Crystallization Technique Quiz
1) What is the purpose of a crystallization?
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The main purpose of a crystallization in an organic chemistry procedure is to purify the desired compound. Crystallization also may be used to isolate an solid organic compound from a mixture of compounds in a step in the work-up of a reaction mixture. After this initial isolation (usually by a quick crystallization), the compound is crystallized (or, recrystallized) carefully to purify it.
2) You want to purify 10 g of Compound A that has been contaminated with 0.2 g of Compound B. Solubilities in water of the two compounds are given in the following table.
| Compound |
Solubility at
20° C (g/10 mL) |
Solubility at
100°C (g/10 mL) |
| Compound A |
0.029 |
0.680 |
| Compound B |
0.22 |
6.67 |
- What volume of boiling water is needed to dissolve the 10 g of Compound A?
- How much Compound A will crystallize after cooling to 20°C?
- Will any Compound B crystals also form?
- What is the maximum amount of Compound A that can be recovered in the first crop of this recrystallization?
- Will the Compound A be pure?
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- Set up a proportion:
0.68 g/10 mL = 10 g/x mL
Solve, and x = 147 mL
- You need to determine how much Compound A will be soluble at 20°C in 147 mL of water. Set up a proportion:
0.029 g/10 mL = x g/mL
Solve, and x = 0.43 g. Since you started with 10.0 g, 9.6 g of Compound A will crystallize.
- You need to determine how much Compound B will be soluble in 147 mL of at 20°C. Set up a proportion:
0.22 g/10mL = x g/147 mL
Solve, and x = 3.2 g. Since you only had 0.2 g Compound B, all of it will be soluble when the solution is chilled.
- From (b), 9.6 g.
- From (c), yes, since all of the Compound B will remain in solution.
3) Answer the previous questions for 10 g of Compound A that has been contaminated with 10 g of Compound B. (Consider the table given in question 2).
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- Set up a proportion:
0.68 g/10 mL = 10 g/x mL
Solve, and x = 147 mL
- You need to determine how much Compound A will be soluble at 20°C in 147 mL of water. Set up a proportion:
0.029 g/10 mL = x g/mL
Solve, and x = 0.43 g. Since you started with 10.0 g, 9.6 g of Compound A will crystallize.
- You need to determine how much Compound B will be soluble in 147 mL of at 20°C. Set up a proportion:
0.22 g/10mL = x g/147 mL
Solve, and x = 3.2 g. Since you had 10 g Compound B, 6.8 g will not be soluble and it will form crystals.
- From (b), 9.6 g.
- From (c), no, since 6.8 g of Compound B will co-crystallize with the Compound A.
4) You have a sample of 0.1 g of compound C (solubility properties given below) which is contaminated with a compound D.
Solubility at
25° C (g/mL) |
Solubility at
100°C (g/mL) |
| 0.01 |
0.1 |
- If compound D is completely insoluble in water and 2 mg of compound D is present, how could you purify Compound C?
- If compound D has the same solubility behavior as C and 2 mg of this compound is present, how could you purify compound C? Would one recrystallization produce absolutely pure C?
- Assume compound D has the same solubility behavior as C and 25 mg of this compound is present. Would one crystallization produce absolutely pure C? How many crystallizations would be needed to produce pure C? How much C would have been recovered when the crystallizations have been completed? If you include multiple crystallizations in your answer, use just enough hot solvent in each step to completely dissolve compound C.
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- Since D is completely insoluble in water, it will be a solid when 0.1 g of compound C is dissolved in 1 mL of water. It could be removed from the solution by hot filtration.
- If only 2 mg (or 0.002 g) of compound D is present, it would be totally soluble in the 1 mL of cool solvent. Therefore, one crystallization would produce absolutely pure C.
- If 25 mg (or 0.25 g) of D is in the 1 mL of cool solution, 0.01 g of this will precipitate out and contaminate the 0.09 g of compound C, so one crystallization would not produce pure C. Therefore, you would now need to crystallize again, starting with the mixture of 0.09 g of C and 0.01 g of D. Dissolve this mixture in 0.9 mL of hot solvent. When cooled, 0.009 of C would remain in solution, and 0.081 g would crystallize. For D, 0.01 g would remain in solution, and therefore the crystals of C would not contain any D. Two crystallizations would be required and 0.081 g of compound C would be isolated.
5) Listed below are solubility vs temperature data for an organic compound in water.
| Temp. (°C) |
Sol. in
10 mL water |
| 0 |
0.15 g |
| 20 |
0.30 g |
| 40 |
0.65 g |
| 60 |
1.10 g |
| 80 |
1.70 g |
- Using the data in the above table, graph the solubility of the compound vs temperature. Connect the data points with a smooth curve.
