Question

The graph shows van Deemter curves for n-nonane at . in the 3.0-m-long microfabricated column in Box 24-2 with a -thick stationary phase.

van Deemter curves. [Data from G. Lambertus, A. Elstro, K. Sensenig, J. Potkay, M. Agah, S. Scheuening, K. Wise, F. Dorman, and R. Sacks, "Design, Fabrication, and Evaluation of Microfabricated Columns for Gas Chromatography," Anal. Chem. 2004, 76, 2629.]

(a) Why would air be chosen as the carrier gas? What is the danger of using

air as carrier gas?

(b) Measure the optimum velocity and plate height for air and for carrier

gases.

(c) How many plates are there in the 3 -m-long column for each carrier gas at

optimum flow rate?

(d) How long does unretained gas take to travel through the column at

optimum velocity for each carrier gas?

(e) If stationary phase is sufficiently thin with respect to column diameter, which of the two mass transfer terms (23-40 or 23-41) becomes negligible?

Why?

(f) Why is the loss of column efficiency at high flow rates less severe for

than for air carrier gas?

Short answer

(a.) better resolution can be achieved with the application of ${\mathrm{H}}_2$. Using air as carrier gas can degrade the stationary phase because it contains traces of ${\mathrm{O}}_2,{\mathrm{H}}_2\mathrm{O}$, and organic compounds. The ideal carrier gas must be of high quality and should not contain the compounds mentioned.

$$\begin{gathered} {\mathrm{V}}_{{\mathrm{H}}_2}=7\mathrm{mm}\times \frac{50\mathrm{cm}/\mathrm{s}}{16.5\mathrm{mm}} \\ \left(\mathrm{b}.\right).=21.21\mathrm{cm}/{\mathrm{sV}}_{\mathrm{Air}}=3\mathrm{mm}\times \frac{50\mathrm{cm}/\mathrm{cm}}{16.5\mathrm{mm}} \\ =9.09\mathrm{cm}/\mathrm{s} \\ {\mathrm{N}}_{{\mathrm{H}}_2}=\frac{\mathrm{L}}{{\mathrm{H}}_{\mathrm{Air}}} \\ \left(\mathrm{c}\right)=\frac{3\mathrm{m}}{0.04\mathrm{cm}\times {\displaystyle \frac{1\mathrm{m}}{100\mathrm{cm}}}} \\ =7500 \\ {\mathrm{t}}_{{\mathrm{H}}_2}=\frac{3\mathrm{m}}{21.21{\displaystyle \frac{\mathrm{cm}}{\mathrm{s}}}\times {\displaystyle \frac{1\mathrm{m}}{100\mathrm{cm}}}} \\ \left(\mathrm{d}.\right)=14.14{\mathrm{st}}_{\mathrm{Air}}=\frac{3\mathrm{m}}{9.09{\displaystyle \frac{\mathrm{cm}}{\mathrm{s}}}\times {\displaystyle \frac{1\mathrm{m}}{100\mathrm{cm}}}} \\ 33.0\mathrm{s} \\ {\mathrm{C}}_{\mathrm{m}}=\frac{1+6\mathrm{k}+11{\mathrm{k}}^2}{24{(\mathrm{k}+1)}^2}\times \frac{{\mathrm{r}}^2}{{\mathrm{D}}_{\mathrm{m}}} \\ \left(\mathrm{e}.\right) \\ {\mathrm{C}}_{\mathrm{s}}=\frac{2\mathrm{k}}{3{(\mathrm{k}+1)}^2}\times \frac{{\mathrm{d}}^2}{{\mathrm{D}}_{\mathrm{m}}} \end{gathered}$$

(f.) The loss of column efficiency at high flowrates is less severe for than Air because the former can be run much faster than its optimal velocity with small depreciation in resolution. This is because solutes can diffuse more rapidly in ${\mathrm{H}}^2$than in Air.

