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IGCSE Chemistry Paper 4 (Theory / Extended)

Topic 12: Experimental Techniques and Chemical Analysis -- Mock Exam 2
1 hour 15 minutes
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75:00
0620

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Question 1 -- Measuring, Collecting and Dissolving
Total: 12 marks
A student in Kuala Lumpur carries out several experiments and must select apparatus, collect a gas and describe solutions correctly.
(a) [4]
Name the measuring instrument used for each task.

(i) recording the temperature change when acid is added to alkali [1]
(ii) measuring the mass of solid lost as a gas escapes from a flask [1]
(iii) measuring the time for a cross drawn under a flask to disappear [1]
(iv) measuring the volume of gas produced during a reaction [1]
Model Answer -- 1(a)
(i) thermometer [1]
(ii) balance [1]
(iii) stopwatch / stop-clock [1]
(iv) gas syringe [1]
⚠ If you missed marks here: Read what quantity is being measured — temperature, mass, time or gas volume — and give the one instrument that measures it directly. A measuring cylinder for gas volume is only acceptable if the gas is collected over water; a gas syringe is the direct answer the examiner expects here.
(b) [4]
The student prepares ammonia gas, which is less dense than air and very soluble in water.

(i) Name the collection method used for ammonia and explain why it is suitable. [2]

(ii) Explain why ammonia cannot be collected over water, and why it cannot be dried with concentrated sulfuric acid. [2]
Model Answer -- 1(b)
(i) upward delivery / collect in an inverted container [1]
(i) suitable because ammonia is less dense than air, so it rises and fills the container [1]
(ii) it is very soluble, so it would dissolve in the water and be lost [1]
(ii) ammonia is a base and reacts with (is absorbed by) concentrated sulfuric acid [1]
⚠ If you missed marks here: Two different properties of ammonia block the two rejected techniques: its solubility rules out collection over water, and its basic character rules out an acidic drying agent. Quoting "less dense than air" as the reason for either of these loses the mark — density only explains the direction of delivery.
(c) [4]
The student stirs potassium nitrate into 50 cm³ of warm water until some solid remains undissolved.

(i) In this mixture, name the solute and the solvent. [2]

(ii) What word describes a solution that contains the maximum amount of dissolved solute at that temperature? State the evidence for this in the student’s beaker. [2]
Model Answer -- 1(c)
(i) the solute is potassium nitrate [1]
(i) the solvent is water [1]
(ii) the solution is saturated [1]
(ii) evidence: undissolved solid remains at the bottom however long it is stirred (at that temperature) [1]
⚠ If you missed marks here: The definition of saturated must be tied to a stated temperature, because warming the water would let more solid dissolve. The visible evidence — excess undissolved solute that stirring cannot remove — is what turns the definition into an observation worth a mark.
Question 2 -- Titration with Sulfuric Acid
Total: 12 marks
A student titrates dilute sulfuric acid against 25.0 cm³ portions of 0.100 mol/dm³ aqueous sodium hydroxide, using methyl orange as the indicator.
(a) [3]
(i) State the colour of the methyl orange in the conical flask at the start of the titration. [1]

(ii) State the colour change observed at the end-point. [1]

(iii) Explain why only a few drops of indicator are added. [1]
Model Answer -- 2(a)
(i) yellow, because the flask contains alkali [1]
(ii) it changes from yellow to red (orange accepted at the exact end-point) [1]
(iii) the indicator is itself a weak acid/reacts slightly, so a large amount would affect the titre; a few drops give a clear colour without changing the result [1]
⚠ If you missed marks here: Work out which solution is in the flask before answering — here the alkali is in the flask, so methyl orange starts yellow and turns red as acid is added. Reversing the colours describes a titration set up the other way round and loses both observation marks.
(b) [4]
The titration results are shown.

titration1 (rough)234
titre / cm³23.1022.6522.5522.60
(i) Explain which titres are concordant. [2]

(ii) Calculate the mean titre. [2]
Model Answer -- 2(b)
(i) titres 2, 3 and 4 are concordant [1]
(i) because they all lie within 0.10 cm³ of one another, while the rough titre does not [1]
(ii) mean = (22.65 + 22.55 + 22.60) ÷ 3 [1]
(ii) = 22.60 cm³ [1]
⚠ If you missed marks here: Check the spread: 22.55 to 22.65 is exactly 0.10 cm³, so all three accurate titres qualify as concordant. The mean must be quoted to two decimal places like every burette value — writing 22.6 cm³ drops the final mark for wrong precision.
(c) [5]
The equation for the reaction is H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O.

Use the mean titre of 22.60 cm³ to calculate the concentration of the sulfuric acid in mol/dm³. Give your answer to three significant figures.

