Question

A chromatographic procedure separatesof unknown mixture on a column with a length ofand a diameter of. (a) What size column would you use to separateof the same mixture? (b) If the flow ison the small column, what volume flow rate should be used on the large column? (c) The mobile phase occupiesof the column volume. Calculate the linear velocity for the small column and the large column.

Short answer

a) The size of the column would be 40cmwith a diameter of 4.25cm

b) The volume flow rate = 5.5mL/min

c) The linear velocity = 1.11cm / min

Step-by-step solution

Define Retention Factor:

The retention factor, also known as the capacity factor (k), in column chromatography is defined as the ratio of time an analyte is retained in the stationary phase to time the analyte is retained in the mobile phase.

Given Information:

${\mathrm{m}}_1=4\mathrm{mg},\mathrm{I}=40\mathrm{cm},{\mathrm{d}}_1=0.85\mathrm{cm}$

To find the size of the two columns:

To compare the size of the two columns are

$$\begin{gathered} \frac{{\mathrm{m}}_2}{{\mathrm{m}}_1}={\left(\frac{{\mathrm{d}}_2}{{\mathrm{d}}_1}\right)}^2 \\ \frac{100\mathrm{mg}}{4\mathrm{mg}}={\left(\frac{{\mathrm{d}}_2}{0.85\mathrm{cm}}\right)}^2 \\ 25=\frac{{\mathrm{d}}_2^2}{0.7225{\mathrm{cm}}^2} \\ {\mathrm{d}}_2^2=18.06{\mathrm{cm}}^2 \\ {\mathrm{d}}_2^2=\sqrt{18.06{\mathrm{cm}}^2} \\ {\mathrm{d}}_2=4.25\mathrm{cm} \end{gathered}$$

The size of the column would be 40cmwith a diameter of 4.25cm

To find the volume flow rate:

If the volume flow is 0.22mL /min on the smaller column, to determine the volume flow rate should be used on the larger column

$$\begin{gathered} \frac{{\mathrm{m}}_2}{{\mathrm{m}}_1}=\frac{{\mathrm{u}}_2}{{\mathrm{u}}_1} \\ \frac{100\mathrm{mg}}{4\mathrm{mg}}=\frac{{\mathrm{u}}_2}{0.22\mathrm{mL}/\min } \\ 25=\frac{{\mathrm{u}}_2}{0.22\mathrm{mL}/\min } \\ 25\times 0.22\mathrm{mL}/\min ={\mathrm{u}}_2 \\ 5.5\mathrm{mL}/\min ={\mathrm{u}}_2 \end{gathered}$$

To find the volume of a small column and linear flow rate:

To find the volume of a small column as

$$\begin{gathered} {\mathrm{V}}_1={\mathrm{\tau \tau }}^2\times \mathrm{I} \\ {\mathrm{V}}_1=\mathrm{\tau \tau }\times {\left(0.85/2\mathrm{cm}\right)}^2\times 40\mathrm{cm} \\ {\mathrm{V}}_1=22.7\mathrm{mL} \end{gathered}$$

Volume of a mobile phase as

$$\begin{gathered} {\mathrm{V}}_{\mathrm{m}}=0.35\times 22.7\mathrm{mL} \\ {\mathrm{V}}_{\mathrm{m}}=7.94\mathrm{mL} \end{gathered}$$

To calculate the linear flow

$$\begin{gathered} {\mathrm{u}}_{\mathrm{x}}=\frac{\mathrm{I}}{{\mathrm{V}}_{\mathrm{m}}\times {\mathrm{u}}_1} \\ {\mathrm{u}}_{\mathrm{x}}=\frac{40\mathrm{cm}}{7.94\mathrm{mL}\times 0.22\mathrm{mL}/\min } \\ {\mathrm{u}}_{\mathrm{x}}=1.11\mathrm{cm}/\min \end{gathered}$$