Purity, formulations & chromatography (RP6) – GCSE Chemistry exam questions with mark scheme

1Grade 1–3Foundation3 marks

A student has a sample of a white solid labelled 'pure aspirin'. Pure aspirin melts at 136 °C.

(a) What is meant by a pure substance in chemistry? [1]

(b) The sample melts over the range 128–133 °C. Is the sample pure? Give two reasons from the data. [2]

(a) a single element or compound, not mixed with any other substance(1)
(b) no / it is impure and it melts below 136 °C / lower melting point(1)
(b) it melts over a range of temperatures (not at a sharp, single temperature)(1)
Answer: (a) a single element or compound not mixed with anything else (b) impure — melts below 136 °C and over a range
Examiner tip: In everyday language 'pure' means natural; in chemistry it means one substance only.

2Grade 4–53 marks

A paint is a formulation.

(a) What is a formulation? [2]

(b) Give one other example of a formulation. [1]

(a) a mixture (that has been designed as a useful product)(1)
(a) components in measured quantities / carefully measured proportions (so the product has the required properties)(1)
(b) any one from: fuels, cleaning agents, alloys, fertilisers, medicines / drugs, foods; allow a named example e.g. toothpaste, shampoo, washing-up liquid, brass / steel, sun cream; do not accept paint (given in the question) or a pure substance such as water(1)
Answer: (a) a mixture designed as a useful product, made by mixing components in measured quantities (b) e.g. medicines
Examiner tip: Both ideas are needed for 2 marks: a mixture, and measured quantities.

3Grade 6–75 marks

A student uses paper chromatography to analyse a food colouring, X, and three dyes, A, B and C. The diagram shows the chromatogram.

The solvent front moved 8.0 cm from the start line. The upper spot of X moved 5.6 cm.

(a) Calculate the R_f value of the upper spot of X. [2]

(b) Which dyes are in X? [1]

(d) Explain why different substances move different distances up the paper. [1]

(a) R_f = 5.6 ÷ 8.0(1)
(a) 0.70; allow 0.7(1)
(b) A and C (both needed; do not accept if B is included)(1)
(c) pencil does not dissolve in the solvent / ink would dissolve and move (and mix with the spots)(1)
(d) different solubility in the solvent / different attraction to the paper (stationary phase); allow different distribution between mobile and stationary phase(1)
Answer: (a) 0.70 (b) A and C (c) pencil is insoluble in the solvent (d) they have different solubilities in the solvent / attraction to the paper
Examiner tip: R_f is always less than 1 — if you get a value above 1 you have divided the wrong way round.

4Grade 1–3Foundation4 marks

A carton of orange juice has the label “100% pure orange juice — nothing added”.

(a) Explain why a chemist would not describe the orange juice as a pure substance. [1]

A student heats four colourless liquids, A, B, C and D, and records how each one boils.

Liquid	Observation when heated
A	boils at 100 °C; the temperature stays at 100 °C until all the liquid has boiled away
B	starts to boil at 101 °C; the temperature rises to 104 °C as it boils
C	boils at 78 °C; the temperature stays at 78 °C until all the liquid has boiled away
D	starts to boil at 80 °C; the temperature rises to 86 °C as it boils

(b) Which two liquids are pure substances? [1]

(d) Liquid B is a sample of water with something dissolved in it. Pure water boils at 100 °C.

Use the data to describe one way the dissolved substance changes how water boils. [1]

(a) it is a mixture / it contains more than one substance / contains water, sugars, acids etc. allow it is not a single element or compound; ignore ‘it has things added’ unless linked to mixture(1)
(b) A and C (both needed, either order)(1)
(c) they boil at a specific / fixed / single temperature (rather than over a range of temperatures) allow temperature stays constant while boiling; do not accept ‘boil at 100 °C’ alone(1)
(d) either: it raises the boiling point / water starts to boil above 100 °C or: it makes the water boil over a range of temperatures (101–104 °C) allow temperature rises as it boils(1)
Answer: (a) it is a mixture of different substances, not a single element or compound (b) A and C (c) they boil at a fixed (single) temperature (d) it raises the boiling point above 100 °C / it boils over a range
Examiner tip: In chemistry “pure” means one single substance — saying the juice is “natural” or “has chemicals in it” scores nothing; you must say it is a mixture.

5Grade 4–55 marks

(a) Which one of these is a formulation? Tick (✓) one box. [1]

☐ brass ☐ copper ☐ nitrogen ☐ sodium chloride

A hand-sanitising gel is made by mixing four substances. The table shows the percentage by mass of each substance and why it is added.

