Fall 07 HW3 solution

Fall 07 HW3 solution - Homework Assignment#3 Name...

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Homework Assignment #3 Name____________________________ Laboratory Section (circle one) MW1.30 MW6.30 TTh7 1. a) Chapter 3, Problem 24, 34 (parts a, b, and c ONLY), 38, 44, 54, 62, 64, 101, 103 3.24 The formula, from the figure, is (C 3 H 5 ) 2 S, and the molar mass is 114.21 g/mol. a) Grams allyl sulfide = ( ) 114.21 g allyl sulfide 1.63 mol allyl sulfide 1 mol allyl sulfide = 186.1623 = 186 g allyl sulfide b) C atoms = ( ) 23 3 5 2 3 5 2 3 5 2 3 5 2 1 mol (C H ) S 6 mol C 6.022 x 10 C atoms 4.77 g (C H ) S 114.21 g (C H ) S 1 mol (C H ) S 1 mol C     = 1.5090591 x 10 23 = 1.51 x 10 23 C atoms 3.34 a) C 4 H 8 = M.F. CH 2 = E.F. E.F. mass = 14.03 g/mol b) C 3 H 6 O 3 = M.F. CH 2 O = E.F. E.F. mass = 30.03 g/mol c) P 4 O 10 = M.F. P 2 O 5 = E.F. E.F. mass = 141.94 g/mol 3.38 a) 0.039 mol Fe 0.039 mol Fe = 1 0.052 mol O 0.039 mol Fe = 1.3333 The formula is Fe 1 O 1.3333 , which in whole numbers (x 3) is Fe 3 O 4 b) ( ) 1 mol P 0.903 g P 30.97 g P = 0.029157 mol P ( ) 1 mol Br 6.99 g Br 79.90 g Br = 0.087484 mol Br 0.029157 mol P 0.029157 mol P = 1 0.087484 mol Br 0.029157 mol P = 3 The empirical formula is PBr 3 . c) Assume a 100 g sample and convert the masses to moles. ( ) 79.9% C 1 mol C 100 g 100% 12.01 g C = 6.6528 mol C ( ) (100 79.9)% H 1 mol H 100 g 100% 1.008 g H - = 19.940 mol H 6.6528 mol C 6.6528 mol C = 1 19.940 mol H 6.6528 mol C = 3 The empirical formula is CH 3 . 3.44 Plan: Assume 100 grams of cortisol so the percentages are numerically equivalent to the masses of each element. Convert each of the masses to moles by using the molar mass of each element involved. Divide all moles by the lowest number of moles and convert to whole numbers to determine the empirical formula. The empirical formula mass and the given molar mass will then relate the empirical formula to the molecular formula. Solution: Moles C = ( ) 1 mol C 69.6 g C 12.01 g C = 5.7952 mol C Moles H = ( ) 1 mol H 8.34 g H 1.008 g H = 8.2738 mol H

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Moles O = ( ) 1 mol O 22.1 g O 16.00 g O = 1.38125 mol O 5.7952 mol C 1.38125 mol O = 4.20 8.2738 mol H 1.38125 mol O = 6.00 1.38125 mol O 1.38125 mol O = 1.00 The carbon value is not close enough to a whole number to round the value. The smallest number that 4.20 may be multiplied by to get close to a whole number is 5. (You may wish to prove this to
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