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A student tests a mixture M by thin layer chromatography on a plate coated with silica gel…

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Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A student tests a mixture M by thin layer chromatography on a plate coated with silica gel. Beside M, she spots two pure compounds, X and Y, on the same base line. She stands the plate in a jar with a little solvent and takes it out when the solvent has risen most of the way up. The figure shows the plate after the run.

A thin layer chromatography plate after the run, 10.5 cm tall. Three lanes are marked M, X and Y on the base line, which is drawn 1.2 cm above the bottom edge. The solvent front is a line 9.7 cm above the bottom edge. Lane M, the mixture, shows two spots: P, whose centre is 7.3 cm above the bottom edge, and Q, whose centre is 3.5 cm above it. Lane X, a known compound, shows one spot at 3.5 cm. Lane Y, another known compound, shows one spot at 5.4 cm. Each height is marked with an arrow from the bottom edge of the plate.

Every height in the figure is measured from the bottom edge of the plate, not from the base line. The mixture gives two spots, P and Q. The retardation factor, $R_f$ , of a compound is the distance it moves from the base line divided by the distance the solvent moves from the base line.

How far did the solvent move from the base line?
What is the $R_f$ value of P?
What is the $R_f$ value of Q?
What is the $R_f$ value of Y?
The run is repeated on a longer plate of the same kind with the same solvent. It is stopped when the solvent front is $17.0\ \text{cm}$ above the base line. How far apart are the centres of P and Q now?
In this second run, how far above the base line would a spot of Y be?

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Given. From the figure: base line $1.2\ \text{cm}$ , solvent front $9.7\ \text{cm}$ , P $7.3\ \text{cm}$ , Q and X $3.5\ \text{cm}$ , Y $5.4\ \text{cm}$ , all above the bottom edge.

Plan. Change every reading to a distance from the base line. Then divide by the solvent distance.

Step 1. Solvent. $9.7\ \text{cm} - 1.2\ \text{cm} = 8.5\ \text{cm}$ .

Step 2. P and Q. P moved $7.3 - 1.2 = 6.1\ \text{cm}$ , so $R_f = \dfrac{6.1\ \text{cm}}{8.5\ \text{cm}} = 0.72$ . Q moved $2.3\ \text{cm}$ , so $R_f = \dfrac{2.3\ \text{cm}}{8.5\ \text{cm}} = 0.27$ .

Step 3. Y. Y moved $4.2\ \text{cm}$ , so $R_f = \dfrac{4.2\ \text{cm}}{8.5\ \text{cm}} = 0.49$ . X stands level with Q, so X has the same value as Q, $0.27$ . So M contains X, and no spot of M matches Y.

Step 4. The longer plate. The $R_f$ values stay the same, so each distance is $R_f$ times the new solvent distance. P and Q were $3.8\ \text{cm}$ apart, so now the gap is $\dfrac{3.8}{8.5} \times 17.0\ \text{cm} = 7.60\ \text{cm}$ . For Y, $\dfrac{4.2}{8.5} \times 17.0\ \text{cm} = 8.40\ \text{cm}$ .

Answer: (a) $8.5\ \text{cm}$ . (b) $0.72$ . (c) $0.27$ . (d) $0.49$ . (e) $7.60\ \text{cm}$ . (f) $8.40\ \text{cm}$ .

Check. The second plate is the first one stretched by $\dfrac{17.0}{8.5} = 2.000$ . The gap of $3.8\ \text{cm}$ becomes $3.8 \times 2.000 = 7.60\ \text{cm}$ . Q, with the smallest $R_f$ , is the component held most strongly by the silica gel.

Watch out. Dividing the heights from the bottom edge gives $\dfrac{7.3}{9.7} = 0.75$ for P, which is wrong. The spots started on the base line, so every distance is measured from there.

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