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 41.8 s , 96.3 s , and 143.8 s , respectively.

What is the selectivity factor (relative retention), α , for this column?

α=

Calculate the hypothetical retention times for caffeine and aspartame if their retention factors were reduced by 28.0% . Assume the mobile phase retention time (tm=41.8 s) remains constant.

tcaffeine=

staspartame=

Select the statements that correctly describe the outcome if the partition coefficient for caffeine (Kc) were to decrease.

The retention time of caffeine would increase.

The retention time of caffeine would decrease.

Caffeine would spend a greater amount of time in the mobile phase.

The retention factor of caffeine would decrease.

The retention factor of caffeine would increase.

The retention time and retention factor of caffeine would not change.

Answer 1

Solution:

What is the selectivity factor (relative retention), α , for this column?

Selectivity factor [α]

The ability of chromatographic system to chemically distinguish between two components of the sample.

It can be determined by following formula

Selectivity factor [α] = [tR2 – t0]/ [tR1 – t0]

Where, tR2 = Retention time of component 2; tR1 = Retention time of component 1 and

t0 = A non-retained compound has no affinity for the stationary phase and elutes with the solvent front at a time t0, which is also known as the ‘hold-up time’ or ‘dead time’.

We have all the information to determine the Selectivity factor [α]

tR2 = ta = Retention time of aspartame = 143.8 s

tR1 = tc = Retention time of caffeine = 96.3 s

t0 = tm = retention time of MeOH = 41.8 s

Substituting these value in formula of selectivity factor [α]

Selectivity factor [α] = [ta – tm]/ [tc – tm] = [143.8 – 41.8]/ [96.3 – 41.8] = 102/54.5 = 1.87

Thus selectivity factor [α] for column is 1.87

Retention (or capacity) factor (k)

The retention (or capacity) factor (k) is a means of measuring the retention of an analyte

on the chromatographic column.

It can be determined by following general formula

Retention (or capacity) factor (k) = [tR – t0]/ t0

Retention (or capacity) factor (k) for aspartame = [tR2 – t0]/ t0

and

Retention (or capacity) factor (k) for caffiene = [tR1 – t0]/ t0

We have all the information to determine the Retention (or capacity) factor (k) for aspartame and caffeine.

tR2 = ta = Retention time of aspartame = 143.8 s

tR1 = tc = Retention time of caffeine = 96.3 s

t0 = tm = retention time of MeOH = 41.8 s

Substituting these value in formula of Retention (or capacity) factor (k)

Retention (or capacity) factor (k) for aspartame = [tR2 – t0]/ t0 = [143.8 – 41.8]/ 41.8 = 2.44

Retention (or capacity) factor (k) for caffiene = [tR1 – t0]/ t0 = [96.3– 41.8]/41.8 = 1.3

Thus, Retention (or capacity) factor (k) for aspartame and caffeine is 2.44 and 1.3, respectively.

Now, calculate the hypothetical retention times for caffeine and aspartame if their retention factors were reduced by 28.0%. Assume the mobile phase retention time (tm=41.8 s) remains constant.

Since, Retention (or capacity) factor (k) for aspartame and caffeine is 2.44 and 1.3, respectively. Now we have to calculate new Retention (or capacity) factor (k) for aspartame and caffeine which is reduced by 28 %.

New Retention (or capacity) factor (k) for aspartame = 2.44 – [2.44 * (28/100)] = 2.44 - 0.68 = 1.75

New Retention (or capacity) factor (k) for caffeine = 1.3 – [1.3 * (28/100)] = 1.3 - 0.364 = 0.936

We have calculated new Retention (or capacity) factor (k) for aspartame and caffeine, using these Retention (or capacity) factor (k) we can calculate the retention time of aspartame and caffeine using above formula,

Retention (or capacity) factor (k) for aspartame 1.75 = [tR2 – t0]/ t0

                                                                            1.75 = [tR2 – 41.8]/ 41.8

                                                                            [1.75 * 41.8] + 41.8 = tR2

By solving this we got tR2 = 114.95 s

Retention (or capacity) factor (k) for caffeine, 0.936 = [tR1 – t0]/ t0

                                                                        0.936 = [tR1 – 41.8]/41.8

                                                                        [0.936 * 41.8] + 41.8 = tR1

By solving this we got tR1 = 80.92 s

Partition Coefficient: [K]

It is ratio between concentration of analyte stationary phase to concentration of analyte in mobile phase

Partition Coefficient [K] = concentration of analyte stationary phase / concentration of analyte in mobile phase

If partition coefficient K is decreased means concentration of analyte in stationary phase will be less and concentration of analyte in mobile phase will be high.

Because of that the retention time of analyte will decrease therefore retention factor will aslo decrease.

Thus,

“Caffeine would spend a greater amount of time in the mobile phase”, "The retention factor of caffeine would decrease" and "The retention time of caffeine would decrease" options are correct.

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