Ratio of concentration = Ratio of solubility

The added solvent must be more volatile than the desired component.

It must also specifically dissolve the desired component.

The component must have a greater tendency to dissolve in the added solvent than in the solution.

KD= CO = SO =

CH20 SH20

A substance X can be isolated from its plant source by solvent extraction. However, a minor component Y has an appreciable solubility in the solvents that may be used. Given below are the solubilities of X and Y in different solvents:Solvent T, ˚C Solubility in 100g solvent at 28˚C

X Y

Ethyl methyl ketone 80 6 5

Cyclohexane 81 8 2

Benzene 80 5 1.8

CCl4 78 8.75 1.25

Water 100 2 1

a. Which is the best extracting solvent? CCl4 is the best extracting solvent

it dissolves only a little amount of Y relative to the X component needed.

It also dissolves the greatest amount of X among the choices.

It is immiscible with water, which happens to be the solvent in the solution.

◦ Mathematically speaking, the choice relies on the KD and the best result is the one with the highest KD. The KD values with respect to water as calculated are the following:

◦ From this table, it is evident that CCl4 is the yields the highest ratio, ergo the best possible choice as the extracting solvent.

Solvent KD

Ethyl methyl ketone 3

Cyclohexane 4

Benzene 2.5

CCl4 4.375

b. Given a saturated aqueous solution of X and Y and using 100mL of solvent in (1), determine the percent recovery of X in a single extraction.

KD = SCCl4 = 8.75g/100g = 4.375

SH20 2.00g/100g

KD = CCCl4 = Xg/100ml = 4.375

CH20 (2.00g-Xg)/100ml

X = 1.63g

%rec = Xpure x 100% = 1.63g x 100% = 81.4%

Ximpure 2.00g

c. Repeat (b) using 50mL of solvent in each of the two successive extractions. Determine the percent recovery and compare this with (b).

KD = SCCl4 = 8.75g/100g = 4.375 SH20 2.00g/100g First extraction:KD = CCCl4 = Xg/50ml = 4.375 CH20 (2.00g-Xg)/100ml

X = 1.37g 

Second extraction:KD = CCCl4 = Xg/50ml = 4.375

CH20 (0.63g-Xg)/100ml

X = 0.43g %rec = Xpure x 100%= (1.37+0.43)g x 100%= 90.0%

Ximpure 2.00g

 In the double extraction, the percent recovery is comparably

higher, considering the same amount of the solvent added.

d. What is the percent recovery of the minor component in a single extraction using 100mL of the solvent in (a)?

 KD = SCCl4 = 1.25g/100g = 1.25 SH20 1.00g/100g

KD = CCCl4 = Yg/100g = 1.25 CH20 (1.00g-Yg)/100g Y = 0.56

%rec = Ypure x 100% = 0.56g x 100% = 56% Yimpure 1.00g