CHEM 31 Introductory Chemistry EXAM #2 October 16, 2002

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1 CHEM 31 Introductory Chemistry EXAM #2 October 16, 2002 Name: Keye, Onsur SSN: Lab T.A.: INSTRUCTIONS: Read through the entire exam before you begin. Answer all of the questions. For questions involving calculations, show all of your work -- HOW you arrived at a particular answer is MORE important than the answer itself! Circle your final answer to numerical questions. The entire exam is worth a total of 150 points. Attached are a periodic table and a formula sheet jam-packed with useful stuff! Good Luck! Page Possible Points Points Earned TOTAL:

2 1. (10 pts) Caffeine (a stimulant found in coffee ) is 49.5% C, 5.15% H, 28.9 % N and 16.5% O, by mass. Determine the empirical formula for caffeine. In a 100-gram sample: 46.5 g C x 1 mol C = mol C g C 5.15 g H x 1 mol H = mol H g H 28.9 g N x 1 mol N = mol N g N 16.5 g O x 1 mol O = mol O g O Thus, the empirical formula is: C 4 H 5 N 2 O 2. (10 pts) The fizz produced when an Alka-Seltzer tablet is dissolved in water is due to the reaction between sodium bicarbonate (NaHCO g/mol) and citric acid (H 3 C 6 H 5 O g/mol): 3 NaHCO 3 (aq) + H 3 C 6 H 5 O 7 (aq) 3 CO 2 (g) + 3 H 2 O (l) + Na 3 C 6 H 5 O 7 (aq) If 2.00 g of sodium bicarbonate and 3.00 g of citric acid are allowed to react, which is the limiting reactant? 2.00 g NaHCO 3 x 1 mol = mol NaHCO g NaHCO g H 3 C 6 H 5 O 7 x 1 mol = mol H 3 C 6 H 5 O g H 3 C 6 H 5 O 7 How many mol NaHCO 3 are needed to react with all of the H 3 C 6 H 5 O 7? mol H 3 C 6 H 5 O 7 x 3 mol NaHCO 3 = mol NaHCO 3 1 mol H 3 C 6 H 5 O 7 Since: mol NaHCO 3 needed = > mol NaHCO 3 actual, NaHCO 3 is the limiting reagent 2

3 3. (5 pts each) Write balanced net ionic equations for the precipitation reactions that may occur in each of the following cases. If no reaction occurs, then you only need write NR. a. Na 2 CO 3 (aq) + MgSO 4 (aq) NaSO 4 (aq) + MgCO 3 (s) Mg 2+ (aq) + CO 3 2- (aq) MgCO 3 (s) b. NaOH (aq) + LiClO 4 (aq) no reaction NR c. AgNO 3 (aq) + (NH 4 ) 2 S (aq) Ag 2 S(s) + 2NH 3 NO 3 (aq) 2Ag + (aq) + S 2- (aq) Ag 2 S(s) 4. (2 pts each) Determine the oxidation number for the indicated element in each of the following compounds: a. Sn in SnCl 4 : +4 b. Cr in Cr 2 O 7 2- : +6 c. N in N 2 O: +1 d. N in NO 2 : +4 e. N in NO 3 - : (10 pts) Write a balanced molecular equation for the reaction that occurs when solid calcium carbonate (the active ingredient in Tums ) reacts with an aqueous solution of nitric acid (Hint: you saw this reaction as a demo in class!). CaCO 3 (s) + 2HNO 3 (aq) H 2 O(l) + CO 2 (g) + Ca(NO 3 ) 2 (aq) 3

4 6. (3 pts each) For the following redox reactions, indicate which element is oxidized and which element is reduced. a. Ni(s) + Cl 2 (g) NiCl 2 (s) Oxidized: Ni Reduced: Cl b. 3Fe(NO 3 ) 2 (aq) + 2 Al(s) 3Fe(s) + 2 Al(NO 3 ) 3 (aq) Oxidized: Al Reduced: Fe c. 14KMnO 4 (s) + 4C 3 H 5 (OH) 3 (l) 7K 2 CO 3 (s) + 7Mn 2 O 3 (s) + 5CO 2 (g) + 16 H 2 O(l) Oxidized: C Reduced: Mn 7. (5 pts) How many moles of HNO 3 are present in 45.0 ml of a 1.68 M solution of nitric acid? 45.0 ml x 1 L x 1.68 mol HNO 3 = x 10-2 mol HNO ml L = 7.56 x 10-2 mol HNO 3 8. (6 pts) Calculate the molarity of a solution made by dissolving grams Ca(OH) 2 (MW = g/mol) in enough water to make exactly ml of solution g Ca(OH) 2 x mol x 1000 ml = x 10-2 M ml g Ca(OH) 2 L = x 10-2 M Ca(OH) 2 4

