Question

The following conditions were established for a chromatographic separation of multiple hydrocarbons on a packed column with 1.1 g of liquid stationary phase: inlet pressure 826 torr, room pressure 756 torr, vapor pressure of water at room temperature 17.5 torr; column temperature 110ºC, room temperature 20°C; measured outlet flow rate 20.0 mL/min; retention time for nonretained species 25.3 s, retention time of the first eluting species 1.56 min. Based on these conditions, the pressure drop correction factor j is _______, the average flow rateF in mL/min is __________, and the specific retention volume for the first eluting species Vg in mL/g is _________. Report all values to 3 significant figures.

Answer 1

The pressure drop correction factor j is given by the following expression:

j= 3[p_i-p_o)²-1] / 2[p_i-p_o)³-1] where P_i is pressure of the carrier gas at inlet and P_o is pressure of the carrier gas at outlet of the comumn.

Substituting the values provide we can get,

j= 3[(826/756)²-1] / 2[(826/756)³-1] = 0.9551

Flow rate F is the volume of the mobile phase passing throgh the column at unit time. In the given data we got F=20 ml/min.

The mobile phase flow rate at ambient temperature Fa = F x (1-p_w/p_a) where p_w is the partial pressure of the water vapour and p_a is room pressure.

Hence Fa = 20 x [1-17.5/756)] = 19.537 ml/min

mobile phase flow rate at column temperature Fc = Fa x(Tc/Ta)

where Tc is column temp= 110 deg centigrade= 110+273=383 K

Ta= Ambient temperature=20 deg centigrade= 20+273=293 K

Fc=19.537 x (383/293) = 25.538 ml/min

The specific retention volume is given by the following equation:

Vg =(273 x Vn) / (W x Tc )

Where Vn = net retention volume, W is the mass of the stationary phase

Vn =J Vr^' =j( Vr-Vm) where Vr^' = adjusted retention volume, Vr = Total retention volume, Vm= hold up volume

we can calculate, Vr= Fc x t_r where t_r is total retention time =1.56 min

Vr= 25.538 x 1.56 = 39.83928 ml

similarly, Vm= Fc x t_m where t_m is retention time of the nonretained species = 25.3 s = 0.4216 min

Vm= 25.538 x 0.4216 = 10.7668 ml

Now Vn = j( Vr-Vm) = 0.9551 x ( 39.83928- 10.7668) = 27.7671 ml

Substituting all values we get, Vg =(273 x Vn) / (W x Tc )

= (273 x 27.7671) / (1.1 x 383) = 17.9929 ml/g