Question

If 1.2 g of aspirin were dissolved in 100 mL of water, how much aspirin might you be able to recover if you were to extract the aqueous solution with 100 mL of dichloromethane? Also calculate the percent recovery.

If instead of extracing the aqueous layer with 100 mL of dichloromethane, what would have been teh percent recovery if you had used two sucessive 50 mL protions of dichloromethane?

Answer 1

To answer this question we need to know the solubility of aspirin in both water and dichloromethane.

The data shows the solubility of aspirin is 3 mg/mL in water at ambient conditions.

and the solubilty in dichloromethane = 59 mg/mL (These solubilty data need to be verified from reliable scientific source)

By definition, the partion coefficient K = S_DCM/S_Water

By substituting the solubilty values,

we get K = 59/3 = 19.67

Now, 1.2 g dissolved in 100 mL water and extracted with 100 mL DCM.

If the distribution of aspirin in water is Aw and in DCM is Ad

then Ad/Aw=19.67

and Ad+Aw=1.2, or 19.67Aw+Aw=1.2 or 20.67Aw=1.2

So, Aw=1.2/20.67 = 0.058 g

Therefore Ad = (1.2-0.058) g = 1.142 g will be recovered by single extraction with 100 mL of Dichloromethane.

So the recovry = 1.142/1.2*100% = 95.16%

If the extraction was done in two 50 mL portions then,

1st extraction with 50 mL DCM and 100 mL water

Ad/Aw=19.67/2=9.835

Again Ad+Aw=1.2 g or 9.835Aw + Aw =1.2 g.

Aw = 1.2/10.835 g = 0.11 g

Ad = 1.2- 0.11 = 1.09 g

For the 2nd extraction with 50 mL of DCM

Ad/Aw=19.67/2=9.835

Ad+Aw = 0.11 g (Remaining part in aqueous)

9.835Aw + Aw = 0.11, or Aw = 0.11/10.835 = 0.01 g

Ad = 0.10 g

So, the total recovery = 1.09 + 0.10 =1.19 g or 1.19/1.2*100% = 99.16%