.
Level: Chemistry grades 9-10 (GCSE · IGCSE · O Level · Class 9-10) · Curriculum: GCSE C8
Problem 1 Pure or impure
1. A and C: each melts at one sharp temperature.2. B melts over a range (63 to 74 °C) and below the sharp melting points typical of pure solids, which is the sign of impurities.
3. A formulation is a mixture designed with exact amounts of components for a useful purpose, for example paint, fuel or fertiliser.
Problem 2 Simple Rf values
1. Spot 1: 3.0 ÷ 6.0 = 0.50. Then spot 2: 4.5 ÷ 9.0 = 0.50. Spot 3: 1.2 ÷ 8.0 = 0.15.2. It is a ratio of two distances in the same unit, so the units cancel.
3. Yes: they have the same Rf in what is presumably the same solvent.
Problem 3 Flame test colours
1. Also, lithium crimson; sodium yellow; potassium lilac; calcium orange-red; copper(II) blue-green.2. Lithium, Li⁺.
Problem 4 Gas tests
1. Also, hydrogen: lit splint gives a squeaky pop. Then oxygen: glowing splint relights. Carbon dioxide: limewater turns cloudy. Also, chlorine: damp litmus paper is bleached white.2. Carbon dioxide.
Problem 5 Reading a chromatogram
1. In addition, the sweet extract has two spots.2. Dye 1 and dye 2: the spots match their distances.
3. 4.8 ÷ 8.0 = 0.60.
4. No: no spot at 6.4 cm in the sweet lane.
Problem 6 Rf for two spots
1. 1.8 ÷ 7.5 = 0.24; 4.5 ÷ 7.5 = 0.60.2. The spot with Rf = 0.24 is held more strongly by the paper (or dissolves less in the solvent), so it travels less far.
Problem 7 Using an Rf value
1. 0.35 × 12.0 = 4.2 cm.2. Front = 5.1 ÷ 0.60 = 8.5 cm.
Problem 8 Method questions
1. Otherwise the sample would dissolve into the solvent in the beaker and nothing would travel up the paper.2. The lid keeps the air saturated with solvent vapour so the solvent does not evaporate from the paper.
3. The front position is needed for the Rf calculation and it must be marked before it evaporates.
4. Rf changes, because the substances dissolve differently.
Problem 9 Melting graph
1. The pure sample stays flat at 80 °C while melting.2. Moreover, the impure curve does not have a flat part; it rises slowly through a range, about 70 to 77 °C, and begins melting lower.
3. Impurities disturb the regular arrangement of particles, so the solid melts at a lower temperature and over a range.
Problem 10 Sodium hydroxide results
1. a: Cu²⁺. B: Al³⁺. C: Ca²⁺ (orange-red flame).2. Mg²⁺, which gives a white precipitate that stays in excess and no flame colour.
Problem 11 Ionic equations
1. Fe³⁺(aq) + 3OH⁻(aq) → Fe(OH)₃(s).2. Fe²⁺(aq) + 2OH⁻(aq) → Fe(OH)₂(s).
3. Mg²⁺(aq) + 2OH⁻(aq) → Mg(OH)₂(s).
4. Fe(OH)₃ is red-brown; Fe(OH)₂ is green. Charges balance in each equation.
Problem 12 Identifying negative ions
1. X contains carbonate; the gas, carbon dioxide, turns limewater cloudy.2. Y: chloride (white). Z: iodide (yellow).
3. Ag⁺(aq) + Cl⁻(aq) → AgCl(s).
Problem 13 Why acidify first
1. Barium carbonate is also a white precipitate, so carbonate would give a false positive.2. In addition, the acid reacts with carbonate and removes it as carbon dioxide gas.
3. Moreover, a white precipitate of barium sulfate remains.
Problem 14 Gas from a metal
1. Mg + 2HCl → MgCl₂ + H₂.2. Also, a lit splint held at the mouth of the tube makes a squeaky pop.
3. P is oxygen, Q is carbon dioxide, R is chlorine.
Problem 15 Paint formulation
1. Pigment 0.40 × 5.0 = 2.0 kg; binder 0.35 × 5.0 = 1.75 kg.2. solvent = 5.0 − 2.0 − 1.75 = 1.25 kg, which is 25 %.
3. A formulation is designed, so exact amounts give the right colour, thickness and drying time every time.
Problem 16 Identify a white solid
1. Lilac flame: potassium; white precipitate with barium: sulfate. X is potassium sulfate, K₂SO₄.2. Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s).
3. M(K₂SO₄) = 2 × 39 + 32 + 64 = 174; n = 0.87 ÷ 174 = 0.0050 mol. 1 : 1 ratio, so BaSO₄ = 0.0050 mol; M = 233. Mass = 0.0050 × 233 = 1.17 g.
Problem 17 Matching dyes
1. 5.4 ÷ 12.0 = 0.45; 7.2 ÷ 12.0 = 0.60; 8.4 ÷ 12.0 = 0.70.2. Dye B (0.45) and dye C (0.70).
3. The spot at Rf 0.60 matches none of the reference dyes, so the extract contains a different dye (so at least three dyes).
Problem 18 Identify a blue solution
1. Cu²⁺ and Cl⁻: copper(II) chloride, CuCl₂.2. Cu²⁺(aq) + 2OH⁻(aq) → Cu(OH)₂(s) and Ag⁺(aq) + Cl⁻(aq) → AgCl(s).
3. M(CuCl₂) = 134.5; n = 1.35 ÷ 134.5 = 0.0100 mol. Cl⁻ = 0.0201 mol, so AgCl = 0.0201 mol; M(AgCl) = 143.5; mass = 0.0201 × 143.5 = 2.9 g.
Problem 19 Faulty notes
1. All four are wrong.2. (a) The wire is cleaned in hydrochloric acid, which gives volatile chlorides. (b) This shows aluminium; calcium gives a precipitate that stays. (c) Rf cannot exceed 1: the substance cannot outrun the solvent front, so the measurement is wrong. (d) The acid is added first to remove carbonate, then barium chloride.
Problem 20 Plan to separate four powders
1. Plan: (i) add dilute acid to a little of each: the two carbonates fizz. (ii) Flame test: lilac flame is K₂CO₃; orange-red flame is CaCO₃. (iii) For the other two, dissolve and add acidified barium chloride to one portion and acidified silver nitrate to another.2. Na₂SO₄: white precipitate with barium chloride, yellow flame. NaCl: white precipitate with silver nitrate, yellow flame. Moreover, the yellow flame shows sodium in both.
