Mr Cole Chemistry

Gas Chromatography Questions

Gas Chromatography

Organic Analysis Worksheet

Fundamentals & Setup

1. State the definition of the term retention time.
Answer

Retention time is the length of time it takes for the compound/sample to reach the end of the coiled tube (column) and be detected.

2. A carrier compound is used in the process of gas chromatography. Suggest a suitable compound for this process.
Answer

Inert Gases are suitable, such as Helium (He), Argon (Ar), or Nitrogen (N2).

3. A sample thought to contain aldehydes and ketones is being analysed. Two stationary phases are being considered: propanol coated beads and hexane coated beads. Which one is the better choice of stationary phase, and why?
Answer

Propanol is the better choice.

Reason: Aldehydes and ketones are polar molecules. To separate them effectively, the stationary phase should have a similar polarity (“like dissolves like”) so that the components interact/dissolve in the stationary phase. Hexane is non-polar and would interact too weakly with the polar analytes.

Chromatogram Analysis

4. An aldehyde, an alkane and a carboxylic acid, all of similar volatility, are mixed together. The mixture is then analysed in a gas chromatograph where the stationary phase was a liquid alcohol. The gas chromatogram produced is shown. State which peak is which compound and explain how you know that. Gas Chromatogram peaks X Y Z
Answer
  • Peak X (Alkane): Has the shortest retention time (least soluble in the stationary phase). Alkanes are non-polar and cannot form hydrogen bonds; they interact weakly via Van der Waals forces with the polar alcohol stationary phase.
  • Peak Y (Aldehyde): Has the second longest retention time. Aldehydes are polar and can accept hydrogen bonds from the alcohol stationary phase, making them more soluble than alkanes but less than carboxylic acids.
  • Peak Z (Carboxylic Acid): Has the longest retention time (most soluble). Carboxylic acids can form strong hydrogen bonds (both donors and acceptors) with the liquid alcohol stationary phase, slowing their passage through the column.

Quantitative & Structural Analysis

5. A cosmetic product containing four esters, J, K, L and M, is analysed by gas chromatography and mass spectrometry.

The chromatogram is shown below (J, K, L, M peaks from left to right).

Chromatogram of J K L M

The numbers by the peaks are the relative molar proportions of the compounds in the mixture.

Relative molar proportions table
a) The concentration of ester K in the cosmetic product is 9.13 × 10−2 g dm−3. Using the results, calculate the concentration, in mol dm−3, of ester M in the cosmetic product. Give your answer to two significant figures.
Answer

1. Calculate concentration of K in mol dm-3:
Assuming Mr of K is 166 (derived from structural data usually provided in context):
9.13 × 10-2 g dm-3 ÷ 166 g mol-1 = 5.50 × 10-4 mol dm-3

2. Determine Molar Ratio (M : K):
From table: M = 5.9, K = 4.3
Ratio = 5.9 ÷ 4.3 = 1.372…

3. Calculate concentration of M:
5.50 × 10-4 × 1.372… = 7.5 × 10-4 mol dm-3 (2 s.f.)

b) A general structure for esters J, L and M is shown below.
General ester structure
Use the mass spectrometry results to deduce possible structures for esters J, L and M.
Answer

Ester J: Based on the mass spectrum (lowest mass), J corresponds to:

Structure of Ester J

Esters L and M: These have the same Mr, so they are isomers. The two possible structures are:

Structures of isomers L and M

Identification based on retention time:

  • L (Shorter Retention): This corresponds to the Branched Isomer. Branching reduces the surface area for Van der Waals forces, lowering the boiling point (increasing volatility). Higher volatility leads to a shorter retention time.
  • M (Longer Retention): This corresponds to the Straight Chain Isomer. The straight chain allows for better packing and stronger Van der Waals forces, resulting in a higher boiling point (lower volatility) and thus a longer retention time.
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