DEPARTMENT OF CHEMISTRY AND CHEMICAL TECHNOLOGY GENERAL CHEMISTRY 202-NYA-05 21, 22 TEST 1 24-SEPT-2012 INSTRUCTOR: I. DIONNE Print your name: Answers INSTRUCTIONS: Answer all questions in the space provided. 1. Duration of this test is 75 minutes. 2. No books or extra paper are permitted. 3. Answer the questions in ink in order to preserve the right to grieve. 4. In order to obtain full credit for your answers, you must clearly show your work. 5. Calculators may not be shared. Programmable calculators are not permitted. 6. Your attention is drawn to the College policy on cheating. This policy will be enforced. 7. A Periodic Table with constants is provided. Problem 1: /8 Problem 6: /2 Problem 2: /2 Problem 7: /6 Problem 3: /3 Problem 8: /2 Problem 4: /3 Problem 9: /2 Problem 5: /2 Problem 10: /6 Total: /36
PROBLEM 1 (8 marks) Name or write the formula for each of the following compounds. (a) Na 2 SO 4 sodium sulphate (b) iron(ii) nitride Fe 3 N 2 (c) lithium carbonate Li 2 CO 3 (d) CoO cobalt(ii) oxide (e) HBrO 4 (aq) perbromic acid (f) hydroiodic acid HI (aq) (g) disulfur decafluoride S 2 F 10 (h) PbCl 2 3H 2 O lead(ii) chloride trihydrate PROBLEM 2 Copper has two naturally occurring isotopes, 63 Cu and 65 Cu. The 65 Cu isotope has a 30.83 % abundance and an isotopic mass of 64.9278 amu. If copper has an atomic mass of 63.5460 amu, what is the isotopic mass of the other isotope? The 63 Cu isotope has an abundance of: 100.00 30.83 = 69.17% 63.5460 amu = (0.6917)(isotopic mass of 63 Cu) + (0.3083)(64.9278 amu) 63.5460 amu = (0.6917)(isotopic mass of 63 Cu) + 20.017241 amu 63.5460 amu 20.017241 amu = (0.6917)(isotopic mass of 63 Cu) 43.528759 amu = (0.6917)(isotopic mass of 63 Cu) isotopic mass of 63 Cu = 43.528759 / 0.6917 = 62.930113 amu The 63 Cu isotope has an isotopic mass of 62.9301 amu. Page 2
PROBLEM 3 (3 marks) A monatomic ion has a charge of +3. The nucleus of the ion has a mass number of 35. The number of neutrons in the nucleus is equal to the number of electrons in S 2 ion. Write the A symbol for the ion and indicate the number of protons, neutrons, and electrons. Z X 35 Cl +3 17 symbol # protons = 17 # neutrons = 18 # electrons = 14 Mass number = # protons + # neutrons The # neutrons = # electrons in S 2 ion = 18 The # protons = mass number # neutrons = 35 18 = 17 Charge is +3, it therefore contains 14 electrons PROBLEM 4 (3 marks) Fill in the blanks: (a) The atomic number of the lightest alkaline earth metal is. 4 (b) The symbol of the heaviest metalloid in Group 4A is. Ge (c) Group 1B consists of the coinage metals. The name of the coinage metal whose atoms have the fewest electrons is. copper (d) The atomic mass of the halogen in Period 5 is. 126.9 amu (e) The name of the monatomic ion formed when hydrogen gains an electron is the hydride ion. (f) The members of Group 8A are known as the. noble gases Page 3
PROBLEM 5 One molecule of the antibiotic known as penicillin G has a mass of 5.342 x 10 21 g. What is the molar mass of penicillin G? Molar mass = g/mol one molecule has a mass of 5.342 x 10 21 g One mole contains 6.022 x 10 23 molecules molar mass is: 5.342 x 10 21 g /molecules x 6.022 x 10 23 molecules/mol = 3.217 x 10 3 g/mol The molar mass of penicillin G is 3217 g/mol. PROBLEM 6 Narceine is a narcotic in opium. It crystallizes from water solution as a hydrate that contains 10.8% water by mass. If the molar mass of narceine hydrate is 499.52 g/mol, determine x in narceine xh 2 O. 10.8% = [(x)(mass of H 2 O) / (mass of compound) x 100 0.108 = (x) (18.016 g/mol) / 499.52 g/mol x = 0.108 x 499.52 g/mol / 18.016 g/mol = 2.994 x is 3. Page 4
