Question

A student is extracting a caffeine/water solution with dichloromethane. The K value is 4.6. The student starts with a total of 70 mg of caffeine in 2 mL of water and decides to use three extracts. How much caffeine will be in each dichloromethane extract? What is the percent recovery of caffeine?

Answer 1

Partition coefficeint = k = (concentration of solute in solvent 1)/(concentration of solute in solvent 2) = 4.6
                           = C1/C2 (where C1 = concentration of caffeine in dichloromethane and C2 = concentration of caffeine in water)

70 mg of caffeine dissolved in 2.0 mL of water.
Suppose, we have extracted three times with 1.0 mL of dichloromethane.

First Extraction:

k = 4.6 = C1/C2 = {(70-x1)/1.0} mg/mL / (x1/2.0 ) mg/mL

70 - x1 = 2.3 x1
x1 = 21.21 mg
Therefore, after first extraction, we have 21.21 mg of caffiene remaining in water and (70-21.21) mg = 48.79 mg of caffeine present in dichloromethane fraction.

Second extraction:

4.6 = C1/C2 = {(21.21-X2 )/1.0} mg/mL / (X 2 /2.00) mg/mL

21.21-x2 = 2.3 x2
x2 = 6.43 mg remaining in water.
In the second extraction dichloromethane contains (21.21 - 4.43) mg = 14.78 mg of caffeine..

Third fraction:

4.6 = C1/C2 = {(6.43-X3 )/1.0} mg/mL / (X 3 /2.00) mg/mL

6.43-x3 = 2.3 x3
x3 = 1.95 mg remaining in water.
In the third extraction dichloromethane contains (6.43 - 1.95) mg = 4.48 mg of caffeine.

The percent of recovery = (total caffiene extracted in dichloromethane x 100)/total amount of caffeine = (48.79+14.78+4.48)*100/70 = 97.2 %