Question

This is a post-lab question of the Recrystallization of Aspirin experiment.

This is related to organic chemistry. Please be detailed, thank you!

Given the table for solubility of compounds X and Y in solvent Z. If 2.00 grams of impure compound X (contaminated by 12.0% Y by weight) is recrystallized from 25.0mL of solvent Z, what will be the purity of recrystallized X?

Solubility per 100 g of Solvent Z
Compound	20℃	80℃
X	1.0 g	8.0 g
Y	0.70 g	6.0 g

Density of solvent Z – 0.90 g/mL

Answer 1

Mass of impurity Y = 2.00 g x 12.0/100 = 0.240 g

Therefore, the mass of X = (2.00 - 0.240) g = 1.76 g

Volume of solvent Z used (V) = 25.0 mL

Density of solvent Z (d) = 0.90 g/mL

Thus, the mass of solvent used = V x d = 25.0 mL x 0.90 g/mL = 22.5 g

Now, at 80 ^oC, 100 g of solvent Z dissolves 8.0 g of compound X

Therefore, at 80 ^oC, 22.5 g of solvent Z dissolves 8.0 g x 22.5 g/100 g = 1.80 g of compound X

Hence, the compound X completely dissolves in 22.5 g of solvent at 80 ^oC.

At 20 ^oC, 100 g of solvent Z dissolves 1.0 g of compound X

Thus, at 20 ^oC, 22.5 g of solvent Z dissolves 1.0 g x 22.5 g/100 g = 0.225 g of compound X

Hence, the amount of recrystallized compound X = (1.76 - 0.225) g = 1.54 g

Now, at 80 ^oC, 100 g of solvent Z dissolves 6.0 g of compound Y

Therefore, at 80 ^oC, 22.5 g of solvent Z dissolves 6.0 g x 22.5 g/100 g = 1.35 g of compound Y

Hence, the compound Y completely dissolves in 22.5 g of solvent at 80 ^oC.

At 20 ^oC, 100 g of solvent Z dissolves 0.70 g of compound Y

Therefore, at 20 ^oC, 22.5 g of solvent Z dissolves 0.70 g x 22.5 g/100 g = 0.158 g of compound Y

Hence, the amount of recrystallized Y = (0.240 - 0.158) g = 0.0820 g

Combined mass of the recrystallized solid = (1.54 + 0.0820) g = 1.62 g

Hence, the purity of recrystallized compound X = (1.54 g x 100)/1.62 g = 95.06 %