Fall 07 HW3 solution

Fall 07 HW3 solution - Homework Assignment #3

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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 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...
View Full Document

Page1 / 4

Fall 07 HW3 solution - Homework Assignment #3

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online