Topic 12: Experimental Techniques and Chemical Analysis -- Cambridge Challenge
1 hour 15 minutes
80
7
75:00
0620
Instructions
Answer all questions in the spaces provided.
Show all working for calculations.
Use appropriate scientific terminology.
Your answers will be automatically graded when you submit.
Question Navigation
Question 1 -- The Four Unlabelled Bottles
Total: 12 marks
A storeroom flood in a school in Lisbon washes the labels off four bottles of soluble salts, A, B, C and D. The salts are known to be chromium(III) sulfate, ammonium iodide, calcium nitrate and potassium carbonate, in some order. A technician dissolves a sample of each salt in water and records the results below.
test
A
B
C
D
flame test on the solid
no distinctive colour
no distinctive colour
orange-red
lilac
add aqueous sodium hydroxide, then excess; warm
green precipitate, dissolves in excess
no precipitate; on warming, a pungent gas turns damp red litmus blue
white precipitate, insoluble in excess
no precipitate
add dilute hydrochloric acid to the solid
no effervescence
no effervescence
no effervescence
effervescence; gas turns limewater milky
acidify with dilute nitric acid, add aqueous silver nitrate
no precipitate
yellow precipitate
no precipitate
--
acidify with dilute nitric acid, add aqueous barium nitrate
white precipitate
no precipitate
no precipitate
--
(a)[4]
Identify the cation present in each of A, B, C and D. For each cation, state the evidence from the table that supports your answer.
Model Answer -- 1(a)
A contains Cr³⁺ (chromium(III)) because it gives a green precipitate with sodium hydroxide which dissolves in excess [1]
B contains NH₄⁺ (ammonium) because warming with sodium hydroxide releases ammonia, which turns damp red litmus blue [1]
C contains Ca²⁺ (calcium) because of the orange-red flame colour and the white precipitate insoluble in excess sodium hydroxide [1]
D contains K⁺ (potassium) because of the lilac flame colour [1]
⚠ If you missed marks here: In a deduction grid each cation must be pinned by evidence, not by elimination from the list of names. Chromium(III) and aluminium both give precipitates soluble in excess sodium hydroxide, but the green colour identifies Cr³⁺; and the ammonium result only counts if you say the gas is ammonia and that it turns damp red litmus blue — "a pungent gas forms" repeats the question without earning the mark.
(b)[4]
Identify the anion present in A, B and D, giving the evidence for each. Explain why the anion in C cannot be identified directly from the table, and suggest what it must be.
Model Answer -- 1(b)
A contains sulfate, SO₄²⁻, because acidified barium nitrate gives a white precipitate [1]
B contains iodide, I⁻, because acidified silver nitrate gives a yellow precipitate [1]
D contains carbonate, CO₃²⁻, because dilute acid produces effervescence and the gas (carbon dioxide) turns limewater milky [1]
C gives no precipitate with silver nitrate or barium nitrate and no gas with acid, so its anion must be nitrate, NO₃⁻, which none of these tests detects [1]
⚠ If you missed marks here: The precipitate colour with silver nitrate is the whole identification: white means chloride, cream means bromide and yellow means iodide, so writing just "a halide" for B is not enough. The nitrate mark needs the negative logic spelled out — C responds to none of the anion tests in the table, and nitrate is the one common anion those tests cannot detect.
(c)[4]
(i) Give the full names of salts A, B, C and D. [2]
(ii) Describe a test the technician could carry out on solution C to confirm the presence of the nitrate ion, and give the positive result. [2]
Model Answer -- 1(c)
(i) A is chromium(III) sulfate and B is ammonium iodide [1]
(i) C is calcium nitrate and D is potassium carbonate [1]
(ii) add aqueous sodium hydroxide and aluminium foil, then warm [1]
(ii) ammonia gas is given off, which turns damp red litmus paper blue [1]
⚠ If you missed marks here: The nitrate test has three essential ingredients — sodium hydroxide, aluminium (foil or powder) and warming — and the positive result is ammonia turning damp red litmus blue. Leaving out the aluminium turns it into the ammonium test, which solution C would fail, and quoting the brown-ring test earns nothing because it is not on the 0620 syllabus.
