what is the minimum volum of water that you would use to recrystallize 50 g of l-glutamic acid
Crystallization Technique Quiz
1) What is the purpose of a crystallization?
The main purpose of a crystallization in an organic chemical science process is to purify the desired compound. Crystallization as well 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. Afterward this initial isolation (normally by a quick crystallization), the compound is crystallized (or, recrystallized) advisedly to purify it.
2) You desire to purify x 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 post-obit table.
| Compound | Solubility at xx° C (g/ten mL) | Solubility at 100°C (g/10 mL) |
| Chemical compound A | 0.029 | 0.680 |
| Chemical compound B | 0.22 | 6.67 |
- What volume of boiling water is needed to dissolve the 10 1000 of Compound A?
- How much Chemical compound A will crystallize after cooling to xx°C?
- Volition whatever Compound B crystals also course?
- What is the maximum amount of Compound A that tin be recovered in the first ingather of this recrystallization?
- Will the Chemical compound A be pure?
- Fix a proportion:
0.68 yard/10 mL = 10 g/x mL
Solve, and x = 147 mL - Y'all need to determine how much Compound A volition be soluble at xx°C in 147 mL of water. Set up a proportion:
0.029 1000/x mL = x g/mL
Solve, and ten = 0.43 chiliad. Since you started with x.0 g, ix.6 g of Compound A volition crystallize. - You need to determine how much Compound B will be soluble in 147 mL of at 20°C. Gear up a proportion:
0.22 g/10mL = x g/147 mL
Solve, and ten = 3.ii g. Since you but had 0.two g Compound B, all of it will be soluble when the solution is chilled. - From (b), 9.6 yard.
- From (c), yep, since all of the Compound B will remain in solution.
3) Answer the previous questions for 10 g of Chemical compound A that has been contaminated with 10 g of Compound B. (Consider the table given in question ii).
- Gear up a proportion:
0.68 g/x mL = 10 grand/x mL
Solve, and x = 147 mL - You need to make up one's mind how much Compound A will be soluble at twenty°C in 147 mL of water. Set up upwardly a proportion:
0.029 g/10 mL = 10 g/mL
Solve, and 10 = 0.43 g. Since you started with 10.0 thousand, ix.6 g of Chemical compound A will crystallize. - Y'all need to determine how much Compound B will be soluble in 147 mL of at xx°C. Ready a proportion:
0.22 g/10mL = x 1000/147 mL
Solve, and ten = 3.2 grand. Since you had 10 g Chemical compound B, half dozen.viii m volition not be soluble and it volition grade crystals. - From (b), nine.vi g.
- From (c), no, since 6.eight one thousand of Compound B will co-crystallize with the Compound A.
four) Yous take a sample of 0.1 yard of compound C (solubility properties given below) which is contaminated with a chemical compound D.
| Solubility at 25° C (chiliad/mL) | Solubility at 100°C (g/mL) |
| 0.01 | 0.ane |
- If chemical compound D is completely insoluble in water and two mg of compound D is nowadays, how could y'all purify Compound C?
- If compound D has the same solubility behavior as C and 2 mg of this chemical compound is present, how could you purify compound C? Would one recrystallization produce absolutely pure C?
- Presume compound D has the same solubility beliefs every bit C and 25 mg of this compound is present. Would one crystallization produce admittedly pure C? How many crystallizations would be needed to produce pure C? How much C would accept been recovered when the crystallizations take been completed? If you include multiple crystallizations in your answer, utilize just plenty hot solvent in each pace to completely dissolve compound C.
- Since D is completely insoluble in h2o, it will exist a solid when 0.1 yard of chemical compound C is dissolved in 1 mL of h2o. It could exist removed from the solution by hot filtration.
- If only 2 mg (or 0.002 g) of compound D is present, information technology would be totally soluble in the 1 mL of absurd solvent. Therefore, ane crystallization would produce absolutely pure C.
