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 132: 61b


$$0.0362 mols$$

Work Step by Step

Divide masses provided by the molecular mass of the molecule/element to find the mols. $12.4gC_{12}H_{22}O_{11}\times\frac{1mol C_{12}H_{22}O_{11}}{342.30g C_{12}H_{22}O_{11}} = 0.0362mol$
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