Question

The distribution coefficient for caffeine in dichloromethane and water is 4.6

a. Assume that 0.100 g of caffeine are dissolved in 15 mL of water. The aqueous layer is extracted twice with 15 mL of dichloromethane. How much caffeine may be recovered from this extraction?

b. Would the previous extraction be more efficient if the aqueous solution of caffeine (0.100 g dissolved in 15 mL of water) was extracted once with 30 mL of dichloromethane instead?

Please provide step by step solutions and clear answers (I am terrible at algebra). Thank you so much!

Answer 1

W= weight of caffeine extracted

K= distribution coefficien t= 4.6= Concentration of caffeine in dischloromethane/ concentration of caffeine in water

stage 1

4.6= (W/15)/ (0.1-W)/15

4.6 =W/(0.1-W)

4.6*(0.1-W)= W

0.46 = 5.6W

W= 0.46/5.6 =0.082

Caffeine left =1-0.082= 0.018

for the second stage

4.6 = W/15/(0.018-W)/15

4.6 = W/(0.018-W)

4.6*(0.018-W)= W

5.6W = 4.6*0.018

W= 0.015

caffeine remaining =0.018-0.015=0.003

b) when 30ml is used at a stretch

4.6 =(W/30)/ (0.1-W)/15

4.6 = W/(0.2-2W)

4.6*(0.2-2W)= W

4.6*0.2/10.2 =W

W= 0.090

remaining caffeine =0.1-0.090 =0.01 gm

W= 0.46/3.6=0.13