Question

How many grams of compound A will be recovered for the following case: There are 54.6 mL of water containing 4.7 of compound A. You decide to perform 1.8 extractions with 33.0 of dichloromethane. The distribution coefficient is 5.2.

For this I got 4.20 g, but I feel as if I did it wrong.

Answer 1

V = 54.6 mL of water

m = 4.7 g of A

m = 33 g of dichlorometahe

K = 5.2

5.2 = phase 1 / phase 2

The formula:

q = [V1/(V1+K*V2) ] ^n

where

q = fraction of solute presnet after extraction

V1 = phase 1 volume; V2 = phase 2 volume; K patition coefficient, n = number of paritions

then

q = [V1/(V1+K*V2) ] ^n

q = [54.6/(33+5.2*54.6) ] ^1.8

q = 0.04219

fraction left in aqueous phase = 0.04219

mass -> q*M = 4.7*0.04219 = 0.198293 g left in aqueous

mass in organic =4.7- 0.198293 = 4.5017 g in organic