# HW4 - momin(rrm497 – Homework 4 – Sutcliffe –(53120 1...

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Unformatted text preview: momin (rrm497) – Homework 4 – Sutcliffe – (53120) 1 This print-out should have 22 questions. Multiple-choice 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 didemnin-A 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 didemnin-A 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 didemnin-A, 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 didemnin-A 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 didemnin-A 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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HW4 - momin(rrm497 – Homework 4 – Sutcliffe –(53120 1...

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