Chapter 22, Problem 22.4P

(a)

Interpretation Introduction

Interpretation:

Reason behind the formation of more theoretical plates per meter than thick film has to be explained.

Concept Introduction:

Plate height can be calculated by van Deemter equation as follows:

    $\text{H}\approx \text{A}+\frac{\text{B}}{\text{u}}+\left(\text{C}\right)\left(\text{u}\right)$

Here $\text{H}$ is the plate height.

$\text{A}$ term is eddy diffusion constant.

$\text{B}$ term is longitudinal diffusion constant.

$\text{C}$ term is mass transfer constant.

$\text{u}$ is flow rate.

The term $\text{A}$ is eddy diffusion constant and it is related to particle size of packing material and uniformity of the flow around the particles. It is independent of the flow rate.

The term $\text{B}$ is longitudinal diffusion and it is related to diffusion of the analyte in mobile phase and on stationary phase.

The term $\text{C}$ is mass transfer constant and it is related to linear velocity and square of the particle size.

Number of theoretical plates is calculated as follows:

    $\text{N}=\frac{\text{L}}{\text{H}}$

Here,

$\text{N}$ is number of theoretical plates.

$\text{L}$ islength of the column.

$\text{H}$ isplate height.

Explanation of Solution

Particle size has a significant impact on the analyte band. For smaller sized particles at very low flow rate the second term in the van Deemter equation is high but other two terms are small so plate height is not high enough. At high flow rate, the third term cannot play significant role as it is proportional to the square of the particle size thus, the value of third term is not high enough. So the summation of all three terms is small. Hence plate height is low.

For lager sized particles at very low flow rate, second term in the van Deemter equation will be predominant, hence at low flow rate plate height is high. And at high flow rate, third term in the equation is predominant as it is proportional to the square of the particle size. So plate height is high.

Hence it can be concluded that in general pate height of thin films is greater than that of thick films because particle size in thin film is smaller than that of thick film.

As number of theoretical plates is inversely proportional to plate height hence number of theoretical plates is higher in case of this film.

(b)

Interpretation Introduction

Interpretation:

Mass of stationary phase in each column has to be calculated.

Concept Introduction:

Formula to calculate volume of stationary phase is calculated as follows:

    $\text{Volume}\text{of}\text{stationary}\text{phase}=\left[\begin{array}{c}\left(\text{π}\right)\left(\text{Plate}\text{height}\right)\left(\text{Column}\text{diameter}\right)\\ \left(\text{Thickness}\text{of}\text{stationary}\text{phase}\right)\end{array}\right]$

Explanation of Solution

For narrow bore column mass of stationary phase is $\text{15}\text{.7}\text{ng}$ and for wide bore column mass of stationary phase is $\text{5}\text{.55}\text{ng}$.

For narrow bore open tubular gas chromatography plate height is calculated as follows:

    $\begin{array}{c}\text{Plate}\text{height}=\frac{\text{Length}\text{of}\text{column}}{\text{Number}\text{of}\text{plates}}\\ =\frac{\text{1}\text{m}}{5000}\\ =2.0\times {10}^{-4}\text{m}\end{array}$

The volume of stationary phase is calculated as follows:

    $\begin{array}{c}\text{Volume}\text{of}\text{stationary}\text{phase}=\left[\begin{array}{c}\left(\text{π}\right)\left(\text{Plate}\text{height}\right)\left(\text{Column}\text{diameter}\right)\\ \left(\text{Thickness}\text{of}\text{stationary}\text{phase}\right)\end{array}\right]\\ =\left(\text{π}\right)\left(2\times {10}^{-4}\text{m}\right)\left(0.250\times {10}^{-3}\text{m}\right)\left(0.1\times {10}^{-6}\text{m}\right)\\ =15.7\times {10}^{-15}{\text{m}}^{\text{3}}\left(\frac{{10}^6\text{mL}}{\text{1}{\text{m}}^{\text{3}}}\right)\\ =15.7\times {10}^{-9}\text{mL}\end{array}$

Mass of stationary phase is calculated as follows:

    $\begin{array}{c}\text{Mass}\text{of}\text{stationary}\text{phase}=\left(\text{Density}\right)\left(\text{Volume}\right)\\ =\left(\text{1}\text{.00}\text{g}{\text{mL}}^{-1}\right)\left(15.7\times {10}^{-9}\text{mL}\right)\\ =15.7\times {10}^{-9}\text{g}\\ =\text{15}\text{.7}\text{ng}\end{array}$

For wide bore open tubular gas chromatography plate height is calculated as follows:

    $\begin{array}{c}\text{Plate}\text{height}=\frac{\text{Length}\text{of}\text{column}}{\text{Number}\text{of}\text{plates}}\\ =\frac{\text{1}\text{m}}{1500}\\ =6.6667\times {10}^{-4}\text{m}\end{array}$

The volume of stationary phase is calculated as follows:

    $\begin{array}{c}\text{Volume}\text{of}\text{stationary}\text{phase}=\left(\begin{array}{c}\left(\text{π}\right)\left(\text{Plate}\text{height}\right)\left(\text{Column}\text{diameter}\right)\\ \left(\text{Thickness}\text{of}\text{stationary}\text{phase}\right)\end{array}\right)\\ =\left(\text{π}\right)\left(6.667\times {10}^{-4}\text{m}\right)\left(0.53\times {10}^{-6}\text{m}\right)\left(5\times {10}^{-6}\text{m}\right)\\ =5.55\times {10}^{-15}{\text{m}}^{\text{3}}\left(\frac{{10}^6\text{mL}}{\text{1}{\text{m}}^{\text{3}}}\right)\\ =5.55\times {10}^{-9}\text{mL}\end{array}$

