A pharmaceutical company uses paper chromatography to check the purity of a newly synthesised drug compound. After running the chromatogram, the solvent front has travelled 9.0 cm. The drug compound produces a single spot that has travelled 6.3 cm. A known pure sample of the same drug, run alongside it, produces a spot that has also travelled 6.3 cm. A technician concludes that the drug sample is pure. Explain how paper chromatography separates the components of a mixture and justify the technician's conclusion that the drug sample is pure.

Paper chromatography separates components by differential adsorption.
1. The solvent (mobile phase) ascends the paper by capillary action, carrying the mixture.
2. Each component interacts differently with the paper (stationary phase) and the solvent; those that bind more strongly to the paper travel more slowly.
3. The Rf value is calculated as the distance travelled by the spot divided by the distance travelled by the solvent front.
4. For the drug sample Rf = 6.3 cm ÷ 9.0 cm = 0.70. The pure reference gives the same Rf (0.70), showing they are the same substance.
5. Only one spot appears for the drug sample, indicating no other substances are present. Hence the sample is pure.

examiner_tips:["Show the Rf calculation explicitly.","Explain why different affinities give different distances.","State that a single spot means no impurities.","Use the exact wording ‘mobile phase’ and ‘stationary phase’."],
common_mistakes:["Calculating Rf incorrectly (e.g. using 9.0/6.3).","Forgetting to mention the role of affinities.","Claiming purity without noting the single spot.