Skip to content
MarkSchemeMarkScheme
Sign up — no card

9701 · 37.1

Thin-layer chromatography — practice questions

Practice and worked examples for 9701 Thin-layer chromatography. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

A TLC experiment was performed using a silica plate and hexane as the mobile phase. The solvent front moved 8.5 cm from the baseline. A single spot was observed at a distance of 5.1 cm from the baseline. Calculate the RfR_f value for this compound.

Show solution outline
  1. Identify the given distances:
    • Distance travelled by solvent front = 8.5 cm
    • Distance travelled by the spot = 5.1 cm
  2. State the formula for the RfR_f value: Rf=distance travelled by the spotdistance travelled by the solvent frontR_f = \frac{\text{distance travelled by the spot}}{\text{distance travelled by the solvent front}}
  3. Substitute the values and calculate: Rf=5.1 cm8.5 cm=0.60R_f = \frac{5.1 \text{ cm}}{8.5 \text{ cm}} = 0.60
  4. Final Answer: The RfR_f value is 0.60 (to 2 significant figures). Note that RfR_f values are unitless and are usually quoted to two decimal places.

Worked example 2

A student suspects a sample of aspirin is impure and contains paracetamol. They run a TLC plate with the sample (S), pure aspirin (A), and pure paracetamol (P). The mobile phase is a mixture of ethyl acetate and hexane. After development and visualisation under UV light, the solvent front had moved 9.0 cm. The chromatogram showed the following spots:

  • Aspirin (A): one spot at 7.2 cm
  • Paracetamol (P): one spot at 4.5 cm
  • Sample (S): two spots, one at 4.5 cm and another at 7.2 cm.

(a) Calculate the RfR_f value for aspirin. (b) What do these results indicate about the student's sample (S)?

Show solution outline

(a) Calculate RfR_f for aspirin:

  • Distance for aspirin spot = 7.2 cm
  • Distance for solvent front = 9.0 cm
  • Rf(aspirin)=7.2 cm9.0 cm=0.80R_f(\text{aspirin}) = \frac{7.2 \text{ cm}}{9.0 \text{ cm}} = 0.80

(b) Analyse the sample (S):

  • The sample (S) shows two spots.
  • One spot has moved 7.2 cm, which corresponds to the same height and therefore the same RfR_f value (0.80) as pure aspirin.
  • The second spot has moved 4.5 cm. The RfR_f for this spot is 4.59.0=0.50\frac{4.5}{9.0} = 0.50. This corresponds to the same height and RfR_f value as pure paracetamol.
  • Conclusion: The sample (S) is impure. It is a mixture containing both aspirin and paracetamol.
Mark a 9701 question on this topicFull lessonOpen in course studio