3.The partition coefficient of Compound A is 7.5 in dichloromethane (a.k.a. meth

Question

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

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

c.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?

Explanation / Answer

Solution

Formula to calculate the amount of the compound extracted in each extraction is as follows

Kd = (x/ml of dichloromethane)/((5-x)/ml of water)

Lets put the values in the formula

First extraction with 25 ml

7.5 = (x/25)/((5-x)/100)

Solving for x we get

x= 3.26 g

after removing 3.26 g remaining compound in 100 ml water is 5g – 3.26 g =1.74 g

calculating amount extracted in second extraction

Kd = (x/ml of dichloromethane)/((5-x)/ml of water)

7.5= (x/25)/((1.74-x)/100)

X= 1.135 g

So the amount left in water is 1.74 g – 1.135 g = 0.605 g

Third extraction

Kd = (x/ml of dichloromethane)/((5-x)/ml of water)

7.5= (x/25)/((0.605-x)/100)

X= 0.3945 g

Amount left in water = 0.605 g – 0.3945 g = 0.2105 g

Fourth extraction

Kd = (x/ml of dichloromethane)/((5-x)/ml of water)

7.5= (x/25)/((0.2105-x)/100)

X= 0.137 g

Amount extracted in each extraction is as follows

Extraction 1 = 3.26 g

Extraction 2 = 1.135 g

Extraction 3 = 0.3945 g

Extraction 4 = 0.137 g

Total amount extracted = 3.26 g +1.135 g + 0.3945 +0.137 = 4.927 g

So the percent of the compound recovered =(amount recovered / total amount) *100%

                                                                                = (4.927 g / 5.0 g)*100%

                                                                                = 98.54 %

Part C volume of dichloromethane needed to extract 4.927 g compound in 1 extraction

Kd = (x/ml of dichloromethane)/((5-x)/ml of water)

7.5 = (4.927 /ml of dichloromethane)/((5-4.927)/100 ml)

X= 900 ml

Thereofre the amount of the cichloromethane needed to extract 4.927 g compound in one extraction is 900 ml dichloloromethane

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