Substance	Percentage by mass	Purpose
ethanol	70%	kills microorganisms
water	27%	dilutes the ethanol
glycerol	2%	stops skin drying out
perfume	1%	gives a pleasant smell

(b) Use the information in the table to explain why the hand gel is a formulation. [2]

Calculate the total mass of gel in the batch. [2]

(a) brass (only one box ticked)(1)
(b) it is a mixture (of substances) made with measured / fixed / exact quantities of each substance allow each substance is present in a set percentage(1)
(b) each substance / component has a (specific) purpose / job, so the mixture is designed as a useful product ignore ‘it is useful’ alone(1)
(c) 315 ÷ 0.70 or 315 × 100 ÷ 70 or 315 ÷ 70 = 4.5 (g per 1%)(1)
(c) 450 (g) allow 450 g with no working for 2 marks(1)
Answer: (a) brass (b) mixture with measured amounts of each component, and each component has a purpose (c) 450 g
Examiner tip: In (b) quote the table: “fixed percentages” earns one mark and “each has a purpose” earns the other — saying only “it is a mixture” gets neither.

6Grade 6–76 marks

A food scientist wants to find out whether a green food colouring contains a dye called Yellow 9.

She has a sample of the green food colouring and a sample of pure Yellow 9.

Describe a method she could use to find out whether the green food colouring contains Yellow 9, using paper chromatography.

Your answer should include how she would use the results to reach a conclusion. [6]

draw a start line in pencil about 1–2 cm from the bottom of the chromatography paper(1)
use a capillary tube / dropper to put a small spot of the green food colouring on the start line(1)
put a spot of pure Yellow 9 on the start line, separate from the first spot(1)
pour solvent (e.g. water) into a beaker to a depth below the start line(1)
stand / hang the paper in the beaker with the bottom of the paper in the solvent(1)
put a lid on the beaker (to stop the solvent evaporating)(1)
leave the solvent to move up the paper; remove the paper before the solvent reaches the top(1)
mark the solvent front in pencil and let the paper dry(1)
if the green colouring has a spot at the same height / same R_f value as the Yellow 9 spot, it contains Yellow 9(1)
measure the distance from the start line to the centre of each spot and to the solvent front(1)
calculate R_f = distance moved by substance ÷ distance moved by solvent and compare(1)
repeat with a different solvent: if the spots still match, the evidence that it contains Yellow 9 is stronger(1)
a pencil line is used because pencil does not dissolve in the solvent (ink would)(1)
the start line must be above the solvent so the spots do not dissolve into the solvent in the beaker(1)

Level	Marks	What it looks like
Level 3	5–6	A coherent, logically ordered method that could be followed to give valid results, including spotting both samples on a pencil line above the solvent level, and a clear explanation of how comparing spot positions or R_f values shows whether Yellow 9 is present.
Level 2	3–4	A method with the main steps in a sensible order but some detail missing (e.g. no reference to the solvent level or the lid), and some attempt to say how the results are compared.
Level 1	1–2	Some relevant steps of chromatography are given, but they are not linked into a workable method and there is little or no indication of how to reach a conclusion.
No relevant content	0	No relevant content.

Answer: Pencil start line; small spots of colouring and pure Yellow 9 on the line; paper in solvent with solvent below the line; lid; remove before solvent reaches top; mark front; dry; compare spot heights / calculate and compare R_f values; matching spot = contains Yellow 9; confirm in a second solvent.
Examiner tip: Answers that never say how to decide (same height / same R_f as the pure Yellow 9 spot) cannot reach Level 3, however good the method is.

7Grade 6–7Higher5 marks

A chemist checks the purity of a sample of a medicine, compound M, using paper chromatography.

(a) In water, the R_f value of compound M is 0.62. The solvent front moved 8.5 cm from the start line.

Calculate the distance moved by the spot of compound M. Give your answer to 2 significant figures. [2]

(b) With water as the solvent, the sample gave one spot.

The chemist repeated the chromatography using propanone as the solvent. This time the sample gave two spots.

What do these results show about the sample? Explain your answer. [2]

Explain why, in terms of the phases in chromatography. [1]

(a) distance = R_f × distance moved by solvent = 0.62 × 8.5 (= 5.27)(1)
(a) 5.3 (cm); an unrounded answer of 5.27 scores the first mark only(1)
(b) the sample is impure / is a mixture / contains (at least) two substances do not accept ‘pure’(1)
(b) a pure compound gives one spot in all / every solvent; (the two substances had the same / similar R_f in water so did not separate) allow the impurity only separates in propanone(1)
(c) compound M has a different solubility in / attraction to propanone than water, so it distributes differently between the mobile phase (solvent) and the stationary phase (paper) allow spends a different proportion of time in the mobile phase(1)
Answer: (a) 5.3 cm (b) the sample is impure — a pure compound gives a single spot in all solvents, so two spots in propanone shows there are at least two substances (c) M has a different solubility in each solvent, so it is distributed differently between the mobile and stationary phases
Examiner tip: One spot in one solvent does not prove purity — examiners want “a pure compound gives one spot in all solvents”.

