Question

You are given a mixture containing two compounds, A and B. Both compounds have a solubility of 1 g/ 100 mL of solvent at 20 °C and 16 g/ 100 mL of solvent at 100 °C. The sample is composed of 3.5 g of A and 10 g of B. At 100 °C the entire sample just dissolves in a minimum amount of solvent. The solution is cooled to 20 °C and crystals are collected.

Calculate the composition of the crystals and the yield of the process.

What is the composition of the mother liquor?

If the crystals obtained in question (1) are recrystallized from 100 mL of solvent, what will be the yield and composition of the crystals obtained?

Answer 1

First, calculate the minimum volume of solvent needed to dissolve both at 100 oC. Because B has the greater mass, the volume needed will be dictated by its solubility.

16 g B/100 mL = 10 g B / x mL
x mL = (10/16)*(100) = 62.5 mL

"The solution is cooled to 20oC and crystals are collected. "

At 20 oC we will have in solution:

1 g A/100 mL = x g A/62.5 mL

x g A = (1/100)*62.5 = 0.625 g A in solution

1 g B/100 mL = x g B/62.5 mL

x g B = (1/100)*62.5 = 0.625 g B in solution

"The sample is composed of 3.5 g of A and 10 g of B"

That means we will have these amounts of solid A and B in the crystals isolated:

3.5 g of A - 0.625 g A in solution = 2.875 g of solid A

10 g of B - 0.625 g B in solution = 9.375 g of solid B

(1) the composition of the crystals

Total solid isolated = 2.875 g of A + 9.375 g of B = 12.25 g of solid

% A = 100% * (2.875 g A/12.25 g) = 23.47 % A

% B = 100% * (9.375 g of B/12.25 g) = 76.53%B


(2) the yield of the process.

12.25 g of solid was isolated. I'm not sure whether that's the answer wanted.

% yield = 100% * (12.25 g of solid)/(3.5 g + 10 g) = 90.74 % yield

(3) "What is the composition of the mother liquor

From above,1 g A/100 mL = x g A/62.5 mL

x g A = (1/100)*62.5 = 0.625 g A in solution

1 g B/100 mL = x g B/62.5 mL

x g B = (1/100)*62.5 = 0.625 g B in solution

So the mother liquor contains 0.625 g of A and 0.625 g of B, both 1% (w/v) (the original given, 1 g/100 mL at 20oC = 1% (w/v))