(a) You wish to use a wide gradient from 5 vol % to 95 vol %B for the first separation of a mixture of small molecules to decide whether to use gradient or isocratic elution. What should be the gradient time, ${\mathbf{t}}_{\mathbf{G}}$,for a $\mathbf{15}\mathbf{\times }\mathbf{0}\mathbf{.}\mathbf{46}\mathbf{-}\mathbf{cm}$column containing $\mathbf{3}\mathbf{-}\mathbf{\mu m}$particles with a flow of $\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{mL}\mathbf{/}\mathbf{min}$

You optimized the gradient separation going from 20vol%to 34 vol% Bin 11.5min at 1.0mLFind$\mathbf{k}\mathbf{*}$for this optimized

separation. To scale up to a $\mathbf{15}\mathbf{\times }\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{cm}$column, what should be the gradient time and the volume flow rate? If the sample load on

the small column was 1mgwhat sample load can be applied to the large column? Verify that $\mathbf{k}\mathbf{*}$is unchanged.

t_G the gradient time (the time from the start to the end of the gradient), V_M the column hold-up volume and F the mobile phase flow rate. The gradient steepness parameter will vary with the S-value, which is only roughly constant for similar compounds.

We assume thatk* is 5 and S is 4 .The volume at which solvent front appear is:

$$\begin{gathered} {\mathrm{V}}_{\mathrm{m}}=\frac{\mathrm{L}.{\mathrm{d}}_{\mathrm{c}}^2}{2} \\ {\mathrm{V}}_{\mathrm{m}}=\frac{15\mathrm{cm}.{\left(0.46\mathrm{cm}\right)}^2}{2} \\ {\mathrm{V}}_{\mathrm{m}}=1.587\mathrm{mL} \end{gathered}$$rr

We can calculate${\mathrm{t}}_{\mathrm{G}}$using the following formula:

${\mathrm{t}}_{\mathrm{G}}$is

role="math" localid="1665031412999" $$\begin{gathered} \mathrm{k}*=\frac{{\mathrm{t}}_{\mathrm{G}}.\mathrm{F}}{∆\mathrm{\Phi }.{\mathrm{V}}_{\mathrm{m}}.\mathrm{S}} \\ {\mathrm{t}}_{\mathrm{G}}=\frac{\mathrm{k}*.∆\mathrm{\Phi }.{\mathrm{V}}_{\mathrm{m}}.\mathrm{S}}{\mathrm{F}} \\ {\mathrm{t}}_{\mathrm{G}}=\frac{5.0.9.1.587\mathrm{mL}.4}{1\mathrm{mL}/\min } \\ {\mathrm{t}}_{\mathrm{G}}=28.6\min \end{gathered}$$

t_G the gradient time (the time from the start to the end of the gradient), V_M the column hold-up volume and F the mobile phase flow rate. The gradient steepness parameter will vary with the S-value, which is only roughly constant for similar compounds.

Volume flow rate is the same as "Flux" and "R". There are two different equations used to express volume flow rate, it can be expressed as Volume/Time.

k* for optimized separation is:

$$\begin{gathered} \mathrm{k}*=\frac{{\mathrm{t}}_{\mathrm{G}}.\mathrm{F}}{∆\mathrm{\Phi }.{\mathrm{V}}_{\mathrm{m}}.\mathrm{S}} \\ \mathrm{k}*=\frac{11.5\min .1\mathrm{mL}/\min }{0.14.1.587\mathrm{mL}.4} \\ \mathrm{k}*=12.9 \end{gathered}$$

If column is scale up to a$15\times 1.0\mathrm{cm}$the column has the same length as the previous column, but the diameter is increased from 0.46 to 1.0cm So, the volume increases by a factor of:

${\left(\frac{1}{0.46}\right)}^2=4.7$

The flow of rate also is increased by the same factor:

The gradient time is unchanged at 11.5min

The sample loading is also increased by factor 4.7:

4.7.1mg = 4.7 mg

Verification that k* is unchanged:

$$\begin{gathered} {\mathrm{V}}_{\mathrm{m}}=\frac{\mathrm{L}.{\mathrm{d}}_{\mathrm{c}}^2}{2} \\ {\mathrm{V}}_{\mathrm{m}}=\frac{15\mathrm{cm}.{\left(1\mathrm{cm}\right)}^2}{2} \\ {\mathrm{V}}_{\mathrm{m}}=7.5\mathrm{mL} \\ \mathrm{k}*=\frac{{\mathrm{t}}_{\mathrm{G}}.\mathrm{F}}{∆\mathrm{\Phi }.{\mathrm{V}}_{\mathrm{m}}.\mathrm{S}} \\ \mathrm{k}*=\frac{11.5\min .4.7\mathrm{mL}/\min }{0.14.7.5\mathrm{mL}.4} \\ \mathrm{k}*=12.9 \end{gathered}$$

k* is the same as in the previous column.