Chemistry · 12 Experimental techniques and chemical analysis

12.3 Chromatography

Before you start

The difference between a pure substance and a mixture.
That a solute dissolves in a solvent to make a solution, and that solubility varies between substances.
Measuring a length in mm with a ruler.

Syllabus objectives covered (4)

Describe how paper chromatography is used to separate mixtures of soluble coloured substances, using a suitable solvent
Interpret simple chromatograms to identify: (a) unknown substances by comparison with known substances (b) pure and impure substances
Describe how paper chromatography is used to separate mixtures of soluble colourless substances, using a suitable solvent and a locating agent
State and use the equation for Rf: Rf = distance travelled by substance / distance travelled by solvent

Part 1 of 4

How the separation actually happens

Chromatography separates the substances in a mixture by how strongly each is attracted to two competing things: the paper, and the moving solvent.
A substance that dissolves well in the solvent and is weakly held by the paper travels a long way up. One that is strongly held by the paper, or barely dissolves, hardly moves.
So the mixture spreads out into separate spots, each at its own height. Different substances, different heights — that is the whole technique.
The method: draw a pencil line near the bottom of the paper, put a small spot of the mixture on it, and stand the paper in a shallow layer of solvent.
Pencil, not pen — and this is a genuine marking point. Ink is itself a mixture of dyes and would run up the paper alongside the sample.
The solvent level must start below the pencil line. If the line is submerged, the sample dissolves into the solvent in the beaker and washes away instead of climbing the paper.
Cover the container. Otherwise the solvent evaporates from the paper as fast as it rises, and the spots never separate properly.
Solvent below the baseline, lid on. Both details are worth marks.
Let the solvent rise until it is near the top, then take the paper out and mark the solvent front in pencil immediately. Once the paper dries you cannot see where the solvent reached, and without it you cannot calculate anything.

Part 2 of 4

Reading a chromatogram

Two questions are asked of every chromatogram: is this substance pure, and what is in this mixture?
Purity. A pure substance gives one spot. More than one spot means more than one substance, so the sample is impure — or is a mixture, which for these purposes is the same observation.
Identification. Run the unknown alongside known reference substances on the same paper. A spot in the unknown at the same height as a known spot suggests that substance is present.
The comparison is only valid on the same paper in the same run. Different solvent, different paper or different temperature moves everything, so comparing across two chromatograms proves nothing.
Be careful with the wording of your conclusion. A matching height is strong evidence, not proof — two different substances can happen to travel the same distance.
A common question: a mixture shows three spots, two of which match knowns. The correct answer is that it contains those two substances plus at least one other, unidentified substance.

Part 3 of 4

Colourless substances and locating agents

The method so far only works if you can see the spots. Amino acids and sugars are colourless, so the chromatogram comes out apparently blank.
The solution is a locating agent — a chemical sprayed onto the dried paper that reacts with the colourless substances to make them visible.
Everything else about the method is unchanged: same pencil line, same solvent below the line, same covered container. Only the final step is added.
The order matters. Let the chromatogram run, mark the solvent front, dry the paper, and only then apply the locating agent.
An alternative for some substances is to view the dried paper under ultraviolet light, which makes them fluoresce.
You are not expected to know the names of particular locating agents. Knowing what one is for, and where it comes in the sequence, is what is examined.

Part 4 of 4

R_f values

A spot's height depends on how long you ran the experiment, so raw distances are not comparable between runs. The $R_{f}$ value fixes this by expressing the distance as a fraction of how far the solvent went.
$R_{f} = \frac{distance travelled by substance}{distance travelled by solvent}$
Measure both distances from the pencil baseline — not from the bottom edge of the paper, and not from the solvent surface. This is where most calculation marks are lost.
For the substance, measure to the centre of its spot. Spots spread as they rise, and the centre is the consistent point.
$R_{f}$ has no units, because it is a distance divided by a distance.
$R_{f}$ is always between 0 and 1. A value above 1 would mean the substance overtook the solvent carrying it, which is impossible — so an answer greater than 1 means you have divided the wrong way round.
A larger $R_{f}$ means the substance travelled further, which means it is more soluble in the solvent, or less strongly attracted to the paper.
Because $R_{f}$ is a ratio, it is the same for a given substance in a given solvent whatever the size of the paper — which is what makes it useful for identification against published values.

