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 | Width of Base
Nonretained 3.1 -
A 5.4 0.41
B 13.3 1.07
C 14.1 1.16
D 21.6 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
UPDATE sorry the top section with the data didn't come out properly heres the data again.
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
Answered this for you in your subsequent post.
To calculate the distribution constant for species A, B, C, and D, you need to use the following formula:
Distribution constant (K) = (Retention time - Nonretained time) / Nonretained time
For species A:
K(A) = (5.4 - 3.1) / 3.1
K(A) = 0.74 (as you have already calculated)
For species B:
K(B) = (13.3 - 3.1) / 3.1
K(B) = 3.29 (as you have already calculated)
For species C:
K(C) = (14.1 - 3.1) / 3.1
K(C) = 5.61 / 3.1
K(C) = 1.81
For species D:
K(D) = (21.6 - 3.1) / 3.1
K(D) = 18.5 / 3.1
K(D) = 5.97
Next, to calculate the plate height for the column, you can use the following formula:
Plate Height (H) = Length of packing (L) / Number of theoretical plates (N)
For species B:
H(B) = 24.7 / 2472.04
H(B) ≈ 0.01 cm
For species C:
H(C) = 24.7 / 2363.97
H(C) ≈ 0.01 cm
To calculate the length of column necessary to separate species B and C with a resolution of 1.5, you can use the following formula:
Length of column = (Resolution)² * (Width of base) / (Number of plates)
Given that the resolution (R) is 1.5, and the width of base for species B is 1.07 and for species C is 1.16, the calculation will be:
Length of column = 1.5² * 1.16 / (2472.04 - 2363.97)
Length of column ≈ 0.02 cm
I hope this helps! Let me know if you have any further questions.
To calculate the distribution constant for species A, B, C, and D, you need to use the equation:
Distribution Constant (K) = (Retention Factor (k) × (Flow Rate (F) × VM) / (L × VS))
For species A:
K_A = (0.74 × (0.313 mL/min × 1.37 mL) / (24.7 cm × 0.164 mL))
For species B:
K_B = (3.29 × (0.313 mL/min × 1.37 mL) / (24.7 cm × 0.164 mL))
For species C:
K_C = (5.96 × (0.313 mL/min × 1.37 mL) / (24.7 cm × 0.164 mL))
For species D:
K_D = (4.44 × (0.313 mL/min × 1.37 mL) / (24.7 cm × 0.164 mL))
Calculate each of these values to find the distribution constants for each species.
Next, to calculate the plate height for the column, you can use the equation:
Plate Height (H) = L / N
Where N is the number of theoretical plates in the column. The number of plates for each peak can be calculated using the equation:
Number of Plates (N) = 16 × (Retention Time (t_r))^2 / (Width of Base (w))^2
For species B and C, you have already calculated the number of plates (a) = 2472.04 and (c) = 2363.97, respectively. Substitute these values into the equation to calculate the plate heights for species B and C.
Finally, to calculate the length of column necessary to separate two species with a resolution (R) of 1.5, you can use the equation:
Length of Column (L) = 16 × ((R + 1)^2 / (R - 1)^2) × ((Retention Time of Later Compound (t_r2) - Retention Time of Earlier Compound (t_r1))^2 / (Width of Base Average (w_avg))^2)
Substitute the values into the equation, where R = 1.5, t_r2 and t_r1 are the retention times of the two species, and w_avg is the average width of the base for the two species.
Use these calculations to find the answers to the remaining questions.