Chemistry: The Central Science (13th Edition)

Published by Prentice Hall
ISBN 10: 0321910419
ISBN 13: 978-0-32191-041-7

Chapter 3 - Chemical Reactions and Reaction Stoichiometry - Additional Exercises: 3.93

Answer

C10H12N2O

Work Step by Step

68.2 % C, 6.86% H, 15.9% N and 9.08 % O Taking a basis of 100g we now have 68.2 g C, 6.86g H, 15.9g N, 9.08g O. Converting to moles we have C is 12 g/mol so 68.2g is 5.68 moles H is 1g/mol so 6.86g is 6.86 moles N is 14g/mol so 15.9g is 1.14 moles O is 16g/mol so 9.08g is 0.568 moles Now we divide each of these by the smallest number, 0.568 5.68/0.568 = 10 6.86/0.568 = 12 1.14/0.568 = 2 0.568/0.568 = 1 The ratio is 10:12:2:1 or C10H12N2O This would have a molar mass of 10*12 = 120 12*1 = 12 14*2 = 28 16 176 g/mol. Therefore, the empirical formula is the molecular formula
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