Worked examples

Example 1. On a chromatogram, the solvent front is $8.0$ cm above the baseline. A spot's centre is $5.2$ cm above the baseline. Calculate the $R_{f}$ value of the substance.

Show the working

Identify the two distances, both measured from the pencil baseline: substance $5.2$ cm, solvent $8.0$ cm.
Substitute into the equation, substance over solvent.
$R_{f} = \frac{5.2}{8.0}$
Work it out: $R_{f} = 0.65$ .
Check it is between 0 and 1, and give no unit — $R_{f}$ is a ratio of two lengths.

Answer: $R_{f} = 0.65$ (no unit).

Example 2. A chromatogram is run with an unknown ink X alongside three known dyes, A, B and C. X produces two spots. One is level with dye A and one is level with dye C. Interpret the chromatogram.

Show the working

Count the spots in the unknown first. Two spots means X is a mixture of two substances, so it is not pure.
Now match them. A spot level with A indicates dye A is present; a spot level with C indicates dye C is present.
State what is absent as well as what is present: there is no spot level with B, so dye B is not present in X.
Word the conclusion carefully — the heights match, which is strong evidence rather than proof, and it holds because all four were run on the same paper in the same solvent.

Answer: X is a mixture of two dyes, not a pure substance. It contains dyes A and C, since its spots are level with theirs on the same chromatogram, and does not contain dye B.

Example 3. Describe how you would use paper chromatography to find out how many amino acids are present in a colourless solution.

Show the working

Set up as usual: draw a pencil baseline near the bottom of the chromatography paper and place a small spot of the solution on it.
Stand the paper in a covered container with the solvent level below the baseline, so the spot is not washed off.
Leave until the solvent has risen near the top, then remove the paper and mark the solvent front in pencil before it dries.
Dry the paper, then spray it with a locating agent, because amino acids are colourless and would otherwise be invisible.
Count the spots that appear. The number of spots is the number of amino acids present.

Answer: Spot the solution on a pencil baseline, run it in a covered container with the solvent below the baseline, mark the solvent front, dry the paper, then spray with a locating agent to make the colourless amino acids visible. The number of spots gives the number of amino acids.

Where people go wrong

Mistake: Drawing the baseline in pen.

Instead: Always use pencil.

Ink is a mixture of dyes and would separate up the paper alongside the sample, ruining the chromatogram.

Mistake: Starting with the solvent above the baseline.

Instead: The solvent must start below the line.

Otherwise the sample dissolves into the solvent in the container and washes away rather than travelling up the paper.

Mistake: Measuring distances from the bottom edge of the paper.

Instead: Measure both distances from the pencil baseline.

The baseline is where every substance started. Measuring from the edge adds the same error to both distances and gives a wrong $R_{f}$ .

Mistake: Measuring to the top or bottom of a spot.

Instead: Measure to the centre of the spot.

Spots spread out as they rise, so only the centre is a consistent, reproducible point.

Mistake: Giving an $R_{f}$ value with a unit, or greater than 1.

Instead: $R_{f}$ has no unit and always lies between 0 and 1.

It is a ratio of two distances, and a substance cannot travel further than the solvent carrying it. A value above 1 means the division was inverted.

Mistake: Forgetting to mark the solvent front before the paper dries.

Instead: Mark it in pencil the moment the paper comes out.

Once dry the front is invisible, and without it no $R_{f}$ can be calculated at all.

