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: 87b

Answer

$C_{4}H_{5}N_{2}O$

Work Step by Step

Assume 100g of the compound. Convert the mass of each constituent element into moles by dividing by its molar mass: $49.48g\ C\times\frac{1mol}{12.01g}=4.120mol\ C$ $5.19g\ H\times\frac{1mol}{1.01g}=5.13mol\ H$ $28.85g\ N\times\frac{1mol}{14.01g}=2.059mol\ N$ $16.48g\ O\times\frac{1mol}{16.00g}=1.030mol\ O$ Formulas have whole number subscripts so divide all the amounts by the smallest and round if the number is really close. $\frac{4.120mol\ C}{1.030}=4$ $\frac{5.13mol\ H}{1.030}=5$ $\frac{2.059mol\ N}{1.030}=2$ $\frac{1.030mol\ O}{1.030}=1$ So the empirical formula for this compound is: $C_{4}H_{5}N_{2}O$
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