Question 2 -- Purity of Soda Ash by Titration
Total: 12 marks
A quality-control chemist in Osaka checks the purity of soda ash, an impure form of anhydrous sodium carbonate. She dissolves 3.00 g of the soda ash in distilled water and makes the solution up to exactly 250.0 cm³ in a volumetric flask. She titrates 25.0 cm³ portions of this solution against 0.200 mol/dm³ hydrochloric acid using methyl orange indicator.
The equation is: Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂. The impurities do not react with the acid. [Mᵣ of Na₂CO₃ = 106]
(a)[3]
Her five titres are shown.
titration
1 (rough)
2
3
4
5
titre / cm³
25.50
24.60
24.95
24.65
24.55
(i) State which titres should be used to calculate the mean, and explain your choice. [2]
(ii) Calculate the mean titre. [1]
Model Answer -- 2(a)
(i) use titrations 2, 4 and 5 (not the rough titration and not titration 3) [1]
(i) because these three are concordant / within 0.10 cm³ of each other [1]
(ii) mean = (24.60 + 24.65 + 24.55) ÷ 3 = 24.60 cm³ [1]
⚠ If you missed marks here: This table hides a trap inside the accurate titres: titration 3 (24.95 cm³) is not the rough run, yet it still lies more than 0.10 cm³ from the others and must be rejected along with titration 1. Averaging all four accurate titres gives 24.69 cm³ and loses both the selection mark and the mean, so always test every titre against the 0.10 cm³ rule before touching the calculator.
(b)[5]
Using the mean titre of 24.60 cm³:
(i) Calculate the number of moles of HCl in the mean titre. [1]
(ii) Use the equation to find the number of moles of Na₂CO₃ in each 25.0 cm³ portion. [2]
(iii) Calculate the number of moles of Na₂CO₃ in the whole 250.0 cm³ of solution. [1]
(iv) Calculate the mass of Na₂CO₃ in the 3.00 g sample of soda ash. [1]
Model Answer -- 2(b)
(i) moles HCl = 0.02460 × 0.200 = 0.00492 mol [1]
(ii) the equation shows 2 mol HCl react with 1 mol Na₂CO₃ (2 : 1 ratio) [1]
(ii) so moles Na₂CO₃ = 0.00492 ÷ 2 = 0.00246 mol [1]
(iii) the 25.0 cm³ portion is one tenth of 250.0 cm³, so moles in flask = 0.00246 × 10 = 0.0246 mol [1]
(iv) mass = 0.0246 × 106 = 2.61 g [1]
⚠ If you missed marks here: Two steps sink most candidates in this chain: dividing by 2 the wrong way round (multiplying gives 0.00984 mol, double the true amount) and forgetting the scale-up from the 25.0 cm³ portion to the full 250.0 cm³ flask. Write the ratio line "2HCl : 1Na₂CO₃" before dividing, and remember the titration only measured one tenth of the dissolved sample.
(c)[4]
(i) Calculate the percentage purity of the soda ash. Give your answer to three significant figures. [2]
(ii) Before filling it with acid, the chemist rinsed the burette with distilled water only. Explain the effect of this mistake on the titre and on the calculated percentage purity. [2]
Model Answer -- 2(c)
(i) % purity = (2.6076 ÷ 3.00) × 100 [1]
(i) = 86.9% (3 significant figures) [1]
(ii) the water dilutes the acid, so a larger volume (titre) is needed to neutralise the alkali [1]
(ii) the calculation still assumes 0.200 mol/dm³, so the moles of acid and therefore the percentage purity come out too high [1]
⚠ If you missed marks here: Percentage purity is always mass of pure substance divided by mass of the whole sample, never the other way up — an answer above 100% is the sign you have inverted the fraction. In (ii) the error chain must be followed to the end: diluted acid means a bigger titre, and because the calculation still uses the original concentration, the final purity is overestimated; stopping at "the titre is bigger" earns only the first mark.
