Question

Suppose you have an organic sample X that is somewhat soluble in water, even though it is

somewhat more soluble in dichloromethane or ether solvents. But if you do a single

extraction, you get only 60% of your material to transfer from the water to the organic layer.

a) How many “washes” would it take to extract over 90% of your organic material from

the water layer?

b) In this case, would it be better to extract with dichloromethane or with ether?

Assume you changed the solvent to ethyl acetate which has a K_D for caffeine of 10. How much caffeine would you extract after three extractions? Show your work.

Answer 1

Extraction of a sample X with an organic solvent is 60%.

Therefore, it would take at least "three washes" with an organic solvent to extract over 90% of the organic compound.

Calculation: First time results in the extraction of 60%. It can be considered that the remaining (40%) is in the aqueous portion.

Second time: Starting point (40% of the initial amount). So, 60% of the remaining 40% is 24%. Therefore, the extractive is 24%. When this is added to the first time product it is approximately 84%.

Third time: Starting point (16% of the initial amount). So, 60% of the remaining 16% is 9.6%. When this is added to the 1st and 2nd extractives, it is 93.6%.

Therefore, it takes at least three washes to extract over 90% of the sample X.

b) It cannot be estimated from the given data as to which organic solvent has higher extraction capacity as the polarity of the sample is not known.

c) The extractive of caffeine by three washes of ethyl acetate solvent would be 30 mol% of caffeine which equals 5.8g