Chapter 25, Problem 25.BE

(a)

Interpretation Introduction

Interpretation:

The retention time for both isomers and the average width of the peak has to be calculated.

Answer to Problem 25.BE

The retention time of R-isomer ( $t_1$) is $2.35\min$.

The retention time of S-isomer ( $t_2$) is $7.11\min$.

The average width ( $w_{av}$) is $0.619\min$.

Explanation of Solution

To find the retention time of R-isomer,

$t_m$ is given as $1.00\min$.

Retention factor ( $k$) is given as $1.35$.

Substitution of the above values in the retention factor equation, the retention time ( $t_1$) can be calculated as shown below,

$\begin{array}{l}k=\frac{t_1-t_m}{t_m}\\ =\frac{t_1-1.00}{1.00}\\ 1.35=\frac{t_1-1.00}{1.00}\\ t_1=1.35+1.00\\ =2.35\min \end{array}$

The retention time of R-isomer is calculated as $2.35\min$.

To find the retention time of S-isomer,

The retention time can be found by using the equation for Relative retention.

Relative retention ( $\alpha$) is given as $4.53$.

Retention time ( $t_1$) for R-isomer is $2.35\min$.

$\begin{array}{l}\text{Relative retention}\text{(α)}\text{=}\frac{t_{r2}^{\text{'}}}{t_{r1}^{\text{'}}}\\ 4.53=\frac{t_2-1.00}{t_1-1.00}\\ =\frac{t_2-1.00}{2.35-1.00}\\ \\ t_2=4.53(1.35)+1.00\\ =6.11+1.00\\ =7.11\min \end{array}$

The retention time of S-isomer is $7.11\min$.

To calculate the average width,

The average width can be calculated form the resolution and the difference in retention time of the two isomers,

Difference in retention time is found as $4.77\min$.

Resolution is given as $7.7$ in the problem statement.

The average width can be calculated as shown below,

$\begin{array}{l}\text{Resolution}\text{=}\frac{\Delta t_r}{w_{av}}\\ 7.7=\frac{4.77}{w_{av}}\\ w_{av}=\frac{4.77}{7.7}\\ =0.619\min \end{array}$

The average width at the base is calculated as $0.619\min$.

(b)

Interpretation Introduction

Interpretation:

The half-width of each peak present in the chromatogram has to be calculated.

Answer to Problem 25.BE

The half-width of peak 1 is $0.181\min$.

The half-width of peak 2 is $0.548\min$.

Explanation of Solution

The plate number can be calculated using the equation given below,

$N=5.54{\left(\frac{t_r}{w_{1/2}}\right)}^2$

From this we infer that retention time is proportional to half-width, if the plate number is a constant.  Hence,

$\begin{array}{l}\frac{w_{1/2}(\text{peak}\text{1})}{w_{1/2}(\text{peak}\text{2})}=\frac{t_1}{t_2}\\ =\frac{2.35}{7.11}\\ =0.330\end{array}$

The average width at the base was calculated as $0.62\min$.

We know that for each peak, the width ( $w$) is equal to $4\sigma$ and half-width ( $w_{1/2}$) is $2.35\sigma$.

From the above two equations, $w$ was found as $1.70w_{1/2}$.

The average base width is $0.62\min$.

$\begin{array}{l}{w}_{av}=\frac{1}{2}(w_1+w_2)\\ =\frac{1}{2}(1.70w_{1/2}(\text{peak}\text{}\text{1})+1.70w_{1/2}(\text{peak}\text{}\text{2}))\end{array}$

We know that,

$w_{1/2}(\text{peak}\text{1})=0.330(w_{1/2}(\text{peak}2)$

Substituting this in the average width at the base equation,

$\begin{array}{l}{w}_{av}=\frac{1}{2}(1.70(0.330w_{1/2}(\text{peak}2))+1.70w_{1/2}(\text{peak}\text{}\text{2}))\\ =\frac{1}{2}(0.561w_{1/2}(\text{peak}2)+1.70w_{1/2}(\text{peak}\text{}\text{2}))\\ 1.24=(2.261w_{1/2}(\text{peak}2))\\ w_{1/2}(\text{peak}2)=\frac{1.24}{2.261}\\ =0.548\min \end{array}$

The half-width of peak 2 is $0.548\min$.

From this half-width of peak 1 can be calculated as shown below,

$\begin{array}{l}{w}_{1/2}(\text{peak}\text{1})=0.330(w_{1/2}(\text{peak}2)\\ =0.330\times 0.548\\ =0.18084\min \\ =0.181\min \end{array}$

The half-width of both peaks present in the given chromatogram was calculated.

(c)

Interpretation Introduction

Interpretation:

The relative heights of the two peaks has to be calculated.

Answer to Problem 25.BE

The relative heights of two peaks is $3.0$.

Explanation of Solution

The half-width of peak 1 is $0.181\min$.

The half-width of peak 2 is $0.584\min$.

In the problem statement it is given that the areas under both the peaks are equal.

$\begin{array}{l}{\text{Height}}_{\text{R}}\times w_R={\text{Height}}_{\text{S}}\times w_S\\ \frac{{\text{Height}}_{\text{R}}}{{\text{Height}}_{\text{S}}}=\frac{w_S}{w_R}\\ =\frac{0.548}{0.181}\\ =3.0\end{array}$

The relative heights of the two peaks is calculated as $3.0$.