Chemistry: A Molecular Approach (3rd Edition)

Published by Prentice Hall
ISBN 10: 0321809246
ISBN 13: 978-0-32180-924-7

Chapter 3 - Sections 3.1-3.12 - Exercises - Problems by Topic - Page 133: 88a



Work Step by Step

Assume 100g of the compound. Convert the mass of each constituent element into moles by dividing by its molar mass: $58.80g\ C\times\frac{1mol}{12.01g}=4.896mol\ C$ $9.87g\ H\times\frac{1mol}{1.01g}=9.77mol\ H$ $31.33g\ O\times\frac{1mol}{16.00g}=1.958mol\ O$ Formulas have whole number subscripts so divide all the amounts by the smallest and round if the number is really close. $\frac{4.896mol\ C}{1.958}=2.5\times2=5$ $\frac{9.77mol\ H}{1.958}=5\times2=10$ $\frac{1.958mol\ O}{1.958}=1\times2=2$ So the empirical formula for this compound is: $C_{5}H_{10}O_{2}$
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