Step-by-step-solution

List the given:

$$\begin{gathered} -\mathrm{T}=\mathrm{Temperature}={70}^{\circ } \\ -\mathrm{L}=\mathrm{Column}\mathrm{lengh}=3.0\mathrm{m} \\ -\mathrm{Thickness}\mathrm{of}\mathrm{stationary}\mathrm{phase}=1-2\mathrm{\mu m} \end{gathered}$$

Step-by-step solution

To finding the danger of using air as carrier gas

(a)

Based on the given plot,${\mathrm{H}}^2$ has a higher optimal velocity than Air. Thus, a faster separation can be achieved by using ${\mathrm{H}}^2$as carrier gas than Air. Furthermore, better resolution can be achieved with the application of ${\mathrm{H}}^2$ . Using air as carrier gas can degrade the stationary phase because it contains traces of ${\mathrm{O}}_{2,}{\mathrm{H}}_2\mathrm{O}$, and organic compounds. The ideal carrier gas must be of high quality and should not contain the compounds mentioned.

To Measure the optimum velocity and plate height for air and for carrier gases

(b).

To acquire the optimum velocity, get the minimum point of the curves for${\mathrm{H}}_2\left({\mathrm{V}}_{{\mathrm{H}}_2}\right)$and Air$\left({\mathrm{V}}_{\mathrm{Air}}\right)$then read the corresponding value in the x-axis.

Perform approximations by measuring the length of a known velocity in the graph and solve for the unknown values by measuring the length and applying ratio and proportion.

The length of a retention time

$$\begin{gathered} 50\mathrm{cm}/\mathrm{s}\mathrm{is}16.5\mathrm{mm}\left(=\mathrm{x}\mathrm{in}\mathrm{the}\mathrm{plot}\right).\mathrm{The}\mathrm{distance}\mathrm{of}{\mathrm{V}}_{{\mathrm{H}}_2}\mathrm{and}{\mathrm{V}}_{\mathrm{Air}}\mathrm{from}\mathrm{the}\mathrm{starting}\mathrm{point}\mathrm{are} \\ 7\mathrm{mm}\mathrm{and}3\mathrm{mm},\mathrm{respectively}. \end{gathered}$$

Compute for the corresponding values of optimum velocity of ${\mathrm{H}}_2$and Air by ratio and proportion:

$$\begin{gathered} {\mathrm{V}}_{{\mathrm{H}}_2}=7\mathrm{mm}\times \frac{50\mathrm{cm}/\mathrm{s}}{16.5\mathrm{mm}} \\ =21.21\mathrm{cm}/{\mathrm{sV}}_{\mathrm{Air}}=3\mathrm{mm}\times \frac{50\mathrm{cm}/\mathrm{s}}{16.5\mathrm{mm}} \\ =9.09\mathrm{cm}/\mathrm{s} \end{gathered}$$

To finding plates are there in the 3 -m-long column for each carrier gas at optimum flow rate

(c)

To acquire the number of plates$\left(\mathrm{N}\right)$, first get the plate height$\left(\mathrm{H}\right)$rom the

corresponding values in the y-axis based on the minimum point of the curves for

${\mathrm{H}}_2\left({\mathrm{H}}_{{\mathrm{H}}_2}\right)\mathrm{and}\mathrm{Air}\left({\mathrm{H}}_{\mathrm{Air}}\right).$

The length of a retention time of 0.05 cm is 7 mm. The distances of${\mathrm{H}}_{{\mathrm{H}}_2}\mathrm{and}{\mathrm{H}}_{\mathrm{Air}}$from the starting point are 7 mm and 5 mm, respectively.

Compute for the corresponding values of the plate height of ${\mathrm{H}}_2$and Air by ratio and

proportion:

localid="1655028203697" $$\begin{gathered} {\mathrm{H}}_{{\mathrm{H}}_2}=7\mathrm{mm}\times \frac{0.5\mathrm{cm}}{7\mathrm{mm}} \\ =0.05{\mathrm{cmH}}_{\mathrm{Air}}=5\mathrm{mm}\times \frac{0.05\mathrm{cm}}{7\mathrm{mm}} \\ =0.04\mathrm{cm} \end{gathered}$$

Calculate the number of plates of ${\mathrm{H}}_2\left({\mathrm{N}}_{{\mathrm{H}}_2}\right)$ :

$$\begin{gathered} {\mathrm{N}}_{{\mathrm{H}}_2}=\frac{\mathrm{L}}{{\mathrm{H}}_{{\mathrm{H}}_2}} \\ =\frac{3\mathrm{m}}{0.05\mathrm{cm}\times {\displaystyle \frac{1\mathrm{m}}{100\mathrm{cm}}}} \\ =6000 \end{gathered}$$

