A chemist analyses a mixture of amino acids using paper chromatography. The solvent front travels 9.0 cm. Three spots are produced at distances of 2.7 cm, 4.5 cm and 7.2 cm from the baseline. A reference sample of alanine produces a spot at 4.5 cm under identical conditions. Explain how the chemist can determine whether alanine is present in the mixture and describe what the Rf value of alanine tells us about how alanine behaves in the chromatography system.

OCR A-Level Chemistry B: Salters (H433) — 6.3 Analysis (chromatography, NMR) · Explain · 5 marks · View as Markdown

Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).

Paper chromatography separates mixtures based on the different affinities each component has for the stationary phase and the mobile phase. Each substance travels a characteristic distance relative to the solvent front, expressed as its Rf value.

Model answer (5 marks)

The Rf value is calculated as the distance travelled by the substance divided by the distance travelled by the solvent front. The Rf value of alanine in the mixture is 4.5 cm ÷ 9.0 cm = 0.5. The spot in the mixture at 4.5 cm matches the reference alanine spot, so alanine is present in the mixture. An Rf of 0.5 means alanine has equal affinity for the mobile phase and the stationary phase – it spends roughly equal time in each. Different substances have different Rf values because they have different affinities or interactions with the paper and the solvent, which allows them to be identified and distinguished.

Examiner tips

  • Calculate Rf first, then compare the spot positions; use the exact value 0.5 to show the match.
  • Explain the significance of Rf=0.5 in terms of phase affinity; mention that equal affinity gives a moderate Rf.

Common mistakes

  • Using the wrong distance for the solvent front (e.g. 9.0 cm vs 10 cm).
  • Failing to state that the spot at 4.5 cm matches the reference, so alanine is present.

Mark scheme (5 marks)

  1. The Rf value is calculated as the distance travelled by the substance divided by the distance travelled by the solvent front
  2. The Rf value of alanine in the mixture is 4.5 / 9.0 = 0.5
  3. The spot in the mixture at 4.5 cm matches the reference alanine spot, so alanine is present in the mixture
  4. An Rf of 0.5 means alanine has equal affinity for (spends equal time in) the mobile phase and the stationary phase
  5. Different substances have different Rf values because they have different affinities / interactions with the stationary phase (paper) and mobile phase (solvent), so they can be identified and distinguished

Key terms in this question

Rf value · paper chromatography

Related

More Analysis (chromatography, NMR) questions

▶ Try answering this question with AI marking (free) →