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.)How many steps of crystallization you needed to do to get pure compound A? b.)What is the amount of pure compound A after all crystallization steps?

Answer 1

132 mg of compound A at 100 ^oC 110 mg/mL is soluble

110 mg/mL = 132mg /x mL

x = 132/110 = 1.2 mL

132 mg dissolves in 1.2 mL

When cooled to 25 ^oC the solubility is 14.0 mg/mL

so 14.0 mg/mL = x mg/1.2mL = 14 x 1.2 = 16.8 mg is soluble so the remaining 132-16.8 mg = 115.2 mg cystallizes out

In the same 1.2mL water the impurity B which has same solubility parameter so in 1.2 mL16.8 mg remains soluble so 24 - 16.8 = 7.2 mg crystallizes out in the first round.

In the second step 115.2 mg A and 7.2 mg B are subject to solubility at 100 ^oC

now we need 115.2/110 = 1.05 mL of water

When this is cooled to 25 ^oC the amount that crystallizes out is

115.2 mg-(14 mg/mL x 1.05 mL) = 100.5 mg

Simultaneously compound B 7.2 mg nothing will crytallize out as its solubility in 1.05 mL is 14 x 1.05 = 14.7 mg

So we need two crystallization step to get pure compound A and 100.5 mg of pure compound A is obtained.