Percent Recovery from an Experiment

Question

The solubility of benzoic acid in water is 0.07 g/100 ml at 0oC and 6.8 g/100 ml at 100oC. Answer the following questions if 5.1 grams of benzoic acid were to be heated to boiling in 100 ml of water. What would be the maximum possible percent recovery in a second batch, if the mother liquor was reduced to 40.11 ml, chilled in ice water and vacuumed filtered? Put your answer in 3 significant figures.

Answer 1

Key idea: evaporating the mother liquor removes water, not benzoic acid. The dissolved mass stays 0.07 g; only the volume changes.

First crop: at 100 C, 100 mL holds 6.8 g, so all 5.1 g dissolves. Chilled to 0 C, 100 mL retains 0.07 g, so 5.03 g crystallizes and 0.07 g stays behind in the mother liquor.

Second crop: reduce that mother liquor to 40.11 mL. It still contains 0.07 g. At 0 C it can hold 0.07 g/100 mL x 40.11 mL = 0.0281 g. Everything above that comes out:

0.07 - 0.0281 = 0.0419 g recovered in the second batch.

As a percent of your original 5.1 g: 0.0419 / 5.1 x 100 = 0.822%

If your instructor defines percent recovery for the second crop against what was left in the mother liquor instead, it is 0.0419 / 0.07 x 100 = 59.9%. Check which convention your lab uses, since both appear in textbooks.