Question 3 -- Forensic Ink Chromatography
Total: 12 marks
A forensic scientist in Singapore uses paper chromatography to find out which brand of pen wrote a disputed signature. A sample of ink, S, taken from the signature is run alongside reference inks from three brands, X, Y and Z, using ethanol as the solvent.
(a)[3]
On the chromatogram, the solvent front travels 8.0 cm from the baseline. Ink S produces two spots: one 3.6 cm and one 6.0 cm from the baseline.
(i) Calculate the Rf value of each spot in ink S. [2]
(ii) Explain why an Rf value can never be greater than 1. [1]
Model Answer -- 3(a)
(i) Rf = 3.6 ÷ 8.0 = 0.45 [1]
(i) Rf = 6.0 ÷ 8.0 = 0.75 [1]
(ii) the spot is carried along by the solvent, so it can never travel further than the solvent front; the fraction is therefore always 1 or less [1]
⚠ If you missed marks here: Rf is always distance moved by the spot divided by distance moved by the solvent, both measured from the same baseline — inverting the fraction gives values above 1, which part (ii) tells you is impossible. Quote Rf values as decimals with no units; "3.6 cm" is a distance, not an Rf value.
(b)[4]
The reference inks give these Rf values in the same solvent.
ink
Rf values of spots
brand X
0.45 and 0.62
brand Y
0.45 and 0.75
brand Z
0.75 only
State whether ink S is a pure substance or a mixture, and deduce which brand of pen wrote the signature. Justify your answer fully, including why the other two brands are ruled out.
Model Answer -- 3(b)
S is a mixture because it separates into more than one spot [1]
S matches brand Y [1]
because both of its Rf values (0.45 and 0.75) are the same as brand Y's spots [1]
brand X is ruled out because S has no spot at 0.62, and brand Z is ruled out because it has no spot at 0.45 [1]
⚠ If you missed marks here: A match requires every spot to agree, not just one — brands X, Y and Z all share a spot with S, which is exactly why the examiner asks you to rule the others out explicitly. The elimination mark needs both halves: X has an extra spot S lacks, and Z is missing a spot S has; naming brand Y without this reasoning scores only the identification mark.
(c)[3]
A trainee repeats the experiment but the finished paper is completely blank. His method reads: "I drew the baseline with a ballpoint pen, placed the ink spots on the line, then poured solvent into the tank until it was 1 cm above the baseline."
Identify the two errors in his method, explain why each one caused the blank result, and state how the baseline should have been drawn.
Model Answer -- 3(c)
the baseline was drawn in ink, which dissolves in the solvent and runs up the paper, ruining the chromatogram [1]
the solvent level was above the baseline, so the ink spots dissolved straight into the solvent in the tank instead of moving up the paper [1]
the baseline should be drawn in pencil (graphite is insoluble), with the solvent level kept below the baseline [1]
⚠ If you missed marks here: Each error needs its consequence attached: an ink baseline dissolves and runs, and solvent above the line washes the spots into the tank — which is precisely why the paper came out blank. Simply listing "use pencil" and "keep solvent below the line" without explaining what went wrong answers a different, easier question and drops the explanation marks.
(d)[2]
The laboratory also runs a chromatogram of a seized powder containing colourless substances. Explain why a locating agent is needed and what it does.
Model Answer -- 3(d)
the substances are colourless, so their spots are invisible on the finished chromatogram [1]
the locating agent is a chemical sprayed or applied to the paper that reacts with the spots to form coloured (visible) products [1]
⚠ If you missed marks here: The key word is "reacts" — a locating agent chemically reacts with the colourless spots to make coloured products; it does not "dye", "stain" or "highlight" them. Both halves are needed: why it is necessary (colourless spots are invisible) and what it does (reacts to give visible spots).