- Suppose 0.1 g of this compound is mixed with 1.0 mL of water and heated to 80° C. Would all of the compound dissolve? Explain.
- The solution prepared in (b) is cooled. At what temperature will crystals of the compound appear?
- Suppose the cooling described in (c) were continued to 0° C. How many grams of the compound would come out of solution?
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- Yes, 1 mL of water at 80°C can dissolve up to 0.17g of compound.
- We need a solubility of 0.1g/1mL, or 1g/10mL. Follow along the curve until you reach this point - approximately 55°C.
- The solubility at 0°C is 0.15g/10mL, or 0.015g/mL. 0.015g would be soluble, and the remaining 0.085g would precipitate out.
6) You have 2 g of benzhydrol and have been advised to recrystallize it from hexanes. How much hexanes will you use to recrystallize this product?
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Since you do not know the exact solubility data for benzhydrol in hexanes, you will add small portions of hot hexanes to the benzhydrol, swirling after each addition, until just enough hot solvent is added to completely dissolve the benzhydrol.
7) Suppose you are recrystallizing a compound and boil the solution for so long that a substantial amount of the liquid evaporates. What is likely to happen to some of the solute? What should you do if this occurs?
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If you boil off too much solvent, it is likely that there is no longer enough hot solvent to completely dissolve the compound you are crystallizing, and it will form a precipitate in the hot solvent. If this occurs, add fresh hot solvent to the solution in small portions until all of the compound is dissolved.
8) Suppose you have prepared a compound which is reported in the literature to have a pale blue color. When dissolving the substance in hot solvent prior to recrystallization, the resulting solution is blue. Should you use decolorizing charcoal before allowing the hot solution to cool? Explain your answer.
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Decolorizing charcoal is used to remove unwanted colored impurities. In this case, the compound that you want is reported to be blue, thus, you should NOT add decolorizing charcoal. If you do add decolorizing charcoal, this treatment would remove some of the desired compound, thus decreasing the yield of your compound.
9) Each of the following compounds, A-D, is equally soluble in the three solvents listed. In each case, which solvent would you choose to recrystalliza a slightly impure sample of each compound? Explain.
- Compound A: benzene, acetone, or chloroform
- Compound B: carbon tetrachloride, methylene chloride, ethyl acetate
- Compound C: methanol, ethanol, or water
- Compound D: ethanol, acetone, or diethyl ether
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- Acetone is the best choice for A because it is the least toxic.
- Ethyl acetate is the best choice for B because it is the least toxic. However, methylene chloride is also a good choice because it is lower-boiling and thus easier to remove after the crystallization. It is also non-flammable.
- Water is the best choice for C because it is not at all toxic. However, ethanol is also a good choice because it is lower-boiling and thus easier to remove after the crystallization.
- All have equal health hazard ratings. Diethyl ether is extremely flammable and should not be used, acetone is a little better than ethanol because it is lower boiling. Acetone is probably the best choice.
10) Suggest possible crystallization solvents for the following compounds.
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- Naphthalene is a hydrocarbon, so try a hydrocarbon solvent like hexanes, petroleum ether or toluene.
- 2-Indanol is an alcohol, so try ethanol.
- Phenylacetic acid is a carboxylic acid, so try ethanol.
- 4-Phenylcyclohexanone is a carbonyl compound, so try ethyl acetate or acetone.
- Dimethylfumarate is a carboxylic acid, so try ethanol.
11) A student was recrystallizing a compound. As the hot solution cooled to room temperature, no crystals appeared. The flask was then placed in an ice-water bath. Suddenly a large amount of solid material appeared in the flask. The student isolated a good yield of product, however, the product was contaminated with impurities. Explain.
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During a recrystallization, you should always wait until crystals appear at room temperature before placing on ice. If the cooling is too rapid, crystals of product form so rapidly that they trap impurities in the growing crystal lattice.
12) A student used benzene to recrystallize a compound. As the hot solution cooled to room temperature, very few crystals appeared. The flask was then placed in an ice-water bath. Suddenly a large amount of solid material appeared in the flask. Then, the student filtered the solid with vacuum, but only a few crystals remained on the filter paper. Explain these results.
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Benzene freezes at 5°C. Therefore, when the flask was placed on ice, the benzene froze - this is the solid material that appeared in the flask. During the vacuum filtration (done at room temperature), the solution warmed up enough to melt the benzene, and only crystals of the compound remained on the filter paper.
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