Chapter 23, Problem 23.FE

(a)

Interpretation Introduction

Interpretation:

The retention factor of compound 2 has to be calculated.

Concept introduction:

Retention factor:

The retention factor $\left(\text{k}\right)$ is given by

$\begin{array}{l}\text{k}\text{=}\frac{{\text{t}}_{\text{r}}\text{-}{\text{t}}_{\text{m}}}{{\text{t}}_{\text{m}}}\\ \text{Where,}\\ \text{k}\text{-}\text{retention}\text{factor}\\ {\text{t}}_{\text{r}}\text{-}\text{time}\text{required}\text{to}\text{elute}\text{the}\text{peak}\\ {\text{t}}_{\text{m}}\text{-}\text{time}\text{required}\text{for}\text{mobile}\text{phase}\text{to}\text{pass}\text{through}\text{the}\text{column}\text{.}\end{array}$

Answer to Problem 23.FE

• The retention factor of compound 2 is $\text{5}\text{.511}\text{.}$

Explanation of Solution

Given information,

The value of relative retention $\left(\text{α}\right)$ is $\text{1}\text{.068}$.

The value of retention factor for compound 1 $\left({\text{k}}_{\text{1}}\right)$ is $\text{5}\text{.16}$.

Calculate the value of ${\text{k}}_{\text{2}}$ by using the relationship between $\text{α}\text{and}\text{k}$

$\begin{array}{l}\text{α}\text{=}\frac{{\text{k}}_{\text{2}}}{{\text{k}}_{\text{1}}}\\ {\text{k}}_{\text{2}}\text{=}\text{α}{\text{k}}_{\text{1}}\\ \text{=}\left(\text{1}\text{.068}\right)\left(\text{5}\text{.16}\right)\text{=}\text{5}\text{.511}\end{array}$

The retention factor of compound 2 is $\text{5}\text{.511}\text{.}$

(b)

Interpretation Introduction

Interpretation:

The length $\left(\text{L}\right)$ of the column has to be calculated.

Concept introduction:

Purnell equation:

The relationship between $\text{H}\text{and}\text{α}$ is given by Purnell equation,

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

Answer to Problem 23.FE

• The length of the column is $\text{6}\text{.45}\text{m}\text{.}$

Explanation of Solution

Given information,

The value of $\text{α}$ is $\text{1}\text{.068}$.

The value of ${\text{k}}_{\text{2}}$ is $\text{5}\text{.511}$.

Height of the plate is $\text{0}\text{.520}\text{mm}\text{.}$

Calculate the number $\left(\text{N}\right)$ of plates on column by Purnell equation

$\begin{array}{l}\text{Resolution}\text{=}\frac{\sqrt{\text{N}}}{\text{4}}\frac{\left(\text{α}\text{-}\text{1}\right)}{\text{α}}\left(\frac{{\text{k}}_{\text{2}}}{\text{1}\text{+}\text{}{\text{k}}_{\text{2}}}\right)\\ \text{1}\text{.5}\text{=}\frac{\sqrt{\text{N}}}{\text{4}}\frac{\left(\text{1}\text{.068}\text{-}\text{1}\right)}{\text{1}\text{.068}}\left(\frac{\text{5}\text{.511}}{\text{1}\text{+}\text{}\text{5}\text{.511}}\right)\\ \\ \sqrt{\text{N}}\text{=}\frac{\text{4}\text{×}\text{1}\text{.5}}{\left(\text{0}\text{.8464}\right)\left(\text{0}\text{.0637}\right)}\text{=}\frac{\text{6}}{\text{0}\text{.0539}}\text{=}\text{111}\text{.32}\\ \\ \text{N}\text{=}\text{12392}\text{=}\text{1}\text{.24}\text{×}{\text{10}}^{\text{4}}\text{plates}\end{array}$