Question 4 -- Separating a Three-Component Mixture
Total: 12 marks
A student in Lagos is given a grey powder that is a mixture of iodine, sodium chloride and sand. She must design a scheme to obtain a pure sample of each of the three substances. Iodine sublimes when warmed; sodium chloride is soluble in water; sand is insoluble in water.
(a)[4]
Describe how the iodine should be removed first and collected. Name the apparatus used, state the property of iodine that the method depends on, and explain why the other two substances are unaffected.
Model Answer -- 4(a)
warm the mixture gently in an evaporating dish (or beaker) [1]
iodine sublimes — it turns directly from solid to vapour [1]
the purple vapour re-forms as solid crystals on a cold surface placed above, e.g. a watch glass of ice or an inverted funnel [1]
sodium chloride and sand have very high melting points and do not sublime, so they stay behind [1]
⚠ If you missed marks here: Sublimation is only half a separation without the cold surface — the vapour must be given somewhere to re-solidify or the iodine is simply lost into the air. Say "sublimes" rather than "evaporates" (iodine goes straight from solid to vapour), and remember the justification mark: the method works because the other two components cannot sublime.
(b)[5]
Describe, step by step, how pure dry samples of sand and of sodium chloride crystals are obtained from the remaining mixture. Name each separation technique and each piece of apparatus used.
Model Answer -- 4(b)
add (warm) water and stir so the sodium chloride dissolves but the sand does not [1]
filter using filter paper in a filter funnel; the sand is left as the residue [1]
wash the sand with distilled water and dry it (e.g. in a warm oven) [1]
heat the filtrate in an evaporating dish to the point of crystallisation, then leave to cool so crystals form [1]
dry the crystals between filter papers or in a warm oven [1]
⚠ If you missed marks here: The words residue and filtrate carry marks: sand is the residue on the paper, the salt solution is the filtrate that passes through. Washing the sand removes traces of salt solution clinging to it — skipping this step leaves the sand impure — and "evaporate to the point of crystallisation then cool" is the phrase for the crystallisation mark; boiling to complete dryness is a weaker answer that spits crystals out of the dish.
(c)[3]
Another student suggests a quicker plan: "Heat the whole mixture strongly in an open beaker until only sand is left." Evaluate this plan by giving three reasons why it fails to obtain the three pure substances.
Model Answer -- 4(c)
the iodine vapour escapes into the room and is lost (and it is harmful/toxic), because nothing is provided to collect it [1]
sodium chloride does not vaporise at these temperatures, so it stays mixed with the sand — "only sand is left" is wrong [1]
no substance is obtained pure: the iodine is lost and the salt and sand remain mixed together [1]
⚠ If you missed marks here: Evaluate questions want you to test each claim in the plan against chemistry: sodium chloride melts at about 800 °C and does not boil away over a Bunsen, so the premise "only sand is left" is itself false. The strongest answers attack all three products separately — iodine lost, salt still mixed with sand, hence nothing recovered pure — rather than making one general complaint about safety.
Question 5 -- Spot the Flaws: Collecting Ammonia
Total: 10 marks
A student in Warsaw prepares ammonia gas, which is very soluble in water and less dense than air. His written method contains three mistakes:
"Step 1: dry the gas by bubbling it through concentrated sulfuric acid.
Step 2: collect the gas over water in an inverted measuring cylinder.
Step 3: alternatively, collect the gas by downward delivery into an upright gas jar."
(a)[6]
Identify the mistake in each of the three steps. For each one, explain why it fails for ammonia and state the correction.
