Question

crystallization problem:

A sample contains 132 mg of compound A along with 24.0 mg of impurity B. The solubility of compound A in water at 25 ^oC is 14.0 mg/ml, and at 100 ^oC is 110 mg/ml. The impurity B has the same solubility behavior as compound A in water.

a.)By showing all the calculations, describe how you would purify the sample.

b.)How many steps of crystallization you needed to do to get pure compound A?

c.)What is the amount of pure compound A after all crystallization steps?

d.)Calculate the percent recovery of pure compound A after all the crystallization steps.

Answer 1

basis : 1 ml water.

As temperature is decreases, solubility of A decreases when temperature is decreased.

When temperature is reduced amount of A precipitates out =(110-14) mg/mL=96 mg/ml and all the B gets

precipitated out.

Then there is 132 mg of sample and 132-96= 56 mg of Pure A is left.

1 ml of water crystalizes s 96 mg of A when cooled from 100 deg,c to 25 deg,c

1. First instnace weight 96 gm of sample and cool it from 100 deg,c to 25 deg,c which dissolves all the A and B

2 The remaining A (only A)= 56 mg needs to be dissolved in water.

for crystaliing remaining 56 mg of pure A water reqiured= 56/96=0.583 ml

Two steps of crystallization are required for getting pure A

c) pure compound A 56 mg

d) percent recovery =100*56/132=42.425