Chemistry: The Central Science (13th Edition)

by Brown, Theodore E.; LeMay, H. Eugene; Bursten, Bruce E.; Murphy, Catherine; Woodward, Patrick; Stoltzfus, Matthew E.

Published by Prentice Hall

ISBN 10: 0321910419

ISBN 13: 978-0-32191-041-7

Chapter 3 - Chemical Reactions and Reaction Stoichiometry - Exercises - Page 114: 3.23a

Answer

63$u$

Work Step by Step

The components of the $NH_3$ have the following atomic masses: $H = 1u; N = 14u; O = 16u.$ So, the molecular mass of nitric acid will be the atomic mass of each element multiplied by the number of atoms of each. MM $(NH_3)= (1\times1)+(1\times14)+(3\times16)=63u$

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