Check yourself

Why must the baseline be drawn in pencil and the solvent level start below it?
Answer
Pencil does not dissolve and separate; ink would. If the solvent were above the baseline the spots would dissolve into it and wash away.
Why is the container covered during the run?
Answer
To stop the solvent evaporating from the paper, which would prevent it rising properly and stop the substances separating.
How does a chromatogram show that a substance is pure?
Answer
A pure substance produces a single spot; two or more spots show a mixture.
Why must the unknown and the known reference substances be run on the same piece of paper?
Answer
Because distances depend on the solvent, paper and conditions, so spots can only be compared within one run.
What is a locating agent, and at what point is it used?
Answer
A chemical sprayed onto the dried chromatogram to make colourless substances visible. It is applied after the run, once the solvent front has been marked and the paper dried.
A substance travels $3.6$ cm and the solvent travels $9.0$ cm. Calculate the $R_{f}$ value.
Answer
$R_{f} = 3.6 / 9.0 = 0.40$ , with no unit.
Two substances have $R_{f}$ values of $0.20$ and $0.80$ in the same solvent. Which is more soluble in that solvent, and how do you know?
Answer
The one with $R_{f} = 0.80$ , because it travelled further up the paper, which means it is more soluble in the solvent and less strongly held by the paper.

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. A student uses paper chromatography to investigate the dyes in a sample of green ink. The solvent front travels $10.0$ cm from the baseline. Two spots appear, with centres $4.5$ cm and $7.5$ cm from the baseline.

(a) Explain why the baseline is drawn in pencil. 1 mark No calculator

Mark scheme

B1 pencil does not dissolve in the solvent / ink would separate and travel up the paper

(b) State what the number of spots tells you about the green ink. 1 mark No calculator

Mark scheme

B1 it is a mixture of two dyes / it is not pure

(c) Calculate the $R_{f}$ value of each dye. 3 marks

Mark scheme

C1 Rf = distance moved by substance / distance moved by solvent, used
A1 0.45
A1 0.75

(d) Explain which dye is more strongly attracted to the paper. 2 marks No calculator

Mark scheme

B1 the dye with Rf 0.45 / the one that travelled 4.5 cm
B1 it travelled less far, so it is held more strongly by the paper and/or is less soluble in the solvent, ORA

Question 2. A forensic scientist analyses a colourless solution thought to contain amino acids.

(a) Explain why a locating agent is needed in this analysis. 2 marks No calculator

Mark scheme

B1 amino acids are colourless, so the spots cannot be seen
B1 the locating agent reacts with them to produce coloured spots

(b) Describe the correct order of the final three steps of the experiment. 3 marks No calculator

Mark scheme

B1 remove the paper and mark the solvent front in pencil
B1 allow the paper to dry
B1 spray with the locating agent (accept view under UV light)

Question 3. Four spots are run on one chromatogram: an unknown sample S, and pure substances P, Q and R. S produces three spots. One is level with P and one is level with R.

(a) Deduce what the chromatogram shows about the composition of S. 3 marks No calculator

Mark scheme

B1 S is a mixture of three substances / is impure
B1 it contains P and R
B1 the third substance is not P, Q or R and has not been identified

(b) Suggest why comparing the spots in S with a chromatogram of Q run on a different day using a different solvent would not be valid. 2 marks No calculator

Mark scheme

B1 the distance a substance travels depends on the solvent (and the paper and conditions) used
B1 so spots can only be compared within the same chromatogram / same run, owtte

Words to know

Chromatography: A technique that separates the substances in a mixture by how far each travels up a paper in a moving solvent.
Baseline: The pencil line on which the samples are spotted, and from which all distances are measured.
Solvent front: The furthest point the solvent reaches, marked in pencil as soon as the paper is removed.
Locating agent: A chemical sprayed on a dried chromatogram to make colourless substances visible.
$R_{f}$ value: The distance travelled by a substance divided by the distance travelled by the solvent. A ratio between 0 and 1, with no unit.

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