8Grade 6–74 marks

Pure compound X melts at 80 °C.

A student melted two samples of compound X, A and B, then let each one cool and recorded its temperature every 30 seconds. One sample is pure; the other is impure.

The graph shows the cooling curves.

(a) Which sample is pure compound X? Give two reasons from the graph. [3]

(b) Use the graph to give the temperature range over which sample B turns from a liquid to a solid. [1]

(a) sample A(1)
(a) A turns solid / freezes at a single / fixed temperature / the temperature stays constant (between 2 and 6 minutes) allow A has a flat section on the graph allow B freezes over a range of temperatures / B has no flat section(1)
(a) A freezes at 80 °C, which is the melting point of pure X allow B freezes below 80 °C; do not award this mark for a second statement about the range / flat section(1)
(b) from 78 °C to 72 °C allow ± 1 °C on each value; allow 72–78 °C(1)
Answer: (a) A — it changes state at a single fixed temperature, and that temperature is 80 °C (the melting point of pure X); B solidifies below 80 °C over a range (b) 78 °C to 72 °C
Examiner tip: For a reason ‘from the graph’ you must describe the shape or quote a value — ‘A is pure because it’s the right one’ scores only the identification mark.

9Grade 8–9Higher6 marks

A student wants to separate two dyes, F and G, by paper chromatography.

The table shows the R_f values of the dyes in two different solvents.

Solvent	R_f of dye F	R_f of dye G
water	0.40	0.42
ethanol	0.25	0.70

(a) A chromatogram of a mixture of F and G is run in ethanol. The solvent front moves 8.0 cm from the start line.

Calculate the distance between the centres of the spots of F and G on the chromatogram. [2]

(b) Explain why ethanol is a better solvent than water for separating F and G. [2]

(c) Explain, in terms of the mobile phase and the stationary phase, why dye G moves further up the paper than dye F when ethanol is the solvent. [2]

(a) (0.70 − 0.25) × 8.0 or 5.6 (cm) and 2.0 (cm) seen(1)
(a) 3.6 (cm) (correct answer with no working scores 2)(1)
(b) in water the R_f values are very close / almost the same so the spots would overlap / would not separate allow spots only 0.16 cm apart in water(1)
(b) in ethanol the R_f values are very different so the spots are far apart / clearly separated(1)
(c) G is more soluble in ethanol / more strongly attracted to the mobile phase (than F)(1)
(c) F is more strongly attracted to the paper / stationary phase, so G spends more time in the mobile phase and is carried further allow converse argument in terms of F(1)
Answer: (a) 3.6 cm (b) in water Rf values are almost equal so the spots overlap; in ethanol they are very different so the spots separate well (c) G is more soluble in ethanol (more attracted to the mobile phase) while F is more attracted to the paper (stationary phase), so G spends more time moving with the solvent
Examiner tip: In (c) you must compare the two dyes and name both phases — “G is more soluble” alone scores only one mark.

10Grade 4–55 marks

A student used paper chromatography to analyse the colouring on a sweet. The diagram shows the chromatogram, drawn to scale.

The table gives the R_f values of four food dyes, measured using the same solvent.

Dye	P	Q	R	S
R_f value	0.18	0.32	0.55	0.68

(a) Calculate the R_f values of the two spots that moved up the paper.

Give your answers to 2 significant figures. [3]

(b) Which dyes from the table are in the sweet coating? [1]

(a) R_f = distance moved by substance ÷ distance moved by solvent, e.g. 2.4 ÷ 7.5 or 5.1 ÷ 7.5(1)
(a) 0.32 (no units)(1)
(a) 0.68 (no units); if both values are correct but either is not given to 2 significant figures (e.g. 0.320), award max 2 for part (a)(1)
(b) Q and S (allow ecf from their R_f values in (a))(1)
(c) it is insoluble in the solvent allow it does not dissolve in the solvent / it is attracted much more strongly to the paper (stationary phase); ignore ‘it is pure’(1)
Answer: (a) 0.32 and 0.68 (b) Q and S (c) the substance is insoluble in the solvent
Examiner tip: R_f = spot distance ÷ solvent distance, so it is always less than 1 and has no units — dividing the other way round (e.g. 7.5 ÷ 2.4) loses the calculation marks.

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