Chemistry 30L
Exam 1 Study Questions
Winter 1999


1. An organic chemistry lab book gives the following solubility data for oxalic acid

(IUPAC name - ethanedioic acid)
9.5 g/100mL of water 23.7 g/100 mL of ethanol 16.9 g/100 mL of ether
(a) A scientist wanted to extract 40 g of oxalic acid that was in 1000 mL of water into an organic solvent for further experiments. Which solvent above is the best choice for the extraction? Explain your choice.

Answer: Since ethanol and water are miscible, you must use ether.

(b) Calculate the partition coefficient for your choice. (Be sure to define the terms of your equation.)

Answer: Defining the partition constant Kp, as solubility in ether/solubility in water gives Kp = 16.9/9.5 = 1.78

(c) Calculate the weight of acid remaining in the water phase if you extracted the 40 g/1000 mL of water with 1000 mL of the solvent you chose in part (a).

Answer: Kp = 1.78 = x/(40-x) and x = 25.6 g

(Since the volumes are the same, the ratio of the weights are the same as the ratio of the (wt/vol)ís

(d) Draw the structure of oxalic acid.

Answer: Please see lecture 2 notes.

(e) How could you extract oxalic acid from an organic phase into an aqueous phase? (Note this is a different question than was originally posted.)

Answer: You need to make the acid more soluble in the aqueous phase and less soluble in the organic phase. That is you need to increase the "ionic character" of the acid. Adding a base to the solution will cause the necessary reaction to leave you with the oxalate ion.

2. In contrast to many solutions, benzoic acid is less soluble in benzene at high temperature than at low temperature. Postulate a molecular structural reason based on the principle of "like dissolves like." (Consider the energy of the hydrogen bond.) Answer: An acid functional group makes a compound quite polar and energeticallnot as soluble in a non-polar solvent as in a polar one. On the other hand, the phenyl ring is very similar to benzene, and tends to soubilize the benzoic acid. The acid grouppolarity can be reduced by two compounds hydrogen bonding together with the proton on the acid group hydrogen on one of the molecules bonding to the oxygen on the carbonyl of the other molecule. That is, (-O-H ...O=C). The two molecules come together and form a weakly bonded third ring in the middle.

At high temperature, there is too much motion in the moleceules to mantain these hydrogen bonds and when they break the solubility of the benzoic acid decreases.
 

3. When a compound melts the intermolecular forces between the molecules are broken. The following pairs of compounds have approximately the same molecular weight. Predict which of each pair has the higher melting point. Explain your reasoning. (a) ethanol (C2H6O) and ethyl amine(C2H8N)

Answer: Ethanol. Alcohols form stronger hydrogen bonds than amines do.
(b) acetic acid ethyl ester and propanoic acid (both C4H8O2)

Answer: Propioic acid. Hydrogen bonds are possible here also, but not in the ester.
(c) benzene (C6H6) and cyclohexane (C6H12)

Answer: Benzene. The polarizable pi bonds of the aromatic ring provide stronger van der Waals forces between the molecules than are possible in cyclohexane where sigma bonds exist.

(d) ammonium chloride (NH4Cl) and methyl chloride (CH3Cl)

Answer: Ammonium chloride. Methyl chloride is a covalent molecule. The only intermolecular forces are van der Waals attractions. Ammonium chloride is an ionic salt composed of amonium ions and chloride ions.

4. Explain why heating a sample rapidly during a melting point determination may lead to the erroneous conclusion that (a) the compound is impure

Answer: If the rate of temperture change is faster than the rate of transfer of heat across the capillary tube, then the thermometer, which is measuring the temperature of the environment, not the sample, will register more change than actually occurs in the sample. The result will be a larger temperature range than should be obtained, and the conclusion could be drawn that the compound is more impure that it really is.

(b) the sample is a different compound

Answer: Again the same phenomenon will occur as above. It is possible that the disapperance of the last solid will occur when the temperature of the environment is higher than the temperature of the sample in the tube. Since the impurities depress the melting point, then a final melting point above the expected one, would imply a different compound.

5. Potassium dichromate can frequently be used as an oxidizing agent in place of potassium permanganate. In the redox reaction, the dichromate ion, Cr2O72- is reduced to Cr3+ (a) Write a balanced half reaction for this reduction.

