Chemistry extraction lab question

Question

I recently did an extraction experiment in my organic chemistry class and I need help answering a question. We did a liquid liquid extraction and in the first trial we did, we used 1.6 mL of each water and dichloromethane. Then we did the same experiment but only using 0.7 mL of each water and dichloromethane. my question is, which would produce a better result? I got better results for the trial using 0.7mL each but was wondering if maybe their was just more error in the 1.6mL trial?

Answer 1

Hi Edgar C.,

I'm greatly pleased to find a student thinking about the causes of processes and sources of error in an experiment!

Let's consider. To a first approximation, the partition of the target substance ("analyte") follows the relationship: (K=equilibrium constant however defined, C = concentration)

K(w)

-------------- = constant

K(DCM)

and since your volumes of each solvent are equal in each trial, the C(w) / C (DCM) is also a constant .

(This assumes that the substance is not saturated in either phase, in either trial.)

But, when you say, "better result", exactly what did you mean by that? And what sort of "error" are you thinking about -- pipetting volume error, quantitation of analyte error ....? Can't guess, from your level of detail stated ....

If you're trying to isolate a somewhat volatile analyte, that might co-distill with its solvent (e.g. nicotine in water), obviously that would be a situation in which you'd want to minimize the volume of solvent to be removed.

And, if you're trying to recover the analyte, by a subsequent liquid:liquid partition, definitely minimize the volumes, otherwise you lose it in liters of extracts (been there!).

That's it for a first approximation. Now, to a SECOND approximation, you might need to consider how the solution of the analyte affects the partitioning ability of the solvent. And this can be a rather complex subject. I recall a case, from my research career, in which I determined the solubility of several of the methylxanthines in the presence of each other. Within the group of caffeine, theophylline, and theobromine, one of them enhanced the solubility of one of the others in water, but suppressed the solubility of the third (I forget now which was which, could look it up if you're really interested). Generally this would work through bulk polarity changes in the solvent due to the analyte, but occasionally local effects (hydrogen-bonded dimer aggregation) might operate, too.

i'd be happy to offer further help, if you want.

-- Cheers, -- Mr. d.