Model Answer -- 5(a)
Step 1: concentrated sulfuric acid is an acid and ammonia is an alkaline gas, so the drying agent reacts with (absorbs) the ammonia [1]
correction: dry the gas with a basic drying agent such as calcium oxide [1]
Step 2: ammonia is very soluble in water, so it dissolves and almost no gas is collected [1]
correction: do not collect over water — collect directly by delivery into a dry container [1]
Step 3: ammonia is less dense than air, so it escapes upwards out of an upright jar [1]
correction: collect by upward delivery into an inverted (upside-down) container [1]
⚠ If you missed marks here: Concentrated sulfuric acid is an excellent drying agent for most gases — the flaw here is chemical, not practical: an acid cannot dry an alkaline gas because it neutralises it. Each of the three errors carries two marks, one for the reason it fails and one for the fix, so an answer that only names the corrections without explaining scores at most half. Match density to delivery direction: less dense than air always means upward delivery into an inverted container.
(b)[2]
Describe the test for ammonia gas and its positive result.
Model Answer -- 5(b)
hold a piece of damp red litmus paper in the gas [1]
the litmus paper turns blue [1]
⚠ If you missed marks here: Both the word "damp" and the colour change "red to blue" are required — ammonia is the only common gas that turns litmus blue, so writing "litmus changes colour" without the direction wastes the result mark. Starting with blue litmus proves nothing because it stays blue.
(c)[2]
Explain why the litmus paper used in the ammonia test must be damp.
Model Answer -- 5(c)
the ammonia must dissolve in the water on the paper to form an alkaline solution (containing hydroxide ions) [1]
a dry gas cannot affect dry indicator paper, so no colour change would be seen [1]
⚠ If you missed marks here: Indicators respond to ions in solution, not to gas molecules — the dampness gives the ammonia water to dissolve in, producing the alkaline solution that turns the litmus blue. "So the gas sticks to the paper" is a common wrong guess; the answer is about dissolving and forming an alkali, not adhesion.
Question 6 -- Melting Points and Purity
Total: 12 marks
A pharmaceutical laboratory in Basel receives four solid samples, P, Q, R and S, each claimed to be a pain-relief compound whose data-book melting point is exactly 135 °C. An analyst measures the temperature at which each sample starts and finishes melting.
sample
melting starts / °C
melting finishes / °C
P
135
135
Q
127
133
R
140
140
S
121
129
(a)[2]
State the two features of melting-point data that show a sample is a pure sample of the claimed compound.
Model Answer -- 6(a)
it melts sharply, at a single temperature rather than over a range [1]
the melting point equals the data-book value for the compound [1]
⚠ If you missed marks here: Purity checking needs both criteria: sharpness tells you the sample is pure, and the match with the data-book value tells you it is the right substance. Sample R in this table is designed to punish anyone who only quotes the first criterion — it melts sharply, yet at the wrong temperature.
(b)[4]
Using the data, state a conclusion about each of the four samples. Give the evidence for each conclusion.
Model Answer -- 6(b)
P is a pure sample of the compound: it melts sharply at exactly 135 °C [1]
Q is an impure sample of the compound: it melts below 135 °C and over a range (127–133 °C) [1]
R is pure but is a different substance: it melts sharply, but at 140 °C, not 135 °C [1]
S is the least pure sample: its melting point is depressed furthest and its melting range (121–129 °C) is the widest [1]
⚠ If you missed marks here: Sample R is the trap: a sharp melting point proves purity, but 140 °C is above the data-book value, and impurities can only lower a melting point — so R must be a pure sample of some other compound entirely. Comparing Q and S, the greater the impurity the lower the start of melting and the wider the range, which is the evidence for calling S the least pure.
(c)[3]
Explain the effect of an impurity on the melting point of a solid and on the boiling point of a liquid. Include one everyday or laboratory example.
Model Answer -- 6(c)
an impurity lowers the melting point of a solid [1]
and makes it melt over a range of temperatures instead of sharply [1]
an impurity raises the boiling point of a liquid, e.g. salt water boils above 100 °C (or salt spread on roads lowers the melting point of ice) [1]
⚠ If you missed marks here: The two effects run in opposite directions and candidates regularly swap them: impurities lower a melting point but raise a boiling point. The range mark is separate from the lowering mark — both must appear — and the example must fit the effect you attach it to: salt water boiling above 100 °C illustrates the boiling-point rise, while salting icy roads illustrates melting-point depression.