Calculate the number of plates of Air$\left({\mathrm{N}}_{\mathrm{Air}}\right)$:

$$\begin{gathered} {\mathrm{N}}_{{\mathrm{H}}_2}=\frac{\mathrm{L}}{{\mathrm{H}}_{Air}} \\ =\frac{3\mathrm{m}}{0.04\mathrm{cm}\times {\displaystyle \frac{1\mathrm{m}}{100\mathrm{cm}}}} \\ =7500 \end{gathered}$$

unretained gas take to travel through the column at optimum velocity for each carrier gas

(d).

To compute for the time of unretained ${\mathrm{H}}_2\left({\mathrm{t}}_{{\mathrm{H}}_2}\right)\mathrm{and}\mathrm{Air}\left({\mathrm{t}}_{\mathrm{Air}}\right)$to travel through the

column at optimum velocity, divide the length of the column by the velocity of each

carrier gas:

$$\begin{gathered} {\mathrm{t}}_{{\mathrm{H}}_2}=\frac{3\mathrm{m}}{21.21{\displaystyle \frac{\mathrm{cm}}{\mathrm{s}}}\times {\displaystyle \frac{1\mathrm{m}}{100\mathrm{cm}}}} \\ =14.14{\mathrm{st}}_{\mathrm{Air}}=\frac{3\mathrm{m}}{9.09{\displaystyle \frac{\mathrm{cm}}{\mathrm{s}}}\times {\displaystyle \frac{1\mathrm{m}}{100\mathrm{cm}}}} \\ =33.0 \end{gathered}$$

To finding the  two mass transfer terms (23-40 or 23-41) becomes negligible

(e).

For a thin stationary phase ,($^0.5\mathrm{\mu m}$) ,the mass transfer is dominated by slow diffusion through the mobile phase$(\mathrm{C}{\mathrm{C}}_{\mathrm{m}})$( than through the stationary phase $\left({\mathrm{C}}_{\mathrm{S}}\right)$Thatis, ${\mathrm{C}}_{\mathrm{s}}<<{\mathrm{C}}_{\mathrm{m}}$in Equations 23-40 and 23-41.

The expressions for ${\mathrm{C}}_{\mathrm{m}}$and ${\mathrm{C}}_{\mathrm{S}}$are shown below:

${\mathrm{C}}_{\mathrm{m}}=\frac{1+6\mathrm{k}+11{\mathrm{k}}^2}{24{(\mathrm{k}+1)}^2}\times \frac{{\mathrm{r}}^2}{{\mathrm{D}}_{\mathrm{m}}}$

where:

$$\begin{gathered} -\mathrm{K}=\mathrm{retention}\mathrm{factor} \\ -\mathrm{r}=\mathrm{column}\mathrm{radius} \\ -\mathrm{d}=\mathrm{thickness}\mathrm{of}\mathrm{stationary}\mathrm{phase} \\ -{\mathrm{D}}_{\mathrm{m}}=\mathrm{diffusion}\mathrm{coefficient}\mathrm{of}\mathrm{solute}\mathrm{in}\mathrm{mobile}\mathrm{phase} \\ -{\mathrm{D}}_{\mathrm{s}}=\mathrm{diffusion}\mathrm{coefficient}\mathrm{of}\mathrm{solute}\mathrm{in}\mathrm{stationary}\mathrm{phase} \end{gathered}$$

By examining the expressions, we can observe that ${\mathrm{C}}^{\mathrm{S}}$is directly proportional

to the thickness of the stationary phase. Thus, this term becomes negligible as compared

to ${\mathrm{C}}_{\mathrm{m}}$when the stationary phase is sufficiently thin with respect to column diameter.

Finding  the loss of column efficiency at high flow rates less severe for H2 than for air carrier gas

(f).

The loss of column efficiency at high flowrates is less severe for ${\mathrm{H}}_2$than Air because the former can be run much faster than its optimal velocity with small depreciation in resolution. This is because solutes can diffuse more rapidly in ${\mathrm{H}}_2$than in Air.