Solutions lab lecture practice questions (expt 1) recrystallization PDF

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Solutions lab lecture practice questions (expt 1) recrystallization...

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SOLUTIONS to practice questions from expt 1 lab lecture notes (recrystallization) 1) You attempted to recrystallize 0.50 g of substance X from 50 mL benzene. On cooling the recrystallization solution in an ice-water bath you saw a huge amount of crystalline solid in the flask (MUCH more than you could have expected based on the amount of X you started with). However, nearly all of it seemed to disappear into solution when you attempted to collect it by vacuum filtration. Explain what happened by taking into account the pertinent physical properties of all materials you’re dealing with: substance X water benzene

water solubility insoluble NA insoluble

benzene solubility 1.180, 0.0525 insoluble NA

boiling point NA 100 oC 80 oC

freezing/melting point NA 0 oC 5.5 oC

The solid in the flask was mostly frozen solvent! Benzene freezes at approx. 5 oC and the recrystallization solution is placed in an ice-water bath at a temp. near 0 oC. The benzene crystals then melted when warmed to room temperature during filtration. This illustrates why you need to consider the physical & chemical properties of all reactants, reagents, solvents etc. before you begin an experiment, so there are no surprises like this one. 2) Based on the following solubility data, choose the most appropriate solvent to recrystallize compound X (bp hexanes ≈70 oC; bp chloroform = 61 oC; bp ethanol = 70 oC; bp water = 100 oC). hexanes (0.012, 0.0524, 0.1070) chloroform (755, 8025, 10062) ethanol (0.102, 4.024, 1178) water (0.055, 0.1024, 2599) The best solvent is the one in which the solute is least soluble at low temperatures and totally soluble when heated. Solvent hexanes chloroform ethanol water

Comments Compound X is very sparingly soluble when heated in hexanes; e.g., 1 liter of boiling hexanes is required to dissolve just 1 g of compound X! Compound X is TOO soluble in chloroform at all temperatures. These two solvents are best; water is better:  More compound X dissolves in hot water than in hot ethanol  The amount of compound X recovered upon cooling would be higher in water than in ethanol since it’s less soluble in water at low temp.

3) Based on the given solubility data, determine the number of milliliters of boiling water required to dissolve 25 g of phthalic acid. If the resulting solution were cooled to 14 oC, how many grams of phthalic acid would recrystallize out of solution? Solubility data for phthalic acid (PA) in water: 0.5414, 1899. 18 g PA 100 mL H2O 0.54 g PA 100 mL H2O

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25 g PA X mL H2O X g PA 139 mL H2O

18X = 2500 X = 139 mL of boiling water needed to dissolve 25 g of phthalic acid. 100X = 75 X = 0.75 g of phthalic acid remains dissolved in water at 14 oC.

Therefore, 25 g – 0.75 g = 24.25 g of phthalic acid crystallizes out of water at 14 oC.

4) Prof. Heinz Ketchup wished to purify 2.50 g of crude solid Y by recrystallization and found the following solubility data in a handbook: solubility of Y in water (g/100 mL) temp (oC) 0.0 2.00 20.0 3.00 40.0 8.50 60.0 11.0 80.0 17.0 100.0 20.0 a) What is the minimum amount of boiling water needed to dissolve the entire sample? b) If the entire sample was dissolved in water, how many grams of Y would be recovered if the solution were cooled in an ice-water bath? c) Assuming that 0.50 g of an impurity Z is present in addition to 2.50 g of Y, describe how you would purify Y if Z were completely insoluble in water. d) Prof. Ketchup misread the marking on the graduated cylinder and added 22.5 mL of water instead of the amount calculated in part a. How much Y would be recovered if he cooled the solution to room temperature (20 oC)? What could he do in an attempt to recover more product Y?

a)

2.50 g Y X mL H2O

b)

XgY 12.5 mL H2O

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20.0 g Y 100 mL H2O

2.00 g Y 100 mL H2O

20.0X = 250 X = 12.5 mL of water needed.

100X = 25 X = 0.25 g of Y remains dissolved @ 0 oC. 2.50 g – 0.25 g = 2.25 g Y would be recovered.

c) Gravity filter the hot solution through fluted filter paper in a stemless funnel, using a slight excess of water to prevent precipitation of Y in the funnel.

d)

XgY = 22.5 mL H2O

3.00 g Y 100 mL H2O

100X = 67.5 X = 0.675 g of Y remains dissolved @ 20 oC. 2.50 g – 0.68 g = 1.82 g Y would be recovered.

To recover additional product he could boil away (i.e., evaporate) ≈10 mL of water and cool the recrystallization solution to 0 oC in an ice-water bath after crystals grow at room temperature....