Chemistry • Year 12 • Module 8 • Lesson 5
Chromatography: TLC, Column & HPLC
Apply the principles of Rf calculation, chromatographic separation and HPLC interpretation to real analytical data from Australian industry contexts.
1. Interpret TLC data — Australian wine grape pigment analysis
The Australian Wine Research Institute (AWRI) uses chromatographic methods to analyse anthocyanin pigments in red wine grapes. A chemist performs TLC on extracts from three grape varieties using the same silica plate and the same organic solvent. The table below records the distances measured from the baseline. 8 marks
| Grape variety / standard | Compound spot distance (cm) | Solvent front distance (cm) | Rf value |
|---|---|---|---|
| Anthocyanin standard (cyanidin-3-glucoside) | 3.6 | 6.0 | — |
| Shiraz extract — spot 1 | 3.6 | 6.0 | — |
| Shiraz extract — spot 2 | 5.1 | 6.0 | — |
| Cabernet Sauvignon extract — spot 1 | 3.6 | 6.0 | — |
| Riesling extract — single spot | 2.4 | 6.0 | — |
1.1 Calculate the Rf value for each sample and the standard. Show working for the standard. 3 marks
1.2 Identify which grape variety (or varieties) contain cyanidin-3-glucoside and justify your answer using Rf values. 2 marks
1.3 The Shiraz extract shows two spots. What does this indicate about the composition of the Shiraz anthocyanin extract? 1 mark
1.4 The Riesling has a lower Rf value (Rf = 0.40) than the anthocyanin standard (Rf = 0.60). What does this tell you about the relative affinity of the Riesling compound for the stationary and mobile phases? 2 marks
2. Interpret an Rf comparison bar chart
The bar chart below shows the Rf values of six individual compounds separated on the same TLC plate under identical conditions. Use the chart to answer the questions that follow. 6 marks
Figure 2.1. Rf values for compounds A–F separated on silica TLC with the same organic solvent mobile phase. Data illustrative; based on principles from AWRI pigment analysis methods.
2.1 Which compound has the strongest affinity for the stationary phase? Explain how you can tell from the chart. 2 marks
2.2 If compounds D and E were the only two components in a mixture, would TLC be likely to resolve them into two distinct spots? Justify your answer using the Rf values. 2 marks
2.3 Explain why Rf values have no units. 2 marks
3. Cause-and-effect chain — why HPLC detects paracetamol impurities
Complete each effect box by explaining what logically follows from the cause. The final box is the overall outcome. 5 marks
| Cause | → | Effect (your answer) |
|---|---|---|
| A paracetamol tablet contains a small amount of an impurity alongside the active ingredient. | → | |
| The impurity has different polarity (and therefore different affinity for the HPLC column) compared to paracetamol. | → | |
| The impurity exits the HPLC column at a different retention time from paracetamol. | → | |
| The peak area of the impurity peak is proportional to its concentration. | → | |
| Overall outcome (so…): | ||
4. Case study — PFAS contamination at an Australian Defence Force base
In 2016 and again in 2021, the NSW EPA reported PFAS (per- and polyfluoroalkyl substances) contamination in groundwater near the RAAF Base Williamtown, New South Wales, linked to the use of aqueous film-forming foam (AFFF) in firefighting training. Environmental scientists needed to quantify multiple PFAS compounds in water samples at very low concentrations (parts per trillion). 6 marks
4.1 Explain why HPLC, rather than TLC or gravity column chromatography, is the most appropriate technique for detecting PFAS at parts-per-trillion concentrations in groundwater samples. In your answer, refer to sensitivity and quantitative capability. 3 marks
4.2 The HPLC analysis returns a chromatogram with 14 distinct peaks. What does the presence of 14 peaks tell a scientist about the PFAS contamination in this water sample? 2 marks
4.3 Scientists run a certified PFAS reference standard alongside the contaminated samples. Explain how the retention times from the standard run help them identify the specific PFAS compounds present. 1 mark
5. Compare and contrast — three chromatographic techniques
Complete the comparison table below. The first row is done as an example. 8 marks
| Feature | TLC | Column chromatography | HPLC |
|---|---|---|---|
| Stationary phase | Silica or alumina on a flat plate | Silica or alumina packed in a glass column | Fine-particle silica in a high-pressure steel column |
| Driving force for mobile phase | |||
| Speed | |||
| Sensitivity | |||
| Can collect separated fractions? | |||
| Quantitative? | |||
| Best used for… | |||
| Australian application example |
Q1.1 — Rf calculations
Standard: Rf = 3.6 ÷ 6.0 = 0.60. Shiraz spot 1: 3.6 ÷ 6.0 = 0.60. Shiraz spot 2: 5.1 ÷ 6.0 = 0.85. Cabernet Sauvignon spot 1: 3.6 ÷ 6.0 = 0.60. Riesling: 2.4 ÷ 6.0 = 0.40.