Mass of stationary phase is calculated as follows:

    $\begin{array}{c}\text{Mass}\text{of}\text{stationary}\text{phase}=\left(\text{Density}\right)\left(\text{Volume}\right)\\ =\left(\text{1}\text{.00}\text{g/mL}\right)\left(5.55\times {10}^{-9}\text{mL}\right)\\ =5.55\times {10}^{-9}\text{g}\\ =5.55\text{ng}\end{array}$

For narrow bore column mass of stationary phase is $\text{15}\text{.7}\text{ng}$ and for wide bore column mass of stationary phase is $\text{5}\text{.55}\text{ng}$.

(c)

Interpretation Introduction

Interpretation:

Mass of analyte can be injected has to be calculated.

Concept Introduction:

Refer to part (b).

Answer to Problem 22.4P

For narrow bore column $\text{0}\text{.157}\text{ng}$ analyte can be injected and for wide bore column $\text{0}\text{.0555}\text{ng}$ analyte can be injected.

Explanation of Solution

For narrow bore open tubular gas chromatography plate height is calculated as follows:

    $\begin{array}{c}\text{Plate}\text{height}=\frac{\text{Length}\text{of}\text{column}}{\text{Number}\text{of}\text{plates}}\\ =\frac{\text{1}\text{m}}{5000}\\ =2.0\times {10}^{-4}\text{m}\end{array}$

The volume of stationary phase is calculated as follows:

    $\begin{array}{c}\text{Volume}\text{of}\text{stationary}\text{phase}=\left(\text{π}\right)\left(\text{Plate}\text{height}\right)\left(\text{Column}\text{diameter}\right)\\ \left(\text{Thickness}\text{of}\text{stationary}\text{phase}\right)\\ =\left(\text{π}\right)\left(2\times {10}^{-4}\text{m}\right)\left(0.250\times {10}^{-3}\text{m}\right)\left(0.1\times {10}^{-6}\text{m}\right)\\ =15.7\times {10}^{-15}{\text{m}}^{\text{3}}\\ =15.7\times {10}^{-9}\text{mL}\end{array}$

Mass of stationary phase is calculated as follows:

    $\begin{array}{c}\text{Mass}\text{of}\text{stationary}\text{phase}=\left(\text{Density}\right)\left(\text{Volume}\right)\\ =\left(\text{1}\text{.00}\text{g}{\text{mL}}^{-1}\right)\left(15.7\times {10}^{-9}\text{mL}\right)\\ =15.7\times {10}^{-9}\text{g}\\ =\text{15}\text{.7}\text{ng}\end{array}$

Mass of analyte can be injected is calculated as follows:

    $\begin{array}{c}\text{Mass}\text{of}\text{analyte}\text{injected}=\left(\text{Mass}\text{of}\text{stationary}\text{phase}\right)\left(\frac{1}{100}\right)\\ =\left(\text{15}\text{.7}\text{ng}\right)\left(\frac{1}{100}\right)\\ =\text{0}\text{.157}\text{ng}\end{array}$

For wide bore open tubular gas chromatography plate height is calculated as follows:

    $\begin{array}{c}\text{Plate}\text{height}=\frac{\text{Length}\text{of}\text{column}}{\text{Number}\text{of}\text{plates}}\\ =\frac{\text{1}\text{m}}{1500}\\ =6.6667\times {10}^{-4}\text{m}\end{array}$

The volume of stationary phase is calculated as follows:

    $\begin{array}{c}\text{Volume}\text{of}\text{stationary}\text{phase}=\left(\text{π}\right)\left(\text{Plate}\text{height}\right)\left(\text{Column}\text{diameter}\right)\\ \left(\text{Thickness}\text{of}\text{stationary}\text{phase}\right)\\ =\left(\text{π}\right)\left(6.667\times {10}^{-4}\text{m}\right)\left(0.53\times {10}^{-6}\text{m}\right)\left(5\times {10}^{-6}\text{m}\right)\\ =5.55\times {10}^{-15}{\text{m}}^{\text{3}}\\ =5.55\times {10}^{-9}\text{mL}\end{array}$

Mass of stationary phase is calculated as follows:

    $\begin{array}{c}\text{Mass}\text{of}\text{stationary}\text{phase}=\left(\text{Density}\right)\left(\text{Volume}\right)\\ =\left(\text{1}\text{.00}\text{g}{\text{mL}}^{-1}\right)\left(5.55\times {10}^{-9}\text{mL}\right)\\ =5.55\times {10}^{-9}\text{g}\\ =5.55\text{ng}\end{array}$

Mass of analyte can be injected is calculated as follows:

    $\begin{array}{c}\text{Mass}\text{of}\text{analyte}\text{injected}=\left(\text{Mass}\text{of}\text{stationary}\text{phase}\right)\left(\frac{1}{100}\right)\\ =\left(5.55\text{ng}\right)\left(\frac{1}{100}\right)\\ =\text{0}\text{.0555}\text{ng}\end{array}$

For narrow bore column $\text{0}\text{.157}\text{ng}$ analyte can be injected and for wide bore column $\text{0}\text{.0555}\text{ng}$ analyte can be injected.