Question

Consider a chromatography experiment in which two components with retention factors ${\mathit{k}}_{\mathbf{1}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{00}$and ${\mathit{k}}_{\mathbf{2}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{00}$are injected into column with $\mathit{N}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}$theoretical plates. The retention time for the less-retained component is $\mathit{t}{\mathit{r}}_{\mathbf{1}}\mathbf{=}\mathbf{10}\mathbf{.}\mathbf{0}\mathit{m}\mathit{i}\mathit{n}$.

(a)Calculate ${\mathit{t}}_{\mathbf{m}}$and${\mathit{t}}_{\mathbf{12}}$.Find ${\mathit{w}}_{\frac{\mathbf{1}}{\mathbf{2}}}$(width at half height) and w (at the base) for each peak.

(b)Using graph paper , sketch the chromatogram analogous to figure 23-7,supposing that two peaks have same amplitude(height). Draw the half widths accurately.

(c)Calculate the resolution of the two peaks and compare this value with those drawn in Figure 23-10.

Short answer

(a)$t_m=2.00min,t_{r2}=12.00min$

are 1.265 min and 1.518 min.

$w_{\frac{1}{2}}$(for peak1 ) and $w_{\frac{1}{2}}$(for peak2) are 0.745 min and 0.894 min

(b) Resolution =1.437

Step-by-step solution

Step1: 

Finding the value of, $t_m,t_{r2}$ w and $W_{\frac{1}{2}}$

We know that $k_1=\frac{t_{r1}-t_m}{t_m}$or, $t_m=\frac{t_{r1}}{k_1+1}$

Or, $t_m=\frac{10.00}{4+1}=2.00min$

We know that , $t_{r2}=t_m\left(K_2+1\right)$Or, $t_{r2}=2.00\left(5.00+1\right)=12min$

Now $N=\frac{5.55\times t_r^2}{{w^2}_{{\displaystyle \frac{1}{2}}}}$ Or, $w_{\frac{1}{2}}=\sqrt{\frac{5.55\times t_r^2}{N}}$

For $w_{\frac{1}{2}}=\sqrt{\frac{5.55\times {\left(10\right)}^2}{1.00\times {10}^3}}=0.745min$

$w_{\frac{1}{2}}=\sqrt{\frac{5.55\times {\left(12\right)}^2}{1.00\times {10}^3}}=0.894min$

$N=\frac{16t_r^2}{w}$ Or, $w=\sqrt{\frac{16\times t_r^2}{N}}$

$$\begin{gathered} w_1=\sqrt{\frac{16\times {\left(10\right)}^2}{1.00\times {10}^3}}=1.265min\backslash \\ {w}_2=\sqrt{\frac{16\times {\left(12\right)}^2}{1.00\times {10}^3}}=1.518min \end{gathered}$$

(b)Step 2 :   Sketching the graph :

   finding Resolution:

(c) Resolution =$\frac{∆t}{w_{av}}=\frac{t\text{'}{\text{'}}_r-t{\text{'}}_r}{w_{av}}$

Or, Resolution=$\frac{\left(12-10\right)}{{\displaystyle \frac{1}{2}}\left(1.265+1.518\right)}$ =1.437

For quantitive analysis resolution >1.5 is highly desirable.