Question

5. (13 pts) Use Figure below to do the following: Flow rate: 0.37 ml./min Pressure: 655 psi Column: 4.6 mm i d.) x 250 mm -Inject 0.5 6.5 8.5 40.5 46.5 48.5 42.5 44.5 Time, min 505 52.5 a. Calculate the elution times of the two peaks. b. Calculate the capacity factors of the two peaks.

Answer 1

​​​​​​Given : flow rate = 0.37mL/min.

Pressure= 655 psi

Column = 4.6mm(I.d)×250mm

a) Elution time : The elution time of a solute is the between the start of separation (i.e., the time st which the solute enters the column) and the time at which the solute elutes . Elution time is equal to the retention time in Gas Chromatography.

So the elution time for the above two peaks :

The formula is -

tR = tR' + tm

where tR = Retention or elution time.

tm= time required to elute non retained solute or Mobile phase.

tR' =  time spent between the non retained solute and the retained solute.

For first peak ,

tR ' = 44.5 - 7.4 = 37.1 min.

tm = 7.4min.

So , tR = 37.1+ 7.4 = 44.5 min.

So the elution time for the first peak is 44.5 minutes.

For second peak ,

tR = 43.1+ 7.4= 50.5 min.

So the elution time for the second peak is 50.5 minutes.

b) Capacity factor is the measure of the retention of the peak that is independent of the column geometry or Mobile phase flow rate.It is symbolised by k . It is calculated as :

k = (tR - tm)/tm

For first peak ,

k =( 44.5 - 7.4)/7.4

k = 5.01 is the capacity factor for the first peak.

For second peak,

k =(50.5-7.4)/7.4

k =5.82 is he capacity factor for the second peak.

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