5 9. (10 pts) The distinctive odor of vinegar is due to acetic acid, CH 3 COOH. As you may recall from lab, acetic acid reacts with sodium hydroxide in the following fashion: CH 3 COOH (aq) + NaOH (aq) CH 3 COONa (aq) + H 2 O (l) If a 5.00-mL sample of vinegar requires 37.2 ml of M NaOH to reach the equivalence point in a titration, what is the molarity of acetic acid in the vinegar sample? 37.2 ml NaOH sol n x L x mol NaOH x 1 mol HAc= x ml L 1 mol NaOH mol HAc x 10-3 mol HAc x 1000 ml = M HAc 5.00 ml Sol n L = M HAc 10. (10 pts) The Hindenburg was a famous hydrogen-filled dirigible that burned in If the Hindenburg held 2.0 x 10 5 m 3 of hydrogen gas (H 2 ) at 25. o C and 1.0 atm, what mass (kg) of hydrogen was present? V = 2.0 x 10 5 m 3 x 1000 L = 2.0 x 10 8 L 1 m 3 n = PV = (1.0 atm)(2.0 x 10 8 L) = x 10 6 mol H 2 RT ( L-atm/mol-K)( K) x 10 6 mol H 2 x g H 2 x 1 kg = x 10 4 kg mol H g = 1.6 x 10 4 kg H 2 5

6 11. (10 pts) Calculate the molar mass (g/mol) of a gas if 3.75 g occupies L at 430. torr at 35.0 o C. 430 torr x 1 atm = atm 760 torr n = PV = ( atm)(0.935 L) = x 10-2 mol RT ( L-atm/mol-K)( K) Molar mass = 3.75 grams = g/mol = 179. g/mol x 10-2 mol 12. (10 pts) Analysis of an air bubble trapped in amber formed 80 million years ago revealed that the atmosphere before the dinosaurs became extinct was 30.0% O 2 and 70.0% N 2 (by mass). Calculate the partial pressure (torr) of oxygen in this prehistoric air sample if the total pressure of the air was 1.00 atm g O 2 x mol O 2 = mol O 2 = n O g O g N 2 x mol N 2 = mol N 2 = n N g N 2 n T = n O2 + n N2 = = mol X O2 = n O2 /n T = / = P O2 = X O2 P T = ( )(1.00 atm) = atm In torr: atm x 760 torr = torr 1 atm = 2.07 x 10 2 torr 6

7 13. (5 pts) A balloon made of rubber that is permeable to small molecules is filled with helium to a pressure of 1 atm. This balloon is then placed in a box that contains pure hydrogen (H 2 ), that is maintained at a pressure of 1 atm. Will the balloon expand or contract? Explain. The balloon will EXPAND. The rate of effusion of the H 2 is greater than that of He, due its lower molar mass (recall: r 1 /r 2 = [(M 2 /M 1 )] ½ ), so H 2 will enter the balloon faster than He will leave the balloon and it will expand. 14. Vessel A contains CO gas at 20 o C and 1 atm. Vessel B contains CO 2 gas at 20 o C and 0.5 atm. The two vessels have the same volume. a. (3 pts) Which vessel contains more molecules? Briefly explain. Vessel A Vessel B Same (circle one) At constant Temp and Volume, mol Pressure; vessel A has a higher pressure than vessel B, so it has more molecules of gas. b. (3 pts) In which vessel is the average kinetic energy of the molecules greater? Briefly explain. Vessel A Vessel B Same (circle one) Temperature is a measure of the average K.E. of a gas since the two vessels are at the same temperature, the average K.E. of the molecules in the two vessels must also be the same. c. (3 pts) In which vessel is the rms speed of the molecules greater? Briefly explain. Vessel A Vessel B Same (circle one) If the K.E. of the molecules in the two vessels are the same, then the rms speed of the molecules will be inversely proportional to the square root of the MASS of the molecules. Since CO is lighter than CO 2, CO (in vessel A) will have a greater rms velocity. 7

8 15. A balloon is heated by adding 400 J of heat. It expands, doing 422 J of work on the atmosphere. For this process: a. (4 pts) Calculate the change in internal energy ( E) of the system. E = q + w = +400 J + (-422 J) = -22 J b. (2 pts) This process is: Endothermic or Exothermic (circle one) 16. (5 pts) In the space below, write the name (first and last!) of the T.A. for your lab section you must spell their name correctly in order to receive credit for your response! Rachel Burdge Mike Calichman Sean McCarthy Gerald (Jerry) Love 17. (10 pts) The specific heat of copper metal is J/g-K. How many J of heat are necessary to raise the temperature of a 2.26 kg block of copper from 25.0 o C to o C? m = 2.26 kg x 1000 g = 2.26 x 10 3 g kg T = o C = 75.0 o C = 75.0 K c s = J/g-K q = mc s T = (2.26 x 10 3 kg)(0.385 J/g-K)(75.0 K) = x 10 4 J = 6.53 x 10 4 J 8