Chemistry 30AL Midterm Review Guidelines and Study Questions Key Announcement: Since the theory of distillation and gc will not be on the midterm, the due date for the post-lab report for Assignment 8 is postponed until Nov 1, 2 – after the midterm. Review: The midterm exam will cover the theory and experiments of the course through October 26 except for distillation and GC. You will be allowed to use your lab notebook, hand-written material, lecture notes and a calculator. Spectroscopy tables will be provided. In reviewing the experiments, you should think about why you used each reagent, and what the consequences on your results would have been if you had not used that reagent. In studying the theory of the techniques, review the readings in Landgrebe and the texts as well as the lecture notes. Understanding, not memorization, is the key. Study questions: 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)’s (d) Draw the structure of oxalic acid. CCOOHOHO2. 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. 3. Explain why heating a sample rapidly during a melting point determination may lead to the erroneous conclusion that (a) the compound is impure (b) 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. 4. 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 required 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. 5. 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 weight 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 the 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. Note compound (f) does not contain ammonia, there is no equivalent weight for the compound. 6. 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/ in a 1-gram sample for each of the salts. Since equivalents of acid = equivalents of base, and HCl has one proton/mole or one
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