(i) Calculate the moles of NaOH used. [1]
(ii) Use the mole ratio to find the moles of H₂SO₄. [2]
(iii) Calculate the concentration of the sulfuric acid. [2]
Model Answer -- 2(c)
(i) moles NaOH = 0.0250 × 0.100 = 0.00250 mol [1]
(ii) the ratio of acid to alkali is 1 : 2 [1]
(ii) moles H₂SO₄ = 0.00250 ÷ 2 = 0.00125 mol [1]
(iii) concentration = 0.00125 ÷ 0.02260 [1]
(iii) = 0.0553 mol/dm³ (3 significant figures) [1]
⚠ If you missed marks here: The 1 : 2 ratio means the moles of acid are HALF the moles of alkali — multiplying by 2 instead of dividing is the single most common error and doubles the final answer to 0.221. Keep every figure in the calculator until the end, then round once to three significant figures.
Question 3 -- Chromatography of Amino Acids
Total: 12 marks
A laboratory in Toronto uses paper chromatography to identify the amino acids in a protein sample. Amino acids are colourless.
(a) [4]
Give a reason for each step of the procedure.

(i) the baseline is drawn in pencil [1]
(ii) the beaker is covered with a lid while the chromatogram runs [1]
(iii) the position of the solvent front is marked as soon as the paper is removed [1]
(iv) the dried paper is sprayed with a locating agent [1]
Model Answer -- 3(a)
(i) ink would dissolve and run up the paper with the samples; pencil is insoluble [1]
(ii) the lid stops the solvent evaporating, keeping the air saturated so the solvent rises evenly [1]
(iii) the solvent front becomes invisible once the paper dries, and it is needed for Rf calculations [1]
(iv) amino acids are colourless, so the locating agent reacts with them to form visible spots [1]
⚠ If you missed marks here: Every "why" answer must connect the step to a consequence: pencil because ink runs, lid because solvent evaporates, marking because the front disappears, locating agent because the spots are invisible. Restating the step ("to mark the solvent front") without the reason earns nothing.
(b) [4]
On the chromatogram the solvent front moves 8.0 cm. Spot A is 2.4 cm and spot B is 6.0 cm from the baseline. Reference Rf values in this solvent: glycine 0.30, alanine 0.55, leucine 0.75.

(i) Calculate the Rf values of spots A and B. [2]

(ii) Identify the amino acids responsible for spots A and B. [2]
Model Answer -- 3(b)
(i) Rf of A = 2.4 ÷ 8.0 = 0.30 [1]
(i) Rf of B = 6.0 ÷ 8.0 = 0.75 [1]
(ii) spot A is glycine (Rf 0.30) [1]
(ii) spot B is leucine (Rf 0.75) [1]
⚠ If you missed marks here: Both Rf values use the same solvent distance of 8.0 cm — a common slip is dividing spot B by spot A. Identification marks depend on quoting or clearly using the calculated values against the reference table, so show the comparison, not just the names.
(c) [4]
A technician suspects that a third spot, C, with Rf 0.55, could be either alanine or a different amino acid, serine, which happens to have the same Rf value in this solvent.

(i) Explain why the Rf value alone cannot settle the question. [2]

(ii) Describe what the technician should do to identify spot C with confidence. [2]
Model Answer -- 3(c)
(i) different substances can have the same Rf value in one particular solvent [1]
(i) so a matching Rf shows the substances could be the same, but does not prove it [1]
(ii) run the chromatogram again in a different solvent (with known samples of alanine and serine alongside) [1]
(ii) the amino acid whose spot still matches C in the second solvent is the one present [1]
⚠ If you missed marks here: The logic has two halves: a match is necessary but not sufficient, and only a second solvent can break the tie because it changes the solubility balance differently for each compound. Running the same solvent again, however carefully, reproduces the same coincidence and earns nothing.
Question 4 -- Crystallisation and Distillation
Total: 12 marks
A student in Cape Town prepares hydrated copper(II) sulfate crystals from its solution, and then studies distillation.
(a) [4]
Describe how the student should obtain large, well-formed hydrated copper(II) sulfate crystals from the solution, and explain why the solution must not be evaporated to dryness.
Model Answer -- 4(a)
heat the solution gently until it reaches the point of crystallisation (a small sample forms crystals on cooling) [1]
leave the concentrated solution to cool slowly so that large crystals grow [1]
filter off the crystals and dry them between filter papers [1]
evaporating to dryness would drive off the water of crystallisation (and could decompose the salt), leaving powder instead of hydrated crystals [1]
⚠ If you missed marks here: Slow cooling is what produces large crystals — the speed of cooling is a marking point, not a decoration. The final reason must name the water of crystallisation; "the crystals would burn" is too vague to score.
(b) [4]
The student sets up simple distillation to recover pure water from copper(II) sulfate solution.

