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Unformatted text preview: momin (rrm497) – Homework 4 – Sutcliffe – (53120) 1 This printout should have 22 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. This homework covers the remaining mate rial that will be on Exam 2. Q 3 and 4 are NUMERICAL so you must read the QUEST help menu to see how to enter these. Everyone should have different masses. 001 10.0 points In 1978, scientists extracted a compound with antitumor and antivirus properties from ma rine animals in the Caribbean Sea. A sample of the compound didemninA of mass of 1.756 mg was analyzed and found to have the fol lowing composition: 1.086 mg C, 0.148 mg H, 0.159 mg N, and 0.363 mg O. The molar mass of didemninA was found to be 942 g/mol. What is the molecular formula of didemnin A? 1. C 48 H 78 N 6 O 12 correct 2. C 70 H 70 NO 12 3. C 49 H 75 N 6 O 7 4. C 12 H 40 N 2 O 12 5. C 49 H 78 N 3 O 49 6. C 40 H 65 N 5 O 10 Explanation: For the 1.756 mg sample of didemninA, mmol of C = 1 . 086 mg 12 . 01 g / mol = 0.0904 mmol mmol of H = . 148 mg 1 . 0079 g / mol = 0.147 mmol mmol of N = . 159 mg 14 . 01 g / mol = 0.0113 mmol mmol of O = . 363 mg 16 . 0 g / mol = 0.0227 mmol Dividing by 0.0113 mmol gives 8.0 C : 13.0 H : 1.0 N : 2.0 O, so the empirical formula of didemninA is C 8 H 13 NO 2 , with a MW of 8(12 . 01 g / mol) + 13(1 . 0079 g / mol) 1(14 . 0 g / mol) + 2(16 . 0 g / mol) = 155 . 19 g / mol Multiplying by 6 gives the molecular for mula of didemninA as C 48 H 78 N 6 O 12 with a molar mass of 942 g/mol. 002 10.0 points Balance the equation ?As 2 S 3 +?O 2 → ?As 4 O 6 +?SO 2 , using the smallest possible integers. What is the sum of the coefficients in the balanced equation? 1. 21 2. 20 3. 17 4. 18 correct 5. 16 Explanation: A balanced equation has the same num ber of each kind of atom on both sides of the equation. We find the number of each kind of atom using equation coefficients and com position stoichiometry. For example, we find there are 4 As atoms on the reactant side: ? As atoms = 2 As 2 S 3 × 2 As 1 As 2 S 3 = 4 As The balanced equation is 2 As 2 S 3 + 9 O 2 → As 4 O 6 + 6 SO 2 , and has 4 As, 6 S and 18 O atoms on each side. ? sum coefficients = 2 + 9 + 1 + 6 = 18 003 (part 1 of 2) 10.0 points The solid fuel in the booster stage of the space shuttle is a mixture of ammonium perchlorate and aluminum powder. On ignition, the reac tion that takes place is 6 NH 4 ClO 4 (s) + 10 Al(s) → 5 Al 2 O 3 (s) + 3 N 2 (g) + 6 HCl(g) + 9 H 2 O(g) . momin (rrm497) – Homework 4 – Sutcliffe – (53120) 2 What mass of aluminum is required to react with 1 . 232 kg of NH 4 ClO 4 for this reaction? Correct answer: 471 . 514 g. Explanation: m AgBr = 1 . 232 kg MW NH 4 ClO 4 = 14 . 01 g / mol + 4 (1 . 0079 g / mol) + 35 . 45 g / mol + 4 (16 . 0 g / mol) = 117 . 492 g / mol m Al = (1 . 232 kg NH 4 ClO 4 ) parenleftbigg 1000 g 1 kg parenrightbigg × parenleftbigg 1 mol NH 4 ClO 4 117 . 492 g NH 4 ClO 4 parenrightbigg × parenleftbigg...
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 Spring '07
 Fakhreddine/Lyon
 Chemistry

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