Chemistry and Chemical Reactivity (9th Edition)

Published by Cengage Learning
ISBN 10: 1133949649
ISBN 13: 978-1-13394-964-0

Chapter 2 Atoms, Molecules, and Ions - Study Questions - Page 95i: 131


Empirical and molecular formula: $C_5H_{14}N_2$

Work Step by Step

Atomic weights (g/mol): $C: 12.011, \ H: 1.008,\ N: 14.007$ In 100 g : C: $58.77\ g \div 12.011 \ g/mol=4.89\ mol$ H: $13.81\ g\div1.008\ g/mol=13.70\ mol$ N: $27.40\ g\div14.007\ g/mol=1.96\ mol$ Dividing all these number of moles by the smallest value (1.96) leads to the empirical formula: C: $4.89\div1.96=2.50$ H: $13.70\div1.96=7.00$ $C_5H_{14}N_2$ Molar mass of the empirical formula: $102.18\ g/mol$ Molar mass of the compound: $102.2\ g/mol$ Ratio of the two: 1.00 Molecular formula: $C_5H_{14}N_2$
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