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 (tm), caffeine (tc), and aspartame (ta) were 42.8 s, 93.6 s, and 170.8 s, respectively. B) Calculate the hypothetical retention times for caffeine and aspartame if their retention factors were reduced by 30.0%. Assume the mobile phase retention time (tm = 42.8 s) remains constant.

Answer 1

Given data,

t_m = 42.8 sec

t_c = 93.6 sec

t_a = 170.8 sec

Let us consider retention factor,

K = ( t_r - t_m) / t_m where t_m is retention time for methanol , t_c is the retention time for caffeine ,

= (170.8 - 42.8) / ( 93.6 - 42.8 )

= 128 / 50.8

K = 2.51

Let the hypothetical retention time for aspartame be ta2

2.51 x 0.72 = ( ta2 - 42.8 ) / 42.8

77.34 = ( ta2 - 42.8 )

ta₂ = 120.14

K for caffeine = ( 93.6 - 42.8) / 42.8

= 1.18

Now K is reduced by 30%.

Let the hypothetical retention time for caffeine be tc₂

So, 1.18 * 0.72 = (tc₂ - 42.8 ) / 42.8

36.36 = (tc₂ - 42.8 )

tc₂ = 79.16 sec