The following data are for a liquid chromatographic column:
Length of packing (L) 24.7 cm
Flow rate 0.313 mL/min
VM 1.37 mL
VS 0.164 mL
A chromatogram of a mixture of species A, B, C and D provided the following data
Retention time
Nonretained 3.1 -
A = 5.4
B = 13.3
C = 14.1
D = 21.6
Width of peak
A = 0.41
B= 1.07
C= 1.16
D= 1.72
Calculate the Retention Factor for A,B,C and D.
I've done that...
A= 0.74
B= 3.29
C= 5.96
D= 4.44
Calculate the Distribution constant for A,B,C and D? (having trouble here)
And Calculate the following..
(a) the number of plates from each peak and the mean.
Number of plates = 16*tr^2/w^2
a = 2775.49
b = 2472.04
c = 2363.97
d = 2523.30
mean = 2533.7
(b) the plate height for the column. (having lots of trouble here)
Calculate for species B and C,
The selectivity factor = tr'2/tr'1
which = 3.54/3.29
= 1.07
the length of column necessary to separate the two species with a resolution of 1.5. ( I cant do this one please help)
Thanks, to whoever helps me out with this, I am having so much trouble
3 answers
k=K/Beta
Where
k=0.74
Beta=VM/VS=1.37 mL/0.164 mL
and
K=???
Solve for K:
K=0.74*(1.37 mL/0.164 mL)
K=0.74*8.354
K=6.18
For the number of plates:
Use the following equation:
N=16*(Tr/W)^2
Where
Tr=5.4
W=0.41
and
N=???
Solve for N:
N=16*(5.4/0.42)^2
N=2.64 x 10^3
*** I'll let you solve for the mean.
For the plate height, use the following equation:
N=L/H
Where
N=2.64 x 10^3
L=24.7 cm
and
H=??
Solve for H:
L/N=H
24.7 cm/2.64 x 10^3=H
H=0.00934=9.34 x 10^-3
For the selectivity factor, it looks good, but we have a difference in the last significant figure:
Use the equation below and solve for alpha:
Alpha=tr2/tr1
Where
tr'2=14.1
and
tr'1=13.5
Alpha=14.1/13.5=1.04