Year 12 Chemistry Chapter 13

The mass of caffeine in a particular coffee drink was determined by high-performance liquid chromatography (HPLC).

The calibration curve produced from running standard solutions of caffeine through an HPLC column is shown to the left.

A 5.0 mL aliquot of the coffee drink was diluted to 50.0 mL with de-ionised water. A sample of the diluted coffee drink was run through the HPLC column under identical conditions to those used to obtain the calibration curve.

The peak area obtained for this diluted sample was 2400 arbitrary units.

The HPLC column used has a non-polar stationary phase.

The most suitable solvent for determining the concentration of caffeine in the sample is

• carbon tetrachloride, CCl₄

• methanol, CH₃OH

• octanol, C₈H₁₇OH

• hexane, C₆H₁₄

The mass of caffeine in a particular coffee drink was determined by high-performance liquid chromatography (HPLC).

The calibration curve produced from running standard solutions of caffeine through an HPLC column is shown to the left.

A 5.0 mL aliquot of the coffee drink was diluted to 50.0 mL with de-ionised water. A sample of the diluted coffee drink was run through the HPLC column under identical conditions to those used to obtain the calibration curve.

The peak area obtained for this diluted sample was 2400 arbitrary units.

The HPLC column used has a non-polar stationary phase.

The mass of caffeine, in grams, in 350 mL of the undiluted coffee drink is closest to

• 0.014

• 0.070

• 0.14

• 0.40

Consider the following changes that could be applied to the operating parameters for a chromatogram set up to carry out high-performance liquid chromatography (HPLC) with a polar stationary phase and a non-polar mobile phase:

I decreasing the viscosity of the mobile phase

II using a more tightly packed stationary phase

III using a mobile phase that is more polar than the stationary phase

Which of the changes would be most likely to reduce the retention time of a sugar in the HPLC?

• I only

• I and III only

• III only

• II and III only

A mixture of organic compounds was analysed using high-performance liquid chromatography (HPLC) and produced the result shown in the diagram to the left.

How were the peaks identified?

• A standard database with retention times of all organic compounds was consulted.

• The heights of the peaks indicated the molecular structure of the organic compounds.

• Known compounds were analysed on the same column under identical conditions.

• The distances between the peaks revealed which organic molecules were present.

A mixture of organic compounds was analysed using high-performance liquid chromatography (HPLC) and produced the result shown in the diagram to the left.

Which one of the following statements is incorrect?

• Ethanoic acid has the strongest attraction to the mobile phase.

• Butanol has the weakest attraction to the stationary phase.

• In the sample analysed, butanol was in the highest concentration.

• Using a higher pressure would result in all retentions times being shorter.

A mixture of propan-1-ol and butan-1-ol is injected into an HPLC column that is using a polar solvent. The print-out shown to the left is obtained.

A second sample is passed through the same column and produces only one peak with a retention time of 9.4 minutes and an area of 1200 units.

The second sample contains

• propan-1-ol and butan-1-ol at lower concentrations than the original solution.

• propan-1-ol only, with a concentration that is one-third of the original solution.

• butan-1-ol only, with a concentration that is one-third of the original solution.

• butan-1-ol only, with a concentration that is three times that of the original solution.

A sample of petrol is injected into a high-performance liquid chromatography (HPLC) instrument. The stationary phase used is polar and the mobile phase is non-polar. The chromatogram shown to the left was obtained.

Select the correct conclusion from the following alternatives.

• There are three components only in the sample.

• Molecule A is likely to be the hydrocarbon of lowest molecular mass.

• Molecule C is likely to be the hydrocarbon of lowest molecular mass.

• The concentration of each component is similar.

The chromatogram shown to the leaft was obtained by analysing a sample containing several amino acids.

From this chromatogram, it would be reasonable to conclude that:

• the solvent is polar

• arginine is least strongly adsorbed onto the stationary phase

• the molecule with the shortest retention time is the least polar

• cysteine must be the most abundant amino acid in the sample

HPLC can be used for

• qualitative analysis only.

• quantitative analysis only.

• neither qualitative nor quantitative analyses

• both qualitative and quantitative analyses