Question

Question:The graph shows retention data from a${\mathbf{C}}_{\mathbf{8}}$silica column with an acetonitrile/water mobile phase.

(a) What mobile phase composition provides greatest retention$\mathbf{\left(}\mathit{k}\mathit{\right)}$for the components? Least retention? Coelution $\mathbf{(}\mathbf{equal}\mathbf{}\mathit{k}\mathit{}\mathit{)}$of two components?

(b) Predict the retention time of each peak at $\mathbf{40}\mathbf{\%}\mathbf{}\mathbf{and}\mathbf{}\mathbf{60}\mathbf{\%}\mathbf{}\mathbf{acetonitrite}$. Draw a chromatogram (a "stick diagram" representing each peak as a vertical line) of the separation at each mobile phase composition.

(c) Would$\mathbf{60}\mathbf{\%}$acetonitrile yield adequate resolution?

(d) Assuming Gaussian peaks, does the separation at$\mathbf{60}\mathbf{\%}$acetonitrile have the attributes of a good separation?

Short answer

a) The greatest and least retention of components in the mobile phase in reversed phase column chromatography, increasing the amount of organic solvent increases the mobile phase strength.

Acetonitrile at $40\%$ yields the longest retention time. The retention time is the shortest at $60\%$.

Toluene and benzophenone are co-elutes at $51.6\%$acetonitrile.

b) The chromatogram:

c) $60\%$acetonitrile yield $2.4$resolution.

d) The separation of peaks can be done by $60\%$acetonitrile and it has additional plus point of separation time less than six minutes.

Step-by-step solution

Concept used.

$$\begin{gathered} \mathrm{Resolution}=\frac{\sqrt{\mathrm{N}\left(\mathrm{\alpha }-1\right)}}{\mathrm{\alpha }}\left(\frac{{\mathrm{k}}_2}{1+{\mathrm{k}}_2}\right) \\ \mathrm{Where}, \\ \mathrm{N}-\mathrm{number}\mathrm{of}\mathrm{theoretical}\mathrm{plates} \\ \mathrm{\alpha }-\mathrm{relative}\mathrm{retention}\mathrm{of}\mathrm{the}\mathrm{two}\mathrm{peaks} \\ {\mathrm{k}}_2-\mathrm{retention}\mathrm{factor}\mathrm{of}\mathrm{the}\mathrm{more}\mathrm{retained}\mathrm{component} \\ \mathrm{The}\mathrm{above}\mathrm{equation}\mathrm{shows}\mathrm{that}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{resolution}\mathrm{is}\mathrm{proportional}\mathrm{to}\sqrt{\mathrm{N}} \end{gathered}$$

The greatest and least retention of components in mobile phase. 

a)

The greatest and least retention of components in the mobile phase in reversed phase column chromatography, increasing the amount of organic solvent increases the mobile phase strength.

Acetonitrile at $40\%$yields the longest retention time. The retention time is the shortest at $60\%$.

Toluene and benzophenone are co-elutes at$51.6\%$acetonitrile

Step 3: Draw the chromatogram (a "stick diagram" representing each peak as a vertical line) of the separation at each mobile phase composition.

$$\begin{gathered} b) \\ \mathrm{Each}\mathrm{composition}\text{'}\mathrm{s}\mathrm{mobile}\mathrm{phase}\mathrm{is}\mathrm{separated}. \\ \mathrm{logk}\mathrm{to}\mathrm{retention}\mathrm{time}\mathrm{conversion}. \\ K=\frac{t_f\mathit{-}{t}_m}{t_m} \\ {\mathrm{t}}_{\mathrm{m}}\times (1+\mathrm{K}) \\ {\mathrm{t}}_{\mathrm{m}}\mathrm{can}\mathrm{be}\mathrm{calculated}\mathrm{by}\mathrm{Equation}25-5~ \\ {\mathrm{t}}_{\mathrm{m}}\approx \frac{1{{\mathrm{A}}_{\mathrm{C}}}^2}{2\mathrm{F}} \\ \approx \frac{15\mathrm{cm}\times (0.46{\mathrm{cm}}^{2)}}{2\times 1.0\mathrm{mil}/\min } \\ =1.6\min \\ {\mathrm{t}}_{\mathrm{f}}=1.6\min \times (1+15.1) \\ =25.8\min \\ \mathrm{The}\mathrm{chromatogram}\mathrm{is}\mathrm{given}\mathrm{below}\mathrm{as} \end{gathered}$$

Step 4: Explain if 60%acetonitrile yield adequate resolution.

c)

Benzophenone and toluene, the chemicals with the closest peak, have the lowest resolution. These chemicals' retention factors are reported as

$$\begin{gathered} {10}^{0.13}=1.35 \\ {10}^{0.21}=1.62 \\ \mathrm{The}\mathrm{formula}\mathrm{for}\mathrm{calculating}\mathrm{relative}\mathrm{retention}\mathrm{is}\mathrm{as}\mathrm{follows}: \\ \mathrm{\alpha }=\frac{{\mathit{K}}_{\mathit{2}}}{{\mathit{K}}_{\mathit{1}}} \\ =\frac{1.61}{1.35} \\ =1.20 \\ \mathrm{N}=\frac{3000.1\left(\mathrm{cm}\right)}{{\mathrm{d}}_{\mathrm{p}}\left(1\mathrm{m}\right)} \\ =\frac{3000\times 15}{5} \\ =9000 \\ \mathrm{Resolution} \\ =\frac{\sqrt{N}}{4}\frac{\left(\mathrm{\alpha }-1\right)}{\mathrm{\alpha }}\left[\frac{{\mathrm{k}}_{\mathit{2}}}{1+{\mathrm{k}}_{\mathit{2}}}\right] \\ =\frac{\sqrt{9000}}{4}\frac{\left(1.20-1\right)}{1.20}\left(\frac{1.62}{1+1.62}\right) \\ =2.4 \end{gathered}$$

Step 5: Whether the separation at60%acetonitrile have the attributes of a good separation.

d)

The requirements for good peak separation are as follows:

$$\begin{gathered} 0.5\le \mathrm{K}\le 20 \\ Resoluction\ge 2 \\ Operating\mathit{}pressure\le 15\mathrm{MPa} \\ 0.9\le \mathrm{asymmetry}\mathrm{factor}\le 1.5 \end{gathered}$$

All of the following requirements for excellent peak separation are met by the $60\%$

acetonitrile. Thus,$60\%$acetonitrile may be used to separate peaks, with the added benefit of a separation period of less than six minutes.