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# hw 5 - shah(rps587 Homework 5 mcdevitt(52365 This print-out...

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shah (rps587) – Homework 5 – mcdevitt – (52365) 1 This print-out should have 26 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Find the percentage oxygen by weight in KMnO 4 . 1. 40.5% correct 2. 10.1% 3. 66.7% 4. 14.5% Explanation: Percent = part whole × 100% In this case, the part would be the mass of O in a certain mass of KMnO 4 (the whole). The simplest amount of KMnO 4 to use is the mass of one mole of KMnO 4 . Each mole of KMnO 4 contains 4 mol of O, 1 mol of Mn, and 1 mol of K. We know the atomic masses of each of these elements from the periodic table. Using these atomic masses we calculate the grams of each element in one mole of KMnO 4 : g from O = 4 mol O × 15 . 9994 g O 1 mol O = 63 . 9976 g O g from Mn = 1 mol Mn × 54 . 938 g Mn 1 mol Mn = 54 . 938 g Mn g from K = 1 mol K × 39 . 0983 g K 1 mol K = 39 . 0983 g K To find the mass of one mole of KMnO 4 we add the masses of the component parts: molar mass KMnO 4 = 63 . 9976 g + 54 . 938 g +39 . 0983 g = 158 . 034 g KMnO 4 mol KMnO 4 , so the percentage of O in KMnO 4 is ? % O = 63 . 9976 g O 158 . 034 g KMnO 4 × 100% = 40 . 4961% O . 002 (part 1 of 2) 10.0 points You are asked to identify compound X (a white, crystalline solid), which was extracted from a plant seized by customs inspectors. An elemental analysis of X shows that the mass percentage composition of the compound is 26.86% C and 2.239% H, with the remainder being oxygen. A mass spectrum of X yields a molar mass of 90.0 g/mol. Write the empirical formula of X. 1. C 3 H 2 O 5 2. C 2 HO 2 3. CH 2 O 4. CHO 2 correct 5. C 2 H 2 O Explanation: For 100 g of compound, mol of C = 26 . 86 g 12 . 01 g / mol = 2.221 mol mol of H = 2 . 239 g 1 . 0079 g / mol = 2.221 mol mol of O = 71 . 081 g 16 . 0 g / mol = 4.443 mol Dividing by 2.221 mol gives 1.0 C : 1 H : 2.0 O, so the empirical formula is CHO 2 . 003 (part 2 of 2) 10.0 points Write the molecular formula of X. 1. C 2 H 2 O 4 correct 2. C 2 H 4 O 2 3. CH 3 O 5 4. C 8 H 7 O 5 5. C 6 H 5 O 6

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shah (rps587) – Homework 5 – mcdevitt – (52365) 2 Explanation: Since the molar mass of the empirical for- mula is 45.0 g/mol while the molar mass of X is 90.0 g/mol, the molecular formula is twice the empirical formula or C 2 H 2 O 4 . 004 10.0 points Balance the equation ? As(OH) 3 (s)+? H 2 SO 4 (aq) ? As 2 (SO 4 ) 3 (aq)+? H 2 O(l) using the smallest possible integers. The co- efficients are 1. 2; 3; 2; 3. 2. The equation is balanced as written with the question marks. 3. 3; 2; 1; 6. 4. 4; 6; 2; 6. 5. 2; 3; 2; 12. 6. 2; 3; 1; 6. correct Explanation: A balanced equation must have the same number of each kind of atom on both sides of the equation. We find the number of each kind of atom using equation coefficients and composition stoichiometry. For example, we find there are 12 H atoms on the product side: ? H atoms = 6 H 2 O × 2 H 1 H 2 O = 12 H The balanced equation is 2 As(OH) 3 + 3 H 2 SO 4 As 2 (SO 4 ) 3 + 6 H 2 O and has 2 As, 12 H, 3 S and 18 O atoms on each side. 005 10.0 points Balance the reaction equation CO + O 2 CO 2 .
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