Calculate the length $\left(\text{L}\right)$ of the column

$\begin{array}{l}\text{H}\text{=}\frac{\text{Length}\text{of}\text{the}\text{column}}{\text{Number}\text{of}\text{plates}\text{on}\text{column}}\text{=}\frac{\text{L}}{\text{N}}\\ \text{L}\text{=}\text{H}\text{×}\text{N}\\ \text{=}\text{0}\text{.520}\text{mm}\text{×}\text{1}\text{.24}\text{×}{\text{10}}^{\text{4}}\text{=}\text{6448}\text{mm}\\ \text{=}\text{6}\text{.45}\text{m}\text{.}\end{array}$

The length of the column is $\text{6}\text{.45}\text{m}\text{.}$

(c)

Interpretation Introduction

Interpretation:

The values of ${\text{t}}_{\text{r}}\text{and}{\text{w}}_{\text{1/2}}$ has to be calculated for the given peaks.

Concept introduction:

Plates of column:

The relationship between $\text{N}\text{and}{\text{w}}_{\text{1/2}}$

$\begin{array}{l}\text{N}\text{=}\frac{\text{5}\text{.55}{\text{t}}_{\text{r}}{}^{\text{2}}}{{\text{w}}_{\text{1/2}}{}^{\text{2}}}\\ \text{Where,}\\ \text{N}\text{-}\text{number}\text{of}\text{plates}\text{on}\text{column}\\ {\text{t}}_{\text{r}}\text{-}\text{retention}\text{time}\\ {\text{w}}_{\text{1/2}}\text{-}\text{half-width}\end{array}$

Answer to Problem 23.FE

• The value of ${\text{t}}_{\text{r}}$ for the compound 1 is $\text{12}\text{.32}\text{min}\text{.}$
• The value of ${\text{t}}_{\text{r}}$ for the compound 2 is $\text{13}\text{.02}\text{min}\text{.}$
• The value of ${\text{w}}_{\text{1/2}}$ for the compound 1 is $\text{0}\text{.261}\text{min}\text{.}$
• The value of ${\text{w}}_{\text{1/2}}$ for the compound 2 is $\text{0}\text{.275}\text{min}\text{.}$

Explanation of Solution

Given information,

The value of $\text{α}$ is $\text{1}\text{.068}$.

The value of ${\text{k}}_{\text{1}}$ is $\text{5}\text{.16}$.

The value of ${\text{k}}_{\text{2}}$ is $\text{5}\text{.511}$.

The value of $\text{N}$ is $\text{1}\text{.24}\text{×}{\text{10}}^{\text{4}}$.

Calculate the value of ${\text{t}}_{\text{r}}$ for the compound 1 by using the relation between retention factor and retention time

$\begin{array}{l}{\text{k}}_{\text{1}}\text{=}\frac{\left({\text{t}}_{\text{1}}\text{-}{\text{t}}_{\text{m}}\right)}{{\text{t}}_{\text{m}}}\text{=}\frac{{\text{t}}_{\text{1}}}{{\text{t}}_{\text{m}}}\text{-1}\\ {\text{t}}_{\text{1}}\text{=}{\text{t}}_{\text{m}}\left({\text{k}}_{\text{1}}\text{+}\text{}\text{1}\right)\text{=}{\text{t}}_{\text{m}}\left(\text{6}\text{.16}\right)\text{=}\left(\text{2}\text{.00}\text{min}\right)\left(\text{6}\text{.16}\right)\text{=}\text{12}\text{.32}\text{min}\end{array}$

The value of ${\text{t}}_{\text{r}}$ for the compound 1 is $\text{12}\text{.32}\text{min}\text{.}$

Calculate the value of ${\text{t}}_{\text{r}}$ for the compound 2 by using the relation between retention factor and retention time

$\begin{array}{l}{\text{k}}_{\text{2}}\text{=}\frac{\left({\text{t}}_{\text{2}}\text{-}{\text{t}}_{\text{m}}\right)}{{\text{t}}_{\text{m}}}\text{=}\frac{{\text{t}}_{\text{2}}}{{\text{t}}_{\text{m}}}\text{-1}\\ {\text{t}}_{\text{2}}\text{=}{\text{t}}_{\text{m}}\left({\text{k}}_{\text{2}}\text{+}\text{}\text{1}\right)\text{=}{\text{t}}_{\text{m}}\left(\text{6}\text{.511}\right)\text{=}\left(\text{2}\text{.00}\text{min}\right)\left(\text{6}\text{.511}\right)\text{=}\text{13}\text{.02}\text{min}\end{array}$

