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 133: 91

Answer

$NCl_{3}$

Work Step by Step

In the 6.61 mg sample of chloride, 0.77 mg nitrogen and (6.61-0.77)mg chlorine are present. Moles of N=$\frac{0.77\times10^{-3}g}{14.0067g/mol}=5.5\times10^{-5}\,mol$ Moles of Cl=$\frac{5.84\times10^{-3}g}{35.453g/mol}=1.65\times10^{-4}\,mol$ Since $5.5\times10^{-5}\,mol$ is the smallest value, division by it gives a ratio of 1:3 for N:Cl. The empirical formula is obtained by mentioning the numbers in the ratio after writing the respective symbols. Thus, the empirical formula is $NCl_{3}$.
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