Answer: First balance the masses of the chromium (the coefficient is 2)

 
Then balance the mass of oxygen (7 molecules)

Then balance the mass of hydrogen (14 protons)

Then balance for electrical neutrality (6 electrons on the left need to be added)

Check everything is balanced.

Cr2O72- + 14H+ + 6e- = 7H2O + 2Cr3+


(b) Write a balanced redox equation for the reaction of dichromte with the oxalate ion.

Answer: The half reaction for oxalate is C2O42- = 2CO2 + 2e-

Since the reaction has 2 electrons and the dichromate in part a has 6, the oxalate reaction must be multiplied by 3 so that the sum of the two has both electrical and mass balance.

Cr2O72- + 14H+ + 3C2O42- = 7H2O + 2Cr3+ + 6CO2 (c) What weight of potassium dichromate would be required to react with the oxalate in a 1.00 g sample of potassium tris(oxalate)ferrate(III)?   Answer: One mole of dichromate reacts with 3 moles of oxalate. There are three moles of oxalate/mole of iron compound. Thus, one mole of potassium dichromate is requred for each mole of potassium tris(oxalate)ferrate(III). Calculate the moles of iron salt in 1.00 g of this compound, then multiply this by the molecular weight of potassium dichromate.
6. Calculate the equivalent weight of each of the following salts. (a) Ni(NH3)6Cl2 (b) Ni(NH3)5(H2O)Cl2

(c) Ni(NH3)5(H2O) 3Cl2 (d) Ni(NH3)4(H2O) 2Cl2

(e) Ni(NH3)3(H2O) 3Cl2 (f) Ni(H2O)6Cl2
 

Answer: The equivalent wieght of a compound is that weight of the compound that reacts with one mole of acid or base. Since one mole of acid reacts with each mole of ammonia, the equivalent weight is tthe weight of compound that contains one mole of ammonia. In other words the equivalent weight is the molecular weight divided by the coefficient indicating the moles of ammonia in the compound.
7. What volume of 0.100 M HCl would be required to directly titrate one gram of each of the salts in question 6. Answer: If you know the equivalent weight (g/equivalent) from question 6, you can calculate the equivalents/gram for each of the compounds. Since equivalents of acid = equivalents of base, ans HCl has one proton/mole or one equivalent/mole, then the volume of acid needed = 6M/equivalents. 8. What would be the most distinct difference in the infrared spectra of (a) acetic acid and acetone

Answer: There will be a braod OH stretch at the left of the characteristic region in the acid spectrum that will not be present in the acetone spectrum.

(b) ethanol and ethyl amine

Answer: This is difficult without tables. You will see two absorptions due to the asymmetric and symmetric stretches due to the -NH2 group,whereas there will ony be one in the alcohol spectrum. Hydrogen bonding will tend to broaden the bands in both spectra unless the solutions are very dilute.

(c) acetone and acetic acid methyl ester

Answer: Again you will need tables. the position of the C=O stretch is characteristic of the type of functional group it is in. In ketones, the absorptions occurs between 1725 and 1705 cm-1, in esters it is between 1765 and 1720 cm-1.

9. You studied the possible vibrations for acetic acid and for propionamide. Predict the spectrum in the characteristic region for the compound acetamide. Answer: The acetamide spectrum will be very similar to the propionamide spectrum except you will have fewer CH- vibrations than in propionamide. In this respect, it will be more like acetic acid. 10. In the aspirin experiment you recrystallized your product from an ethanol solution to which you added water. In the iron oxalate experiment you recrystallized the product from aqueous solution to which you added ethanol. Explain why these two recrystallizations were carried out in opposite directions. Answer: In the case of aspirin, you had a saturated ethanol solution for an organic compound. To decrease the solubility you added a more polar miscible solvent - water. In the case of the iron oxalate complex, you had an aqueous solution of the salt. When you added the ethanol, you decreased the polarity of the solvent and therevy the solubility of the salt. 11. The Merck Index lists the solubility of cholesterol in alcohol as 1.29% (w/w) at 20o and 28g/100g at 80o)

(a) What volume of alcohol would you use to recrystallize 5 g of cholesterol?

Answer: You want to dissolve the cholesterol in the minimum volume of hot solution. If 28 g dissolve in 100 g of ehtanol they you willr equire 17.9 g of ethanol to dissolve 5 g of cholesterol. The density of ethanol is 0.79 g/mL. You will need 22.6 mL of ethanol.

(b) What is the maximum weight of product you could recover in a recrystallization?

