Chemistry 12th Edition

Published by McGraw-Hill Education
ISBN 10: 0078021510
ISBN 13: 978-0-07802-151-0

Chapter 3 - Mass Relationships in Chemical Reactions - Questions & Problems - Page 108: 3.42

Answer

Ammonia $(NH_3)$ is the richest source of nitrogen on a mass percentage basis.

Work Step by Step

The mass percent of nitrogen can be calculated using this equation: $$mass \space percentage \space (N) = \frac{molar \space mass (N) \times amount \space of \space N \space in \space 1 \space molecule}{molar \space mass \space (compound)} \times 100\%$$ (a) $ (NH_2)_2CO $ : ( 1.008 $\times$ 4 )+ ( 12.01 $\times$ 1 )+ ( 14.01 $\times$ 2 )+ ( 16.00 $\times$ 1 )= 60.06 g/mol $$\frac{14.01 \times 2}{60.06} \times 100\%= 46.7 \%$$ (b) $ NH_4NO_3 $ : ( 1.008 $\times$ 4 )+ ( 14.01 $\times$ 2 )+ ( 16.00 $\times$ 3 )= 80.05 g/mol $$\frac{14.01 \times 2}{80.05} \times 100\%= 35.0 \%$$ (c) $ HNC(NH_2)_2 $ : ( 1.008 $\times$ 5 )+ ( 12.01 $\times$ 1 )+ ( 14.01 $\times$ 3 )= 59.08 g/mol $$\frac{14.01 \times 3}{59.08} \times 100\%= 71.1 \%$$ (d) $ NH_3 $ : ( 1.008 $\times$ 3 )+ ( 14.01 $\times$ 1 )= 17.03 g/mol $$\frac{14.01 \times 1}{17.03} \times 100\%= 82.3 \%$$
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