Chemistry and Chemical Reactivity (9th Edition)

Published by Cengage Learning
ISBN 10: 1133949649
ISBN 13: 978-1-13394-964-0

Chapter 2 Atoms, Molecules, and Ions - 2-10 Chemical Analysis: Determining Compound Formulas - Review & Check for Section 2-10 - Page 93: 3



Work Step by Step

For a 100.0 g sample: C: $73.14\ g,\ 12.011\ g/mol \rightarrow 6.089\ mol$ H: $7.37\ g,\ 1.008\ g/mol \rightarrow 7.312\ mol$ O: $19.49\ g,\ 15.999\ g/mol \rightarrow 1.218\ mol$ Normalizing by the number of moles of oxygen: C: $6.089/1.218=5.0$ H: $7.312/1.218=6.0$ O: $1.218/1.218=1.0$ The empirical formula is $C_5H_6O$ and its molar mass is $M_u=82.102\ g/mol$ Since the molar mass is $M=164.2\ g/mol$, $M/M_u=2.0$ So, the molecular formula is $C_{10}H_{12}O_2$
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