A mixture of 14compounds was subjected to a reversed-phase gradient separation going from 5%to 100%acetonitrile with

a gradient time of 60min. The sample was injected at t =time. All peaks were eluted between 22and 50min.

(a) Is the mixture more suitable for isocratic or gradient elution?

(b) If the next run is a gradient, select the starting and ending %acetonitrile

and the gradient time.

isocratic elution refers to maintaining a constant concentration in the mobile phase, whereas gradient elution refers to maintaining a varying concentration in the mobile phase.

First we need to calculate$∆\mathrm{t}$:

$$\begin{gathered} ∆\mathrm{t}=\left(50-22\right)\min \\ ∆\mathrm{t}=28\min \end{gathered}$$

Then we calculate$∆\mathrm{t}/{\mathrm{t}}_{\mathrm{G}}$

role="math" localid="1663667217587" $\frac{∆\mathrm{t}}{{\mathrm{t}}_{\mathrm{G}}}=\frac{28}{60}=0.47$

Because the $∆\mathrm{t}/{\mathrm{t}}_{\mathrm{G}}>0.40$the mixture is more suitable for gradient solution.

Question: A mixture of \(14\) compounds was subjected to a reversed- phase gradient separation going from \(5\% \) to \(100\% \)acetonitrile with

a gradient time of \(60\)min. The sample was injected at \(t = \;dwell\;\) time. All peaks were eluted between \(22\)and\(50\)min.

(b) If the next run is a gradient, select the starting and ending % acetonitrile and the gradient time.

Answer:

The portions of the gradient from\(5{\rm{\;to\;}}40{\rm{\% }}\)acetonitrile and \(85{\rm{\;to\;}}100{\rm{\% }}\) acetonitrile were not needed. Therefore, the second run could be made with a gradient from \(40{\rm{\;to\;}}85{\rm{\% }}\) acetonitrile over the same \({t_G} = 60{\rm{min}}.\)

t_G the gradient time (the time from the start to the end of the gradient), V_M the column hold-up volume and F the mobile phase flow rate. The gradient steepness parameter will vary with the S-value, which is only roughly constant for similar compounds.

The next step in developing a gradient method is to spread the peaks out by choosing a shallower gradient. Peak 1 was eluted at 22 min when the solvent was 40% acetonitrile:

\(5 + \frac{{22}}{{60}} \cdot 95{\rm{\% }} = 40{\rm{\% }}\)

Peak \(2\) was eluted at \(50\)min, when the solvent was 85percent acetonitrile

\(5 + \frac{{50}}{{60}} \cdot 95{\rm{\% }} = 85{\rm{\% }}\)

The portions of the gradient from\(5{\rm{\;to\;}}40{\rm{\% }}\)acetonitrile and \(85{\rm{\;to\;}}100{\rm{\% }}\) acetonitrile were not needed. Therefore, the second run could be made with a gradient from \(40{\rm{\;to\;}}85{\rm{\% }}\) acetonitrile over the same \({t_G} = 60{\rm{min}}.\)