Homework 2 Key

Homework 2 Key - CHEMISTRY 201X: Homework #2 Due: Friday,...

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CHEMISTRY 201X: Homework #2 Due : Friday, September 7, 2007 “Complex Stoichiometry” (My apologies, but there is a necessary correction to problem #2b: the yield of solid is 85.00 g .) (1) (5 points) A 0.5000 g sample of an unknown hydrocarbon (compound containing just carbon and hydrogen) was analyzed by thorough combustion and found to produce 1.6477 g carbon dioxide. (a) What is the empirical formula of the hydrocarbon? C m H n + (2m + n/2) O 2 ⎯→ m CO 2 + (n/2) H 2 O Mass fraction of C in CO 2 = (12.011 amu)/(12.011 amu + 2 × 15.999 amu) = 0.2729 Mass of C in C m H n = (0.2729)(1.6477 g) = 0.4497 g C # moles C = (0.4497 g) / (12.011 g/mol) = 0.03744 mol C Mass of H in C m H n = 0.5000 g 0.4497 g = 0.0503 g H # moles H = (0.0503 g) / (1.008 g/mol) = 0.04990 mol H Therefore, m:n = 0.03744 : 0.04990 = 1: 1.333 = 3 : 4 Empirical formula is C 3 H 4 (b) A mass spectrum of the substance shows the largest mass peaks at 15 amu, 75 amu, 90 amu, 105 amu and 120 amu. What is the molecular formula of this compound? Explain
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This homework help was uploaded on 03/27/2008 for the course CHEM 201 taught by Professor Miller during the Fall '07 term at Iowa State.

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Homework 2 Key - CHEMISTRY 201X: Homework #2 Due: Friday,...

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