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 - Exercises: 3.54a

Answer

Empirical Formula: $C_{13}H_{18}O_2$ Molecular Formula: $C_{13}H_{18}O_2$

Work Step by Step

1. Divide the percentages by the molar mass of the atoms: C: 75.69% $\div$ 12 = 6.3075 H: 8.8 $\div$ 1 = 8.8 O: 15.51 $\div$ 16 = 0.969 These are the proportions: $C_{6.3075}H_{8.8}O_{0.969}$ 2. Now we need to transform these numbers to integers numbers: *We can divide all numbers by the minor of them: C : 6.3075 / 0.969 = 6.5 H : 8.8 / 0.969 = 9 O: 0.969 / 0.969 = 1 * To get rid of the .5, we can multiply all numbers by 2 C: 6.5 * 2 = 13 H: 9 * 2 = 18 O: 1 * 2 = 2 The Empirical Formula is: $C_{13}H_{18}O_2$ To find the Molecular Formula, let's calculate the molecular mass of the empirical, and compare: $C_{13}H_{18}O_2$ : 13*12 + 18*1 + 2*16 = 206g/mol Since the molar mass of the empirical is equal to the molar mass given in the problem, we don't need to change anything. Molecular Formula: $C_{13}H_{18}O_2$
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