Answer: At 20 degree C, the solubility is 1.29 g/100g. The difference in solubility between 80 degree and 20 degree is 26.7 g. The percent recovery is therefore, 26.7/28 = 95.4%. In the situation in part (a) you could recover 95.4% of 5 g = 4.7 g.

12. The lone pair of electrons on the nitrogen in pyridine, C5H5N, can bond as a ligand to metals or as a base to protons.

C5H5N + H+ = C5H5NH+

Cobalt forms several pyridine complexes, which have the general formula Co(C5H5N)x(H2O)y2+. The specific complex that forms depends on both the concentration of the pyridine in the solution and the temperature at which the reaction is carried out.

When a complex is made the amount of pyridine in the complex can be determined through an acid-base titration. To ensure that the reaction is quantitative, a sample of the salt is dissolved in a known volume of standard acid and the excess acid is titrated with standard base. The question below deals with an analysis such as this.

(a) A 0.2671-g sample of a cobalt-pyridine salt was dissolved in 30.00 mL of 0.1042 M HCl and back titrated with 8.56 mL of 0.0985 M NaOH to the equivalence point.

Calculate the equivalent weight of the salt.

Answer: There are two sources of base (the NaOH and the ammonia in the comound) and one source of acid (the HCl.)

equivalents of acid = equivalents of base.

0.03000L x 0.1042mol/L = (0.0985mol/L x 0.00856L) + 0.2671g /(EW)

EW = 117g

(b) Which of the following choices is the most plausible formula for the compound analyzed in part (a)? Explain your reasoning. (If you do not have an answer for part (a) assume an equivalent weight of 115 g)

(i) Co(C5H5N)6Cl2

Answer: EW = MW/6 = 604/6 = 101

(ii) Co(C5H5N)5(H2O)1Cl2

Answer: EW = MW/5 = 543/5 = 109

(iii) Co(C5H5N)5(H2O)3Cl2

Answer: Eight ligands will not fit around the cobalt atom - not a plausible compound.

(iv) Co(C5H5N)4(H2O)2Cl2

Answer: EW = MW/4 = 482/4 = 121 - best choice; closest to experimental value

(v) Co(C5H5N)3(H2O)32+

Answer: This is an ion, not a neutral comound - not possible.

(c) If the limiting reagent in the preparation of the complex salt was 15.00 mL of a 1.00 M CoCl2 solution, and 6.20 g was recovered from the reaction, what was the percent yield in the synthesis?

Answer: Maximum yield =0.15 mol X 121 g/mol = 18.15 g

% yield = 6.20/18.15 = 34.2%

13. (a) List three requirements for an extraction to be an effective technique for the purification of the crude mixture of a reaction. Answer: immisicible solvents, differing polarities of compounds to be separated, differing solubiliteis of each compound in the two solvents, differing solubilities of the compounds with respect to each other.

(b) The Merck Index gives the following solubility data for the solubility for 1 g of aspirin or salicylic acid.
 

Aspirin (acetylsalicylic acid) Salicylic acid
300 mL of H2O 460 mL of H2O
12 mL diethyl ether 3 mL of diethyl ether
Calculate the partition constants, K(ether/H2O), for aspirin and for salicylic acid. (Hint: First convert from volumesolvent /g compound to g compound/100 mL solvent .)

Answer: solubility of aspirin in water = 0.33 g/100 mL

solubility of aspirin in ether = 8.33 g/100 mL

solubility of salicylic acid in water = 0.22 g/100 mL

solubility of salicylic acid in ether = 33.3 g/100 mL

aspirin partition coefficient = 8.33/.33 = 25

salicylic acid partition coefficient = 151

(c) Assume your crude product from the aspirin synthesis contained equal weights of aspirin and salicylic acid. If you dissolved 2 g of crude product in 100 mL of water and extracted with 100 mL of ether, how much purification would you achieve? That is, what would be the ratio of weights of the aspirin and salicylic acid in the resulting ether layer?

Answer: The extraction will transfer 0.9934 g of the salicylic acid to the ether layer and 0.9215 g of the aspirin to the ether layer. The ratio of weights will be 1.08/1 or a purification of only 8%.

(d) Would extraction or recrystalliztion be a better purification technique if the crude product contained 95% aspirin and 5% salicylic acid? Explain your answer.

Answer: Because the difference in solubilities is much greater than the difference in partitioning, recrystallization is the better choice for purification.


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