Question

3.The partition coefficient of Compound A is 7.5 in dichloromethane (a.k.a. methylene chloride) with respect to water.

a- If 5 grams of Compound A were dissolved in 100 mL of water, how much of Compound A would be extracted with four 25-mL portions of dichloromethane?

Extraction 1:

Extraction 2:

Extraction 3:

Extraction 4:

b- a.After four extractions, what percentage of the Compound A has been extracted (use percent recovery)?       

c- a.You are tired and want to finish your experiment as quickly as possible. How much dichloromethane would be required in one extraction to remove the same percentage of Compound A that you recovered after four 25-mL portions of dichloromethane?   

Answer 1

                                                             

                                                    solubility in solvent 1

Kc = partition coefficient =      ‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑­

                                          solubility in solvent 2

               

   The partition coefficient Kc   of compound A in dichloromethane = 7.5

Extraction 1 :   Kc = 7.5 = (A g/25 mL Dichloromethane)/((5 –A)g/100 mL water)

                                                7.5 (5-A) = 4 A

                                                37.5 – 7.5A = 4A

                                                A = 37.5/11.5 = 3.26 g     

After first extraction A= 3.26 g of compound A is extracted into dichloromethane.

Extraction 2 : we have remaining compound A in water = 5.0 -3.26 = 1.74 g

Kc = 7.5 = (B g/25 mL Dichloromethane)/((1.74 –B)g/100 mL water)

                                                7.5 (1.74-B) = 4 B

                                                13.05 – 7.5B = 4B

                                                B = 13.05/11.5 = 1.135 g     

In Second extraction B= 1.135 g of compound A is extracted into dichloromethane.

Total compound extracted from first and second extraction = Extraction 1 + Extraction 2 = A+ B

                                                                                                                = 3.26 + 1.135 g    = 4.4 g

Extraction 3 : we have remaining compound A in water = 5.0 -4.4 = 0.6 g

Kc = 7.5 = (C g/25 mL Dichloromethane)/((0.6 –C)g/100 mL water)

                                                7.5 (0.6-C) = 4 C

                                                4.5 – 7.5C = 4C

                                                C = 4.5/11.5 = 0.391 g     

In third extraction C= 0.391 g of compound A is extracted into dichloromethane.

Total compound extracted from three extractions = A+ B+C

                                                                        = 3.26 + 1.135 g + 0.391 g        

= 4.786 g

Extraction 4 : we have remaining compound A in water = 5.0 -4.786 = 0.214 g

Kc = 7.5 = (D g/25 mL Dichloromethane)/((0.214 –D)g/100 mL water)

                                                7.5 (0.214-D) = 4 D

                                                1.605 – 7.5D = 4D

                                                D = 1.605/11.5 = 0.139 g     

In fourth extraction D= 0.139 g of compound A is extracted into dichloromethane.

Total compound extracted from four extractions = A+ B+C+D

                                                                       = 3.26 + 1.135 g + 0.391 g +0.139 g             

= 4.925 g

b) The percent of compound recovered after four extractions =( total recovered / total compound)x 100

                                                                                                                = (4.925 g/5.0 g) x100 = 98.5%

c) now, we need to find the volume of dichloromethane is required to extract 4.925 g compound in a single extraction.

   Kc = 7.5 = (4.925 g/V mL Dichloromethane)/((5 –4.925)g/100 mL water)

                                                7.5 x 0.075 = 4.925 x 100/V

                                                V = 4.925x100/0.5625

                                                V= 875 mL     

The volume required to extract 98.5% of compound is = 875 mL