Chapter 25, Problem 25.EE

(a)

Interpretation Introduction

Interpretation:

The greatest and least retention of components in mobile phase has to be explained.

Explanation of Solution

To explain: The greatest and least retention of components in mobile phase

• Increase in volume of organic solvent increases the mobile phase strength in reversed phase column chromatography.
• $40\%$ of acetonitrile gives highest retention time.  $60\%$ gives least retention time.
• At $51.6\%$ acetonitrile, the co-elutes are toluene and benzophenone.

(b)

Interpretation Introduction

Interpretation:

The mobile phase of each composition separation has to be drawn as chromatogram.

Concept introduction:

$\begin{array}{l}\text{Resolution}\text{=}\frac{\sqrt{\text{N}}}{\text{4}}\frac{\left(\text{α}\text{-}\text{1}\right)}{\text{α}}\left(\frac{{\text{k}}_{\text{2}}}{\text{1}\text{+}\text{}{\text{k}}_{\text{2}}}\right)\\ \text{Where,}\\ \text{N}\text{-}\text{number}\text{of}\text{theoretical}\text{plates}\\ \text{α}\text{-}\text{relative}\text{retention}\text{of}\text{the}\text{two}\text{peaks}\\ {\text{k}}_{\text{2}}\text{-}\text{retention}\text{factor}\text{of}\text{the}\text{more}\text{retained}\text{component}\end{array}$

The above equation shows that the value of resolution is proportional to $\sqrt{\text{N}\text{.}}$

Explanation of Solution

To draw: The mobile phase of each composition separation.

Conversion of log k to retention time.

$\begin{array}{l}\text{K}\text{=}\frac{{\text{t}}_{\text{r}}\text{-}{\text{t}}_{\text{m}}}{{\text{t}}_{\text{m}}}\\ \\ {\text{t}}_{\text{m}}\text{×}\text{(1+}\text{K)}\\ \\ {\text{t}}_{\text{m}}\text{can be calculated by Equation 25-5}\\ \\ {\text{t}}_{\text{m}}\approx \frac{{\text{Ld}}_{\text{c}}{}^{\text{2}}}{\text{2F}}\approx \frac{\text{15}\text{cm}\text{×}\text{(0}{\text{.46cm}}^{\text{2}}\text{)}}{\text{2}\text{×}\text{1}\text{.0}\text{mL/min}}\text{=}\text{1}\text{.6}\text{min}\\ \\ {\text{t}}_{\text{r}}\text{=}\text{1}\text{.6}\text{min}\text{×}\text{(1}\text{+}\text{15}\text{.1)}\text{=}\text{25}\text{.8}\text{min}\end{array}$ ~

The chromatogram is given below as

Figure 1

(c)

Interpretation Introduction

Interpretation:

The $60\%$ acetonitrile is enough or not for resolution has to be explained.

Concept introduction:

$\begin{array}{l}\text{Resolution}\text{=}\frac{\sqrt{\text{N}}}{\text{4}}\frac{\left(\text{α}\text{-}\text{1}\right)}{\text{α}}\left(\frac{{\text{k}}_{\text{2}}}{\text{1}\text{+}\text{}{\text{k}}_{\text{2}}}\right)\\ \text{Where,}\\ \text{N}\text{-}\text{number}\text{of}\text{theoretical}\text{plates}\\ \text{α}\text{-}\text{relative}\text{retention}\text{of}\text{the}\text{two}\text{peaks}\\ {\text{k}}_{\text{2}}\text{-}\text{retention}\text{factor}\text{of}\text{the}\text{more}\text{retained}\text{component}\end{array}$

The above equation shows that the value of resolution is proportional to $\sqrt{\text{N}\text{.}}$

Explanation of Solution

To draw: The $60\%$ acetonitrile is enough or not for resolution.

The compounds with closest peak have very less resolution and those are benzophenone and toluene.  The retention factors of these compounds are given as

$\begin{array}{l}{\text{10}}^{0.13}=1.35\\ \\ {10}^{0.21}=1.62\end{array}$

Relative retention is calculated as

$\text{α}\text{=}\frac{{\text{K}}_{\text{2}}}{{\text{K}}_{\text{1}}}\text{=}\frac{\text{1}\text{.62}}{\text{1}\text{.35}}\text{=}\text{1}\text{.20}$

$\begin{array}{l}\text{N}\text{=}\frac{\text{3000}\text{.}\text{L(cm)}}{{\text{d}}_{\text{p}}\text{(μm)}}\text{=}\frac{\text{3000}\text{×}\text{15}}{\text{5}}\text{=}\text{9000}\\ \\ \text{Resolution}\text{=}\frac{\sqrt{\text{N}}}{\text{4}}\frac{\left(\text{α}\text{-}\text{1}\right)}{\text{α}}\left(\frac{{\text{k}}_{\text{2}}}{\text{1}\text{+}\text{}{\text{k}}_{\text{2}}}\right)\\ \\ \text{=}\frac{\sqrt{\text{9000}}}{\text{4}}\frac{\text{(1}\text{.20}\text{-}\text{1)}}{\text{1}\text{.20}}\left(\frac{\text{1}\text{.62}}{\text{1+1}\text{.62}}\right)\text{=}\text{2}\text{.4}\end{array}$

(d)

Interpretation Introduction

Interpretation:

The $60\%$ acetonitrile can give good separation of peaks or not has to be explained.

Concept introduction:

$\begin{array}{l}\text{Resolution}\text{=}\frac{\sqrt{\text{N}}}{\text{4}}\frac{\left(\text{α}\text{-}\text{1}\right)}{\text{α}}\left(\frac{{\text{k}}_{\text{2}}}{\text{1}\text{+}\text{}{\text{k}}_{\text{2}}}\right)\\ \text{Where,}\\ \text{N}\text{-}\text{number}\text{of}\text{theoretical}\text{plates}\\ \text{α}\text{-}\text{relative}\text{retention}\text{of}\text{the}\text{two}\text{peaks}\\ {\text{k}}_{\text{2}}\text{-}\text{retention}\text{factor}\text{of}\text{the}\text{more}\text{retained}\text{component}\end{array}$

The above equation shows that the value of resolution is proportional to $\sqrt{\text{N}\text{.}}$

Explanation of Solution

To draw: The $60\%$ acetonitrile can give good separation of peaks or not.

For good separation of peaks the criteria are

$\begin{array}{l}\text{0}\text{.5}\le \text{K}\le \text{20}\\ \\ \text{Resolution}\ge \text{2}\\ \\ \text{operating pressure}\le \text{15}\text{MPa}\\ \\ \text{0}\text{.9 }\le \text{ asymmetry factor }\le \text{ 1}\text{.5}\end{array}$

The $60\%$ acetonitrile satisfies all the above criteria for good separation of peaks.  Thus, the separation of peaks can be done by $60\%$ acetonitrile and it has additional plus point of separation time less than six minutes.