(i) State where the thermometer bulb should be positioned, and why. [2]

(ii) State the reading on the thermometer while pure water distils over, and name the process happening in the condenser. [2]
Model Answer -- 4(b)
(i) the bulb is placed at the side-arm / still-head where the vapour leaves the flask [1]
(i) so it measures the temperature of the vapour actually passing to the condenser, not the boiling liquid [1]
(ii) the thermometer reads 100 °C while pure water distils (at standard atmospheric pressure) [1]
(ii) in the condenser the vapour is cooled and condenses back to liquid water [1]
⚠ If you missed marks here: A thermometer dipped into the liquid measures the solution, whose boiling point is raised by the dissolved salt — only the vapour at the side-arm shows the true boiling point of the substance distilling over. The steady 100 °C reading is itself evidence that pure water is being collected.
(c) [4]
Data for two samples of ethanoic acid are compared with the pure compound, which melts at 17 °C and boils at 118 °C. Sample 1 melts at 17 °C and boils at 118 °C, both sharply. Sample 2 melts at 12 °C and boils at 121 °C.

(i) State which sample is pure, and how you know. [2]

(ii) Explain the melting and boiling results of the other sample. [2]
Model Answer -- 4(c)
(i) sample 1 is pure [1]
(i) it melts and boils sharply at exactly the accepted values for the pure compound [1]
(ii) sample 2 contains impurities [1]
(ii) impurities lower the melting point and raise the boiling point [1]
⚠ If you missed marks here: Learn the direction pair as one fact: impurities push melting DOWN and boiling UP. Candidates who remember only "impurities change the values" cannot explain why sample 2 moves in opposite directions at the two ends, and that direction is where the final mark lives.
Question 5 -- Flame Tests and Ammonium Ions
Total: 10 marks
A student in Chennai tests a set of labelled salts to build a reference table of flame colours, then tests a fertiliser for ammonium ions.
(a) [4]
Give the flame test colour produced by each ion.

(i) lithium [1]
(ii) calcium [1]
(iii) potassium [1]
(iv) copper(II) [1]
Model Answer -- 5(a)
(i) lithium: red [1]
(ii) calcium: orange-red [1]
(iii) potassium: lilac [1]
(iv) copper(II): blue-green [1]
⚠ If you missed marks here: The pair that catches most students is lithium red versus calcium orange-red — the syllabus wording must be used exactly, so plain "orange" or "brick red" for calcium risks the mark. Copper’s colour is the two-word blue-green, not simply green.
(b) [3]
Describe how the student can show that the fertiliser contains ammonium ions. Give the reagent, the conditions and the result of the test.
Model Answer -- 5(b)
add aqueous sodium hydroxide to the fertiliser and warm the mixture [1]
ammonia gas is released; test it with damp red litmus paper [1]
the damp red litmus paper turns blue, confirming ammonium ions [1]
⚠ If you missed marks here: The warming step is a genuine marking point — without heat the ammonia is released too slowly to detect. The litmus must be damp and must be red to start with; each detail changes whether the test can work at all.
(c) [3]
(i) Explain why a flame test cannot tell potassium chloride and potassium sulfate apart. [1]

(ii) The student’s first flame test gives a strong yellow colour even though the salt should burn lilac. Suggest the cause, and state how the wire should be prepared to avoid this problem. [2]
Model Answer -- 5(c)
(i) both salts contain the same cation, potassium, and the flame colour comes from the cation only, so both give lilac [1]
(ii) the wire is contaminated, most likely with traces of a sodium compound whose yellow flame masks the lilac [1]
(ii) clean the wire by dipping it in concentrated hydrochloric acid and heating it until the flame shows no colour [1]
⚠ If you missed marks here: Flame colours identify cations only — that single sentence answers (i). In (ii) name the likely contaminant: sodium’s yellow is intense and persistent, which is why the acid-clean-and-reheat cycle must be repeated until the flame stays colourless before each new test.
Question 6 -- Identifying Salt Y
Total: 12 marks
A student in Nairobi is given a white solid, Y, and records these results. Test 1: flame test gives an orange-red colour. Test 2: a solution of Y gives a white precipitate with aqueous sodium hydroxide, insoluble in excess. Test 3: a solution of Y acidified with nitric acid gives no precipitate with aqueous silver nitrate. Test 4: a solution of Y acidified with nitric acid gives a white precipitate with aqueous barium nitrate.
(a) [4]
State what each test shows about Y.