- If 25 mg (or 0.25 thou) 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 ane crystallization would not produce pure C. Therefore, you would now need to crystallize over again, starting with the mixture of 0.09 thou of C and 0.01 one thousand 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 non contain any D. Two crystallizations would exist required and 0.081 g of compound C would be isolated.
5) Listed beneath are solubility vs temperature data for an organic chemical compound in water.
| Temp. (°C) | Sol. in 10 mL water |
| 0 | 0.15 k |
| 20 | 0.30 g |
| 40 | 0.65 g |
| threescore | 1.10 g |
| 80 | 1.70 g |
- Using the information in the higher up table, graph the solubility of the compound vs temperature. Connect the information points with a smooth curve.
- Suppose 0.i g of this compound is mixed with 1.0 mL of h2o and heated to 80° C. Would all of the compound dissolve? Explicate.
- The solution prepared in (b) is cooled. At what temperature will crystals of the chemical compound appear?
- Suppose the cooling described in (c) were continued to 0° C. How many grams of the compound would come out of solution?
-
- Yes, 1 mL of water at 80°C can dissolve upwardly to 0.17g of compound.
- Nosotros 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.
half dozen) You take ii g of benzhydrol and have been advised to recrystallize information technology from hexanes. How much hexanes will you use to recrystallize this product?
Since you do not know the exact solubility information for benzhydrol in hexanes, you will add small portions of hot hexanes to the benzhydrol, swirling later on each add-on, until just enough hot solvent is added to completely deliquesce the benzhydrol.
seven) 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 lot practise if this occurs?
If you lot boil off as well much solvent, it is likely that at that place is no longer plenty hot solvent to completely dissolve the compound you lot are crystallizing, and it will form a precipitate in the hot solvent. If this occurs, add together fresh hot solvent to the solution in small portions until all of the compound is dissolved.
8) Suppose you accept 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 apply decolorizing charcoal earlier allowing the hot solution to cool? Explain your answer.
Decolorizing charcoal is used to remove unwanted colored impurities. In this case, the compound that you want is reported to exist bluish, thus, you should Non add decolorizing charcoal. If y'all do add decolorizing charcoal, this treatment would remove some of the desired compound, thus decreasing the yield of your chemical compound.
9) Each of the following compounds, A-D, is equally soluble in the 3 solvents listed. In each instance, which solvent would you lot choose to recrystalliza a slightly impure sample of each chemical compound? Explain.
- Chemical compound A: benzene, acetone, or chloroform
- Compound B: carbon tetrachloride, methylene chloride, ethyl acetate
- Compound C: methanol, ethanol, or h2o
- Chemical compound D: ethanol, acetone, or diethyl ether
- Acetone is the all-time option for A because it is the to the lowest degree toxic.
- Ethyl acetate is the best choice for B because it is the least toxic. However, methylene chloride is besides a skilful 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 pick because it is lower-boiling and thus easier to remove after the crystallization.
- All have equal wellness hazard ratings. Diethyl ether is extremely flammable and should non exist used, acetone is a petty ameliorate than ethanol because information technology is lower boiling. Acetone is probably the best choice.
10) Suggest possible crystallization solvents for the following compounds.
- Naphthalene is a hydrocarbon, so endeavor 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, and so attempt ethyl acetate or acetone.
- Dimethylfumarate is a carboxylic acid, and then attempt ethanol.
11) A educatee was recrystallizing a chemical compound. As the hot solution cooled to room temperature, no crystals appeared. The flask was then placed in an water ice-water bath. Suddenly a big amount of solid material appeared in the flask. The student isolated a good yield of production, all the same, the product was contaminated with impurities. Explain.
During a recrystallization, you lot should always look until crystals appear at room temperature before placing on ice. If the cooling is too rapid, crystals of production form so quickly 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. Of a sudden a large amount of solid material appeared in the flask. So, the educatee filtered the solid with vacuum, but only a few crystals remained on the filter paper. Explain these results.
Benzene freezes at 5°C. Therefore, when the flask was placed on water 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 chemical compound remained on the filter paper.
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