Question

The following data are for a liquid chromatographic column

$$\begin{gathered} \mathrm{Length}\mathrm{of}\mathrm{packing}24.7\mathrm{cm} \\ \mathrm{Flow}\mathrm{rate}0.313\mathrm{mL}/\min \\ {V}_M\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}1.37\mathrm{mL} \\ {V}_S\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}0.164\mathrm{mL} \end{gathered}$$

A chromatogram of a mixture of species A, B, C, and D provided the following data:

Calculate

(a) the number of plates from each peak.

(b) the mean and the standard deviation for N.

(c) the plate height for the column.

Short answer

(a) Number of theoretical plates for A $=2562$

Number of theoretical plates for B$=2472$

Number of theoretical plates for C$=2263$

Number of theoretical plates for D$=2523$

(b)

$$\begin{gathered} \mathrm{Mean}\mathrm{for}N=2455 \\ \mathrm{Standard}\mathrm{deviation}=115 \end{gathered}$$

(c) The plate height for the column is $0.010\mathrm{cm}$.

Step-by-step solution

Part (a) Step 1: Given information

$$\begin{gathered} \mathrm{Length}\mathrm{of}\mathrm{packing}=24.7\mathrm{cm} \\ \mathrm{Flow}\mathrm{rate}=0.313\mathrm{mL}/\min \\ {V}_M=1.37\mathrm{mL} \\ {V}_S=0.164\mathrm{mL} \end{gathered}$$

A chromatogram of a mixture of species A, B, C, and D provided the following data:

Part (a) Step 2: Number of theoretical plates 

The equation for the number of theoretical plates is as follows:

$N=16{\left(\frac{t_R}{W}\right)}^2.......\left(1\right)$

Here, $t_R$is the retention time, $N$is the number of plates and$W$ is the width of peak base.

Part (a) Step 3: Calculate the number of theoretical plates for A

Substitute $6.2$for $t_R$and $0.49$for $W$in equation (1) as follows:

role="math" localid="1649133270777" $$\begin{gathered} N_A=16{\left(\frac{t_R}{W}\right)}^2 \\ =16{\left(\frac{6.2}{0.49}\right)}^2 \\ =2562 \end{gathered}$$

Part (a) Step 4: Calculate the number of theoretical plates for B

Substitute $13.3$for $t_R$and for$W$ in equation (1) as follows:

$$\begin{gathered} N_B=16{\left(\frac{t_R}{W}\right)}^2 \\ =16{\left(\frac{13.3}{1.07}\right)}^2 \\ =2472 \end{gathered}$$

Part (a) Step 5: Calculate the number of theoretical plates for C

Substitute $15.7$for localid="1649133693149" $t_R$and $1.32$for $W$in equation (1) as follows:

$$\begin{gathered} N_C=16{\left(\frac{t_R}{W}\right)}^2 \\ =16{\left(\frac{15.7}{1.32}\right)}^2 \\ =2263 \end{gathered}$$

Part (a) Step 6: Calculate the number of theoretical plates for D

Substitute $21.6$for $t_R$and $1.72$for $W$ in equation (1) as follows:

$$\begin{gathered} N_D=16{\left(\frac{t_R}{W}\right)}^2 \\ =16{\left(\frac{21.6}{1.72}\right)}^2 \\ =2523 \end{gathered}$$

Part (b) Step 1: Calculate Mean and Standard deviation 

The mean is as follows:

$M=\frac{N_A+N_B+N_C+N_D}{4}.........\left(2\right)$

Substitute the values in the above equation (2) as follows:

$$\begin{gathered} M=\frac{N_A+N_B+N_C+N_D}{4} \\ =\frac{2562+2472+2263+2523}{4} \\ =\frac{9820}{4} \\ =2455 \end{gathered}$$

Part (b) Step 2: Calculate Standard deviation 

Standard deviation will give how the values deviate from mean value of $N$.

$SD=\sqrt{\frac{{\left(N_A-N\right)}^2+{\left(N_B-N\right)}^2+{\left(N_C-N\right)}^2+{\left(N_D-N\right)}^2}{4}}.........\left(4\right)$

Substitute all the value in the above equation (4) as follows:

$$\begin{gathered} SD=\sqrt{\frac{{\left(2562-2455\right)}^2+{\left(2472-2455\right)}^2+{\left(2263-2455\right)}^2+{\left(2523-2455\right)}^2}{4}} \\ =\sqrt{\frac{11449+289+36864+4624}{4}} \\ =115 \end{gathered}$$

Part (c) Step 1: Plate height for the column

The height of the plate is as follows:

$H=\frac{L}{N}.........\left(5\right)$

Here, $N$is the plate count, $L$is the length of the packing in cm and $H$is the height of the column.

Part (c) Step 2: Calculate plate height for the column

Substitute the value in equation (5) as follows:

$$\begin{gathered} H=\frac{L}{N} \\ =\frac{24.7\mathrm{cm}}{2455} \\ =0.10\mathrm{cm} \end{gathered}$$