(i) Test 1 [1]
(ii) Test 2 [1]
(iii) Test 3 [1]
(iv) Test 4 [1]
Model Answer -- 6(a)
(i) the orange-red flame shows Y contains calcium ions [1]
(ii) a white precipitate insoluble in excess sodium hydroxide is consistent with calcium ions [1]
(iii) no precipitate with acidified silver nitrate shows chloride, bromide and iodide are absent [1]
(iv) the white precipitate with acidified barium nitrate shows Y contains sulfate ions [1]
⚠ If you missed marks here: Tests 1 and 2 point at the same cation from two directions — the flame colour identifies calcium and the hydroxide behaviour confirms it. Notice that Test 2 alone would not be conclusive, because the white-insoluble result only narrows the field; that is why examiners give both tests.
(b) [4]
(i) Identify salt Y. [1]

(ii) In Test 4, explain why the solution is acidified before the barium nitrate is added. [2]

(iii) Describe what would have been seen in Test 4 if Y had contained carbonate ions and no acid had been added. [1]
Model Answer -- 6(b)
(i) Y is calcium sulfate [1]
(ii) the acid reacts with and removes any carbonate ions [1]
(ii) carbonate would otherwise also give a white precipitate with barium ions, a false positive [1]
(iii) a white precipitate (barium carbonate) would still have formed, wrongly suggesting sulfate [1]
⚠ If you missed marks here: The acidification logic is identical to the silver nitrate test: carbonates are the impostor ion in both, and the acid destroys them first. Writing "to make the solution acidic" without saying what the acid removes explains nothing and scores nothing.
(c) [4]
A second white solid, Z, is either potassium bromide or potassium iodide.

(i) Describe a test on a solution of Z, and the result for each possible salt, that identifies which one it is. [3]

(ii) State one further test that would confirm the potassium ion, and its result. [1]
Model Answer -- 6(c)
(i) acidify the solution with dilute nitric acid and add aqueous silver nitrate [1]
(i) potassium bromide gives a cream precipitate [1]
(i) potassium iodide gives a yellow precipitate [1]
(ii) a flame test: a lilac flame confirms potassium [1]
⚠ If you missed marks here: The halide colour sequence white-cream-yellow must be attached to the right halides — cream for bromide and yellow for iodide are close enough to confuse under exam pressure, so learn them as a graded series. The test needs the acidification step stated to earn the reagent mark.
Question 7 -- Planning Tests to Distinguish Substances
Total: 10 marks
A laboratory assistant in Mumbai must design simple tests to tell apart pairs of similar-looking substances whose labels have been lost.
(a) [4]
Two pale green solutions contain iron(II) sulfate and iron(III) chloride.

(i) Name one reagent that can distinguish them. [1]
(ii) Give the result for the iron(II) solution. [1]
(iii) Give the result for the iron(III) solution. [1]
(iv) State one other reagent that would work equally well, and why. [1]
Model Answer -- 7(a)
(i) aqueous sodium hydroxide [1]
(ii) iron(II) gives a green precipitate [1]
(iii) iron(III) gives a red-brown precipitate [1]
(iv) aqueous ammonia, because it precipitates the same coloured hydroxides [1]
⚠ If you missed marks here: The hydroxide colours are the whole test: green for iron(II), red-brown for iron(III). Both sodium hydroxide and ammonia work because the coloured precipitates are the same hydroxides either way — a fact worth a mark on its own in part (iv).
(b) [3]
Two colourless solutions contain sodium chloride and sodium sulfate. Describe one test, with the result for each solution, that identifies which is which.
Model Answer -- 7(b)
acidify each solution with dilute nitric acid and add aqueous barium nitrate [1]
the sodium sulfate gives a white precipitate [1]
the sodium chloride gives no precipitate (alternatively, acidified silver nitrate gives a white precipitate only with the chloride) [1]
⚠ If you missed marks here: Either the barium test or the silver test can score full marks, but the reagent must be matched to the ion it detects and both outcomes must be given. A common error is adding barium nitrate and claiming the chloride gives a precipitate too — barium chloride is soluble, so it does not.
(c) [3]
Two white solids are sodium carbonate and sodium chloride. Describe one test, with the result for each solid, that identifies which is which.
Model Answer -- 7(c)
add dilute hydrochloric acid (or another dilute acid) to each solid [1]
the sodium carbonate fizzes and the gas turns limewater milky (carbon dioxide) [1]
the sodium chloride shows no reaction [1]
⚠ If you missed marks here: Observing fizzing is not the complete carbonate test — the gas must be identified with limewater to secure the middle mark. The negative result for the chloride is a marking point too; a distinguishing test is only complete when both behaviours are recorded.

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IGCSE Chemistry 0620 -- Paper 4 Theory (Extended) -- Topic 12: Experimental Techniques and Chemical Analysis -- Mock Exam 2
For examination preparation only. Not an official Cambridge Assessment paper.