HPLC: A Study of Some Basic Variables

Caffeine as an Example

You will be provided with a caffeine stock solution, this compound can be present in several types of samples. The idea of this experiment is to be able to assess and evaluate how changing some chromatographic parameters can reflect on retention time, separation efficiency and other important factors.

Caffeine will be chromatographed and detected at 270 nm, using the conditions mentioned below.

Procedure

  1. Prepare a 20 ppm solution of caffeine from the stock standard solution provided.
  2. Condition your column usingthe mobile phase consisting of 30:70 methanol/ water at 1 mL/min for 10 min or until you get a stable baseline.
  3. Follow instructions of operation of the Agilent 1200 HPLC system and the Chemstation software.
  4. Inject 20 L of caffeineand Collect your chromatogram at a flow rate of 1mL/min, and record the different chromatographic parameters for the compound.

Compound / Peak Area / Peak Height / Ret. Time / K' / W1/2 / N
Caffeine

5. Flush your column at a flow rate of 2 mL/min, till you obtain a stable baseline, then inject 20 L of caffeine and Collect your chromatogram at a flow rate of 2 mL/min, record the different chromatographic parameters.

Compound / Peak Area / Peak Height / Ret. Time / K' / W1/2 / N
Caffeine
  1. Design a flow gradient program as the following:

0min at 1mL/min

1 min at 1 mL/min

4 min at 3 mL/min

5 min at 3 mL/min

6minat 1 mL/min

  1. Collect a chromatogram of a blank injection, and record your chromatogram as your blank.
  2. Inject 20 L of caffeine and Collect your chromatogram using the flow gradient program, and record the different chromatographic parameters.

Compound / Peak Area / Peak Height / Ret. Time / K' / W1/2 / N
Caffeine

7Subtract the blank chromatogram from the chromatogram of caffeine to get your corrected baseline chromatogram

8Repeat steps 3-5 but using a mobile phase composition of 50:50 (methanol water). Record your chromatograms and the different parameters as above:

Compound / Peak Area / Peak Height / Ret. Time / K' / W1/2 / N
Caffeine

10 . Flush your column at a flow rate of 1 mL/min using a 30:70 methanol/water mobile phase, till you obtain a stable baseline.

  1. Design a mobile phase (methanol/water) gradient elution program, at 1 mL/min, as the following:

0 min at 30:70

1min at 30:70

4 min at 70:30

5 min at 40:60

6 min at 50:50

  1. Collect a chromatogram of a blank injection, and record your chromatogram as your blank.
  2. Inject 20 L of caffeine and Collect your chromatogram using the mobile phase gradient program, and record the different chromatographic parameters.

Compound / Peak Area / Peak Height / Ret. Time / K' / W1/2 / N
Caffeine
  1. Subtract the blank chromatogram from the chromatogram of caffeine to get your corrected baseline chromatogram.
  2. Flush your column with a 50:50 methanol water mobile phase for 10 min.
  3. From the software, turn the pump and DAD off, exit the software, and finally switch off the power of the pump and detector.

Calculations:

N = 5.54(tr/W1/2)2 where N is the efficiency of separation, tr is the retention time in min. and W1/2 is the peak width at half height.

To calculate K', use the relationP'AB = AP'A + BP'B to calculate the polarity index of your mobile phase

Where, P'AB is the polarity index of the mobile phase containing A and B are volume fractions of solvents A and B while P'A and P’B are the polarity indices of pure solvents A and B.

log k'2/k'1 = (P'2 – P'1)/2

Where, P’1 and P'2 are the initial and final polarity indices of the mobile phase that will result in bringing the value of the retention factor from k'1 to k'2.

K' = (tr – to)/to

Example

A solute having a retention time of 31.3 min is separated using and mobile phase composition of 30%methanol, 70% water. Find k', if the mobile phase composition that can bring tr to 2.88 min was 73:27 methanol water.

P'AB = AP'A + BP'B

Values of P' can be obtained from tables of polarity indices:

P'AB = 0.30*5.1 + 0.70*10.2 = 8.7

log k'2/k'1 = (P'2 – P'1)/2

log k'2/k'1 = (6.5 – 8.7)/2

log k'2/k'1 = -1.1

log (tr2 – to)/(tr1 – to) = -1.1

log (2.88 – to)/(31.3 – to) = -1.1

(2.88 – to)/(31.3 – to) = 10-1.1

(2.88 – to)/(31.3 – to) = 0.0794

To = 0.43 min

Now you can calculate k'

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