2. The following data apply to a column for liquid chromatography: Length Flow R
ID: 711862 • Letter: 2
Question
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.Explanation / Answer
Answer:
Number of plates form each peak:
N = 16{tR/W}2
Mean
(2775.49 + 2472.04 + 2363.97 + 2523.31)/4
Mean = 2533.7025 (we denote it as ‘x’)
Standard deviation:
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
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 (tRC - tRB)/ (WC + WB)
R = 2 (14.1 – 13.3)/ (1.16 + 1.07)
R = 0.72
The retention factor for B and C are
K’ = (tR – tM) / tM
kB’ = (13.3 – 3.1)/ 3.1 = 3.3
kC’ = (14.1 – 3.1)/ 3.1 = 3.5
The selectivity factor = kC’/ kB
= 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
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