PROBLEM 7 (6 marks) A compound was found to contain only three elements, C, H, and Cl. When a 1.50 g sample of the compound was completely combusted in air, 3.52 g of CO 2 was formed. In a separate experiment the chlorine in a 1.00 g sample of the compound was converted to 1.27 g of AgCl. Determine the empirical formula of the compound. 3.52 g CO 2 / 44.01 g/mol = 0.07999818 mol CO 2 Since all C is converted to CO 2, the 1.50 g sample contains 0.07999818 mol C 0.07999818 mol C x 12.01 g/mol = 0.9606 g C (0.9606 g C / 1.50 g sample) x 100 = 64.04 % C 1.27 g AgCl / 143.35 g/mol = 0.0088594 mol AgCl Since all Cl is converted to AgCl, the 1.00 g sample contains 0.0088594 mol Cl 0.0088594 mol Cl x 35.45 g/mol = 0.31407 g Cl (0.31407 g Cl / 1.00 g sample) x 100 = 31.41 % Cl % H = 100.00 (64.04 + 31.41) = 4.55 % H Assume 100.0 g 64.04 g C / 12.01 g/mol = 5.3322 mol C 31.41 g Cl / 35.45 g/mol = 0.8860 mol Cl 4.55 g H / 1.008 g/mol = 4.5139 mol H smallest ratio C: 5.3322 / 0.8860 = 6.01 Cl: 0.8860 / 0.8860 = 1.00 H: 4.5139 / 0.8860 = 5.09 The empirical formula is C 6 H 5 Cl The empirical formula is C 6 H 5 Cl. Page 5
PROBLEM 8 (a) What is the empirical formula for each of the following compounds? i. C 3 H 6 O 3 CH 2 O ii. Ga 2 (SO 4 ) 3 Ga 2 (SO 4 ) 3 (b) What is the molecular formula of a compound which has a molar mass of 32.05 g/mol and the empirical formula NH 2? The mass of NH 2 is 16.026 g/mol This unit fits: 32.05 / 16.026 = 1.999 times = 2 2 x NH 2 = N 2 H 4 The molecular formula is N 2 H 4. PROBLEM 9 Write balanced equations for each of the following by inserting the correct coefficients in the blanks: (a) (b) P 4 O 10 (s) + H 6 2 O(l) H 4 3 PO 4 (l) CH 3 NH 2 (g) + O 2 (g) CO 2 (g) + H 2 O(g) + N 2 (g) 4 9 4 10 2 Page 6
PROBLEM 10 (6 marks) Phosphorus pentachloride reacts with water to give phosphoric acid and hydrochloric acid. PCl 5 (s) + H 2 O (l) H 3 PO 4 (aq) + HCl (aq) If 191.56 g of PCl 5 is mixed with 51.89 g of water. (a) Calculate the theoretical yield of HCl (in grams). (b) How many grams of the excess reactant are left over after the reaction is complete? (c) What is the percent yield if 125.51 g of HCl are produced? PCl 5 (s) + 4H 2 O (l) H 3 PO 4 (aq) + 5HCl (aq) 191.56 g PCl 5 /208.22 g/mol = 0.91999 mol PCl 5 51.89 g H 2 O / 18.016 g/mol = 2.88022 mol H 2 O If PCl 5 is limiting: 5 x 0.91999 = 4.59995 mol HCl will form. If H 2 O is limiting: 5/4 x 2.88022 = 3.600275 mol HCl will form. Smallest amount of product formed is due to H 2 O being the limiting reagent. Therefore 3.600275 mol x 36.458 g/mol = 131.3 g HCl will form Excess reagent = amount of PCl 5 present at start amount of PCl 5 used during the reaction At start: 0.91999 mol Amount used: 2.88022 mol H 2 O used up x 1/4 = 0.720055 mol PCl 5 used up Excess: 0.91999 mol 0.720055 mol = 0.199935 mol PCl 5 0.199935 mol x 208.22 g/mol = 41.64 g Percent yield: (actual yield / theoretical yield) x 100 (125.51 g / 131.3 g) x 100 = 95.59 % Theoretical yield of HCl Excess reactant left over 131.3 g g 41.64 g g % yield of HCl 95.59 % % Page 7