Elements, Compounds and Mixtures
6 marks · NormalA teacher finds a powdery sample on a lab bench. She wants to find out if it is a pure substance or a mixture. She records the melting point, then runs paper chromatography on the dissolved sample, using water as the solvent. The chromatogram below shows her results.
| Sample X observations |
|---|
| Melting starts at 87°C |
| Melting finishes at 97°C |
The solvent front travelled 8.0 cm from the pencil line.
Explain what the melting point results show about whether sample X is a pure substance or a mixture.
Describe how paper chromatography was carried out in this investigation. Include the key steps.
Use the chromatogram to identify which dyes are present in sample X. State your reasoning.
Calculate the R_f value for the dye in sample X that matches Dye B. Give your answer to two decimal places.
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
Show answer
(a) The sample melts over a range of temperatures (87°C to 97°C), which means it is a mixture, not a pure substance. Pure substances have a sharp melting point. · must show Recognition of melting range, Link to mixture
(b) Draw a pencil line near the bottom of the paper. Add a spot of the sample on the line. Suspend the paper in solvent, making sure the spot is above the solvent and not touching it. Allow the solvent to rise, then remove paper once near the top. Mark the solvent front with pencil. · must show pencil line, spotting sample, placing paper in solvent, removing and marking solvent front
(c) Sample X contains Dye A and Dye B, as the spots are at the same heights (2.0 cm and 6.5 cm) as these dyes in the chromatogram. · must show Match spots by distance, Both dyes identified
(d) 0.81 · must show Rf = 6.5 / 8.0
Common mistakes: Forgetting to compare both start and end of melting for purity; Mixing up solvent front and spot distance for R_f; Incorrectly assigning dyes (using vertical, not distance, on chromatogram)
Show mark scheme and worked solution
Marking points
M1 (a) · 1 mark · States/infers the melting occurs over a range (not sharp).
A1 (a) · 1 mark · Correctly concludes mixture (not pure) because melting is not single temperature.
M2 (b) · 1 mark · Describes a key step: drawing pencil line/spotting sample/placing paper in solvent.
A2 (b) · 1 mark · Mentions marking the solvent front at the end or stopping near the top.
M3 (c) · 1 mark · Correctly identifies both dyes in sample X, with suitable reasoning (matches heights/distances on chromatogram).
M4 (d) · 1 mark · Calculates Rf for dye matching Dye B using R_f = distance moved by spot / distance moved by solvent front; value correct to two decimal places (0.81).
Worked solution
a) The melting starts at 87°C and finishes at 97°C, so melting happens over a range of 10°C. This indicates the sample is a mixture. Pure substances melt at a single sharp temperature.
b) Steps for paper chromatography:
1. Draw a line in pencil 1 cm from the bottom of the chromatography paper.
2. Place a spot of dissolved sample X on the line and let it dry.
3. Suspend the paper in a beaker with a small amount of water (solvent), ensuring the spot is above the level of the solvent.
4. Let the solvent rise near to the top of the paper, then remove the paper and mark the solvent front in pencil.
c) The chromatogram shows two spots for sample X: one at 2.0 cm and one at 6.5 cm. Dye A has a spot at 2.0 cm and Dye B at 6.5 cm (Dye C is at 5.8 cm). Therefore, sample X contains Dye A and Dye B (their spots line up).
For the spot at 6.5 cm (same as Dye B):