2 The following data apply to a column for liquid chromatogr

2. The following data apply to a column for liquid chromatography: Length Flow Rate VM 24.7 cm 0.313 mlmin 1.37 ml 0.164 ml A chromatogram of a mixture of species A, B, C, and D yielded the following data: Peak width at base min) Retention Time (min) 3.1 5.4 13.3 14.1 21.6 Nonretained 0.41 1.07 1.16 1.72 Calculate: a. The number of theoretical plates from each peak b. The mean and the standard deviation for N. c. The plate height for the column d. The capacity factor for each component. c. The partition coefficient for each component f The resolution for B and C g The relative retention for B and C. h. The length of cokumn required to separate B and C with a resohution of 1.5.

Solution

Answer:

Number of plates form each peak:

N = 16{t_R/W}²

Mean

(2775.49 + 2472.04 + 2363.97 + 2523.31)/4

Mean = 2533.7025 (we denote it as ‘x’)

Standard deviation:

x_i

x_i - x

(x_i – x)²

2775.49

241.7875

58461.1951

2472.04

61.6625

3802.2639

2363.97

169.7325

28809.1215

2523.31

10.3925

108.004

(x_i – x)² = 91180.5845

Standard deviation = (91180.5845/2)^1/2

                                                  = ± 213

Plate height for column is L/N

= 24.7/2533.7025

= 0.009748 cm

Resolution for B and C

2 (t_RC - t_RB)/ (W_C + W_B)

R = 2 (14.1 – 13.3)/ (1.16 + 1.07)

R = 0.72

The retention factor for B and C are

K’ = (t_R – t_M) / t_M

k_B’ = (13.3 – 3.1)/ 3.1 = 3.3

k_C’ = (14.1 – 3.1)/ 3.1 = 3.5

The selectivity factor = k_C’/ k_B

= 3.5/3.3 = 1.06

xi	xi - x	(xi – x)2
2775.49	241.7875	58461.1951
2472.04	61.6625	3802.2639
2363.97	169.7325	28809.1215
2523.31	10.3925	108.004
		(xi – x)2 = 91180.5845