chem 31.1 expt 2

1) The fluted filter paper allows more liquid to pass through by increasing the surface area open to the air, the wire can remove air pressure caused by the filtrate’s heat. 2) Premature recrystallization happens when the crystals form before it is separated from the contaminants. This can be avoided by making sure that the mixture is sufficiently hot, but also that the funnel is sufficiently hot, so that the crystals form only after the purified sample is collected. Premature recrystallization reduces the amount of crystals obtained which reduces the recovered pure sample. 3) The solubility of a solute in a solvent is defined as the m aximum amount of solute that dissolves in equilibrium. Since the the object o f hot filtering is to remove solvent insoluble particles, it is necessary that the desired solute be fully soluble in hot solvent, that is—the solubility in hot solvent is greater than the amount of solute. We see that both A and B dissolves fully in hot solvent. We would only need to focus on cold solvent solubility. We need a scenario where one of A or be is fully crystallized or fully dissolved. (grams in solute – grams solubility) = grams crystallized ↓ grams crystallized A

B

ethanol

1.5

3

acetone

(-3) ∴ 0

7.9

1.4

7.5

water

a) With acetone, all of A is dissolved and 7.9g of B is crystallized, therefore we use acetone. b) When separated, the 7.9g of B is 100% pure while the 2g of A is contaminated with 0.1g of B, so its purity is 95 95.23% .23% 4) We see that A, B & C dissolves fully in hot solvent. We would only need to focus on cold solvent solubility. We need a scenario where one o f A or be is fully crystallized or fully dissolved. a)

(grams in solute  – grams solubility) = grams crystallized ↓ solute crystallized A

B

C

2.95

(-1.6) ∴ 0

(-0.2) ∴ 0

A is pure b) After filtering, only 1.05g of A are left in 100mL of EtOH, along with 4g each of B and C. Evaporating half of the solution (but obviously the solubilities remain the same), the solubility of A, B and C are 0.525g/50mL, 2.8g/50mL and 2.1g/50mL respectively. (grams in solute  – grams solubility) = grams crystallized ↓ solute crystallized A

B

C

0.525

1.2

1.9

[0.525/(0.525 + 1.2 + 1.9)]*100% = 14.5% of A [1.2/(0.525 + 1.2 + 1.9)]*100% = 33.1% of B [1.9/(0.525 + 1.2 + 1.9)]*100% = 52.4% of C D.Pasto, C. Johnson and M. Miller, Experiments and Techniques in Organic Chemistry , Prentice Hall, New Jersey, 1992.