A student researcher performed a chromatographic separation of caffeine and aspa

Question

A student researcher performed a chromatographic separation of caffeine and aspartame with methanol as the mobile phase using a C - 18 column. The retention times for methanol (t_m), caffeine (t_c), and aspartame (t_a) were 28.8 s, 94.6 s, and 164.5 s, respectively. A) What is the selectivity factor (relative retention), alpha, for this column? alpha = B) Calculate the hypothetical retention times for caffeine and aspartame if their retention factors were reduced by 25.0%. Assume the mobile phase retention time (t_m = 28.8 s) remains constant. t_caffeine = t_aspartame = Choose the statements that correctly describe the outcome if the partition coefficient for caffeine (K_c) were to decrease The retention factor of caffeine would increase. The retention time of caffeine would decrease. The retention time of caffeine would increase. The retention factor of caffeine would decrease. Caffeine would spend a greater amount of time in the mobile phase. The retention time and retention factor of caffeine would not change.

Explanation / Answer

A)

selective factor ALPHA:

t-methanol = 28.8 s

t-caffeine= 94.6 s

t-aspartame= 164.5 s

t-caff = 94.6-28.8 = 65.8

t-asp = 164.5 -28.8 = 135.7

find selectivity via

alpha = t-aspartame/t-caffeine = 135.7/65.8 = 2.06

alpha = 2.06

B)

find retention factor

k = (tr-tm)/tm

for caffene

k = (tr-tm)/tm

t-retention = t-caffeine= 94.6 s

t-methanol = 28.8

sp

k = (94.6 -28.8)/(28.8) = 2.28

now, for 25% reduction

k = 2.28 - 2.28*0.25 = 1.71

recalcualte

1.71 = (Tr-28.8)/(28.8)

t = 1.71 *28.8 + 28.8 = 78 sec

for aspartame:

K = (164.5 -28.8) / 28.8 = 4.71

then

k = 4.71-4.71*0.25 = 3.53

3.53 = (t-new-28.8)/28.8

tnew = 3.53*28.8 + 28.8 = 130.46

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