(d)[3]
The analyst wants final proof that sample P really is the claimed compound and not a different pure substance that happens to melt at 135 °C. Describe how mixing P with an authentic pure sample of the compound and re-measuring the melting point provides this proof.
Model Answer -- 6(d)
mix (grind) P thoroughly with the authentic sample and measure the melting point of the mixture [1]
if P is the same compound, the mixture still melts sharply at 135 °C [1]
if P is a different substance, each solid acts as an impurity in the other, so the mixture melts below 135 °C and over a range [1]
⚠ If you missed marks here: The clever part of the mixed melting-point method is that a different compound acts as an impurity even though both solids are individually pure — so a depressed, broadened melt exposes the impostor. Full marks need both outcomes stated: unchanged sharp melting at 135 °C proves identity, while lowering and broadening disproves it; describing only one case leaves the logic incomplete.
Question 7 -- Six Gases from Six Observations
Total: 10 marks
Six numbered test tubes in a laboratory in Athens each contain one of these gases: hydrogen, oxygen, carbon dioxide, chlorine, sulfur dioxide and ammonia. A student records one observation for each tube.
(a)[6]
Name the gas in each tube.
(i) Tube 1: a lighted splint gives a squeaky pop. [1] (ii) Tube 2: a glowing splint relights. [1] (iii) Tube 3: the gas turns limewater milky but has no effect on acidified aqueous potassium manganate(VII). [1] (iv) Tube 4: damp blue litmus paper turns red and is then bleached white. [1] (v) Tube 5: acidified aqueous potassium manganate(VII) changes from purple to colourless. [1] (vi) Tube 6: damp red litmus paper turns blue. [1]
Model Answer -- 7(a)
(i) hydrogen [1]
(ii) oxygen [1]
(iii) carbon dioxide [1]
(iv) chlorine [1]
(v) sulfur dioxide [1]
(vi) ammonia [1]
⚠ If you missed marks here: Tube 3 is the discriminating clue: both carbon dioxide and sulfur dioxide turn limewater milky, so the "no effect on potassium manganate(VII)" half of the observation is what forces the answer to be carbon dioxide. Keep the splint tests straight — a lighted splint pops with hydrogen, a glowing splint relights in oxygen — and remember bleaching identifies chlorine while red-to-blue litmus identifies ammonia.
(b)[2]
Explain why the damp blue litmus paper in Tube 4 turns red before it turns white.
Model Answer -- 7(b)
chlorine dissolves in the water on the paper to form an acidic solution, which turns the blue litmus red [1]
chlorine (its solution) is also a bleach, so it then removes the colour, turning the paper white [1]
⚠ If you missed marks here: Two different properties of chlorine act one after the other: it is acidic in solution (litmus goes red) and it is a bleach (colour then destroyed). Answers that only mention bleaching cannot explain the red stage, and answers that only mention acidity cannot explain the white — the sequence needs both.
(c)[2]
A student claims the limewater test alone can distinguish carbon dioxide from sulfur dioxide. Explain why this claim is wrong, and state the test that does distinguish the two gases, with its result.
Model Answer -- 7(c)
both gases are acidic and both turn limewater milky, so limewater cannot tell them apart [1]
use acidified aqueous potassium manganate(VII): sulfur dioxide turns it from purple to colourless, but carbon dioxide leaves it purple [1]
⚠ If you missed marks here: Say "decolourised" or "purple to colourless" for the potassium manganate(VII) result — "goes clear" is not a colour change and is routinely refused. The first mark needs the reason the claim fails: sulfur dioxide also turns limewater milky, so a positive limewater result is ambiguous between the two acidic gases.
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