4
$\begingroup$

In chromatography, a mobile phase with analytes is passed over a stationary phase. A given analyte has different solubility in the two phases and consequently spends different periods of time in them. A particular molecule while in the stationary phase remains immobile, is swept along when present in the mobile phase. $T_m$, the time spent in the mobile phase, is the same for all analytes. The reason (I'm not so sure) must be since all analytes travel the same distance finally, with the same speed as the mobile phase, so take the same time in so doing. But since some of these spend longer times in the stationary phase than others, the total time to reach the end, that is, $T_m + T_r$ is different ($T_r$ is the time spent in the stationary phase).

But I have a bit of a trouble understanding the rate aspect of the process, and what exactly determines the $T_r$ and $T_m$ - are they the average times spent by an arbitrary molecule in the two phases? If so, are their values dependent on the rate of the process, or the solubilities?

$\endgroup$
1
$\begingroup$

Think of it like a solvent extract. Each solvent extraction is a plate.

So at time 1 I do extraction 1 and move top layer to extraction flask 2.

At time 2 I have two extractions to do. I extract flask 2 and move top layer to flask 3 and then I extract 1 again and move top layer to 2 again.

At time 3 I have three extractions to do. and so one.


Another analogy.

Think of the whole thing as binomial trials. Each molecule has a different binomial probability. So the absolute difference in the peak positions between the molecules gets bigger as the number of trials increases.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.