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 - Study Questions: 89

Answer

Empirical formula $C_3H_4$ Molecular formula: $C_9H_{12}$

Work Step by Step

Atomic weights (g/mol): $C: 12.011,\ H: 1.008$ In 100.0 g: C: $89.94\ g \div 12.011 = 7.488\ mol$ H: $10.06\ g\div 1.008 = 9.980\ mol$ Normalizing by the smallest amount of moles: C: $7.488/7.488= 1.0$ H: $9.980/7.488 = 1.333$ To get integer numbers on both proportions, we multiply each by 3 which yields: Empirical formula $C_3H_4$ Empirical formula's molar mass: $M_u=40.06\ g/mol$ $M/M_u=3.0$ Molecular formula: $C_9H_{12}$
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