The value of ${\text{t}}_{\text{r}}$ for the compound 2 is $\text{13}\text{.02}\text{min}\text{.}$

Calculate the value of ${\text{w}}_{\text{1/2}}$ for the compound 1 by using the relationship between $\text{N}\text{and}{\text{w}}_{\text{1/2}}$

$\begin{array}{l}\text{N}\text{=}\frac{\text{5}\text{.55}{\text{t}}_{\text{r}}{}^{\text{2}}}{{\text{w}}_{\text{1/2}}{}^{\text{2}}}\\ {\text{w}}_{\text{1/2}}\text{=}\sqrt{\frac{\text{5}\text{.55}}{\text{N}}}\text{×}{\text{t}}_{\text{r}}\text{=}\sqrt{\frac{\text{5}\text{.55}}{\text{1}\text{.24}\text{×}{\text{10}}^{\text{4}}}}\text{×}\left(\text{12}\text{.32}\text{min}\right)\text{=}\text{0}\text{.261}\text{min}\text{.}\end{array}$

The value of ${\text{w}}_{\text{1/2}}$ for the compound 1 is $\text{0}\text{.261}\text{min}\text{.}$

Calculate the value of ${\text{w}}_{\text{1/2}}$ for the compound 2 by using the relationship between $\text{N}\text{and}{\text{w}}_{\text{1/2}}$

$\begin{array}{l}\text{N}\text{=}\frac{\text{5}\text{.55}{\text{t}}_{\text{r}}{}^{\text{2}}}{{\text{w}}_{\text{1/2}}{}^{\text{2}}}\\ {\text{w}}_{\text{1/2}}\text{=}\sqrt{\frac{\text{5}\text{.55}}{\text{N}}}\text{×}{\text{t}}_{\text{r}}\text{=}\sqrt{\frac{\text{5}\text{.55}}{\text{1}\text{.24}\text{×}{\text{10}}^{\text{4}}}}\text{×}\left(\text{13}\text{.02}\text{min}\right)\text{=}\text{0}\text{.275}\text{min}\text{.}\end{array}$

The value of ${\text{w}}_{\text{1/2}}$ for the compound 2 is $\text{0}\text{.275}\text{min}\text{.}$

(d)

Interpretation Introduction

Interpretation:

The partition coefficient value for compound 1 has to be calculated.

Concept introduction:

Partition coefficient:

The relationship between retention factor and partition coefficient is given by

$\begin{array}{l}\text{k}\text{=}\text{K}\frac{{\text{V}}_{\text{s}}}{{\text{V}}_{\text{m}}}\\ \text{Where,}\\ \text{k}\text{-}\text{retention}\text{factor}\\ \text{K}\text{-}\text{partition}\text{coefficient}\\ {\text{V}}_{\text{s}}\text{-}\text{volume}\text{of}\text{stationary}\text{phase}\\ {\text{V}}_{\text{m}}\text{-}\text{volume}\text{of}\text{mobile}\text{phase}\end{array}$

Answer to Problem 23.FE

• The partition coefficient value for compound 1 is $\text{17}\text{.2}\text{.}$

Explanation of Solution

Given information,

The value of ${\text{V}}_{\text{s}}\text{/}{\text{V}}_{\text{m}}$ is $\text{0}\text{.30}$.

The value of ${\text{k}}_{\text{1}}$ is $\text{5}\text{.16}$.

Calculate the value of ${\text{K}}_{\text{1}}$ for the compound 1 by using the relation between retention factor and partition coefficient

$\begin{array}{l}{\text{K}}_{\text{1}}\text{=}{\text{k}}_{\text{1}}\frac{{\text{V}}_{\text{m}}}{{\text{V}}_{\text{s}}}\\ \text{=}\left(\text{5}\text{.16}\right)\left(\text{1/0}\text{.30}\right)\text{=}\text{17}\text{.2}\end{array}$

The partition coefficient value for compound 1 is $\text{17}\text{.2}\text{.}$