Q1.2 — Identification using Rf
Both Shiraz (spot 1) and Cabernet Sauvignon (spot 1) have Rf = 0.60, which matches the anthocyanin standard (cyanidin-3-glucoside, Rf = 0.60) under the same TLC conditions. This supports the identification of cyanidin-3-glucoside in both varieties. The Riesling compound (Rf = 0.40) does not match and is a different compound.
Q1.3 — Two spots in Shiraz
Two spots indicate the Shiraz extract contains at least two different anthocyanin (or pigment) components, not just one pure compound. It is a mixture.
Q1.4 — Lower Rf for Riesling compound
A lower Rf (0.40 vs 0.60) means the Riesling compound travels a shorter distance up the plate. This indicates it has a stronger affinity for the stationary phase (silica) and/or a weaker affinity for the mobile phase, so it is retained more by the plate and moves less with the solvent.
Q2.1 — Strongest affinity for stationary phase
Compound A (Rf = 0.12) has the strongest affinity for the stationary phase. The lowest Rf value means it travelled the shortest distance relative to the solvent front, indicating the compound spent the most time interacting with (adsorbed to) the stationary phase and the least time in the mobile phase.
Q2.2 — Resolution of compounds D and E
Yes, TLC should resolve them into two distinct spots. Compound D has Rf = 0.62 and compound E has Rf = 0.78, a difference of 0.16. This is a sufficient difference in affinity for the stationary phase to produce two separate, non-overlapping spots on the same plate under the same conditions.
Q2.3 — Why Rf has no units
Rf is a ratio of two lengths (both in centimetres): distance the compound travels ÷ distance the solvent front travels. When you divide a length by a length, the units cancel out, leaving a dimensionless number between 0 and 1.
Q3 — Cause-and-effect chain answers
Row 1: The injection of the tablet extract onto the HPLC column introduces both paracetamol and the impurity into the system simultaneously.
Row 2: The impurity will travel through the column at a different rate to paracetamol, because it interacts differently with the stationary phase.
Row 3: The detector records the impurity as a separate peak at a different time from the paracetamol peak, making it visible on the chromatogram.
Row 4: The chemist can compare the impurity peak area with a calibration standard to quantify exactly how much impurity is present.
Overall outcome: HPLC can detect and quantify even trace levels of impurity in a paracetamol tablet, providing rigorous quality-control evidence before the product is released for sale.
Q4.1 — Why HPLC for PFAS at ppt concentrations
HPLC is appropriate because it has high sensitivity—its instrumental detection (e.g. mass spectrometry or UV detection) can identify compounds present at parts-per-trillion concentrations that would be too low to detect visually on a TLC plate. It is also quantitative: peak area is proportional to concentration, allowing precise measurement of PFAS levels required to compare with regulatory guidelines. TLC and gravity column chromatography lack the sensitivity and quantitative precision needed for trace environmental analysis at this level.
Q4.2 — Fourteen peaks in PFAS chromatogram
Fourteen distinct peaks indicate that the water sample contains a mixture of at least 14 different PFAS compounds (each with a different retention time on the column). Each peak represents one component eluting at a characteristic time, confirming that the contamination is complex and involves multiple individual PFAS molecules, not a single substance.
Q4.3 — Role of reference standard retention times
Under identical HPLC conditions, each pure certified PFAS compound produces a peak at a known, characteristic retention time. By matching the retention times of peaks in the contaminated sample to those of the reference standard, scientists can identify which specific PFAS compounds are present in the groundwater.
Q5 — Comparison table completed
| Feature | TLC | Column chromatography | HPLC |
|---|---|---|---|
| Driving force | Capillary action | Gravity | High pressure (pump) |
| Speed | Fast (minutes) | Slow (hours) | Fast and automated (minutes) |
| Sensitivity | Moderate | Moderate | High |
| Collect fractions? | No | Yes | Yes (with fraction collector) |
| Quantitative? | Semi-quantitative at best | No | Yes (peak area vs concentration) |
| Best used for… | Quick purity check; comparing unknowns to standards | Separating and collecting preparative quantities | High-precision analytical identification and quantification |
| Australian application | TGA pharmaceutical purity spot-check | Synthetic chemistry lab purification | AWRI wine anthocyanin analysis; PFAS environmental monitoring (NSW EPA) |