Chromatography Worked ExamplesIGCSE Chemistry 0620

(a) Mistake 1: the solvent level is above the start line (and the spots). The food colourings would dissolve into the solvent in the beaker instead of travelling up the paper, so no chromatogram forms.

Mistake 2: the start line is drawn in ballpoint pen (ink). The ink is soluble in the solvent, so it would move up the paper and separate into its own dyes, mixing with the spots from the food colourings.

(b) A pencil. Graphite is insoluble in the solvent, so a pencil line stays where it is.

(c) The lid keeps the air in the beaker saturated with solvent vapour, so the solvent does not evaporate from the paper as it rises. The solvent front then moves up steadily and the separation is better.

(a) A pure dye gives one spot. Only P has a single spot.

(b) Read across at the height of P's spot: R and S both have a spot at the same height as P. So the dye in P (the spot about halfway up) is also in R and S.

(c) S has three spots: low, halfway and two-thirds. P supplies the halfway spot and Q supplies the low and two-thirds spots, so S = P + Q.

(d) B has four spots. Q gives the low and two-thirds spots; R gives the halfway and high spots. Together they account for all four spots, so B = Q + R.

Check the alternatives: P + Q gives only three spots (no high spot); P + R gives only two different spots. Neither matches B.

(a) Both distances are measured from the baseline.

(b) Spot 1:

Spot 2:

(c) In the same solvent, a substance always has the same $R_f$ value. Spot 1 matches E ($0.32$) and spot 2 matches G ($0.80$).

Both distances must be measured from the baseline, where the substance and the solvent start together.

The answer is B. A ($\frac{5.4}{8.7}$) adds the $1.5$ cm to the solvent distance only. C ($\frac{6.9}{8.7}$) measures both distances from the bottom edge of the paper. D is the fraction upside down — an $R_f$ value can never be greater than $1$, because no substance travels further than the solvent.

A locating agent is sprayed on a dried chromatogram to make colourless spots visible. It is only needed when the substances being separated cannot be seen.

The answer is B — amino acids are colourless, so their spots are invisible until a locating agent reacts with them.

A, C and D all involve coloured substances (pigments, ink, dyes), which can be seen directly on the paper.

Draw a baseline in pencil about $1$–$2$ cm from the bottom of the chromatography paper.

Put a small, concentrated spot of the fruit juice on the baseline, and spots of glucose, fructose and sucrose alongside it, spaced out.

Stand the paper in a beaker containing a suitable solvent, with the solvent level below the baseline. Put a lid on the beaker.

Leave until the solvent has risen most of the way up the paper, then remove it, mark the solvent front and let the paper dry.

Spray the dry paper with a locating agent, which makes the colourless sugar spots visible.

Compare the positions of the juice spots with the sugar spots (or compare $R_f$ values). A juice spot at the same height (same $R_f$) as a known sugar shows that sugar is present.

If two sugars have the same $R_f$ value in this solvent, they travel the same distance and their spots overlap as one spot. That leaves three spots for four sugars. The answer is C.

A is wrong: an insoluble sugar would not move, but it would still be on the paper and would show as a spot on the baseline. The chromatogram has no spot there.

B is wrong: sugars are not volatile. D is wrong: marking the solvent front is needed to calculate $R_f$ values, but it does not change how many spots appear.

(a) T has two spots. The upper one is at the same height as the spot for aspirin A and the lower one is at the same height as caffeine C. So T contains aspirin and caffeine.

T has no spot level with P, so the tablet does not contain paracetamol.

(b) A pure substance gives one spot. N gives two spots: one matches aspirin, and the other (fainter, lower) matches none of the known substances. N is not pure — it contains aspirin plus an impurity.

(c) How far a substance travels depends on the solvent (and the conditions). Running every sample together means each substance has the same conditions, so equal heights (equal $R_f$ values) really do mean the same substance.

(a) Divide each spot's distance by the solvent's distance.

(b) Match each value with the table: $0.18$ = blue 1, $0.46$ = red 3, $0.75$ = yellow 5.

(c) A substance is carried up the paper by the solvent, so it cannot travel further than the solvent front. The top of the fraction is never larger than the bottom, so $R_f \le 1$.

(a) To calculate $R_f$ you need the distance travelled by the solvent. If the solvent reaches the top edge, you do not know how far it would have gone, and a fast-moving dye may also have run off the paper.

Improvement: remove the paper before the solvent reaches the top, and immediately mark the solvent front in pencil (it evaporates and disappears as the paper dries).

(b) A large drop spreads out, giving large, blurred spots that overlap and whose centres are hard to measure. Use a small spot, let it dry, then add more to the same place to make it concentrated.

(c) The green dye is not a single substance — it is a mixture of (at least) a blue dye and a yellow dye.

(a) From the grid: solvent front $9.0$ cm; spot centres at $1.6$ cm, $4.4$ cm and $7.6$ cm.

(b) Spot 1 ($0.18$) is W; spot 3 ($0.84$) is Z.

(c) Distance travelled = $R_f \times$ distance travelled by solvent.

The two positions are only about $0.4$ cm apart, which is less than the size of a spot (about $0.5$ cm). If both dyes are present their spots overlap into one long spot stretching from about $4.0$ to $4.8$ cm — which is what spot 2 looks like. But a large amount of just one dye could also spread out, so the chromatogram cannot tell us.

(d) Run it again in a different solvent in which X and Y have very different $R_f$ values, with pure X and pure Y spotted alongside D. (Using a much longer piece of paper would also increase the gap between the spots.)

(a) In solvent 1, K, L and M all have spots at the same height — the same $R_f$ value. K matches both L and M, so it could be either (or a mixture of them).

(b) In solvent 2:

This matches M ($0.65$) and not L ($0.30$) or N ($0.55$). K matched M in both solvents, so K is M.

(c) It is supported: a pure substance gives a single spot, and K gave one spot in each solvent. (If K were a mixture of L and M, it would have given two spots in solvent 2.)