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 - 2-10 Chemical Analysis: Determining Compound Formulas - Case Study - Questions: 2


$C_{18}H_{18}O_3N_3As_3$ and $C_{30}H_{30}O_5N_5As_5$ respectively

Work Step by Step

From a 100.0 g sample: C: 39.37 g, 12.011 g/mol --> 3.278 mols H: 3.304 g, 1.008 g/mol --> 3.278 mols O: 8.741 g, 15.999 g/mol --> 0.5463 mols N: 7.652 g, 14.007 g/mol --> 0.5463 mols As: 40.932 g, 74.922 g/mol --> 0.5463 mols Normalizing by the smallest number of mols: C, H: 3.278/0.5463 = 6.0 O, N, As: 0.5463/0.5463 = 1.0 The emprical formula of both compunds is: $C_6H_6ONAs$, which has a molar mass of $M_u=183.042\ g/mol$ For the first compound ($M=549\ g/mol$): $M/M_u=3.0$, so its molecular formula is: $C_{18}H_{18}O_3N_3As_3$ For the second compound ($M=915\ g/mol$): $M/M_u=5.0$ $C_{30}H_{30}O_5N_5As_5$
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