Chemistry 10th Edition

Published by Brooks/Cole Publishing Co.
ISBN 10: 1133610668
ISBN 13: 978-1-13361-066-3

Chapter 2 - Chemical Formulas and Composition Stoichiometrhy - Exercises - Composition Stoichiometry - Page 76: 50

Answer

a) $C_9H_9N$ b) $131.178\ g/mol$

Work Step by Step

Atomic weights: C - 12.011, H - 1.008, N - 14.007 a) Calculate the mass of each element in 100.00 g of this compound: $m_C=100.00\ g\cdot\frac{82.40\ g_C}{100.00\ g}=82.40\ g_C$ $m_H=100.00\ g\cdot\frac{6.92\ g_H}{100.00\ g}=6.92\ g_H$ $m_N=100.00\ g\cdot\frac{(100.00-82.40-6.92)\ g_N}{100.00\ g}=10.68\ g_N$ Calculate the number of moles of each element in this sample: $n=\frac mM$ $n_C=\frac{82.40\ g}{12.011\ g/mol}=6.860\ mol$ $n_H=\frac{6.92\ g}{1.008\ g/mol}=6.87\ mol$ $n_N=\frac{10.68\ g}{14.007\ g/mol}=0.7625\ mol$ Find the ratio of the number of moles of each element by the one with the least number of moles, Nitrogen: $\frac{n_C}{n_N}=\frac{6.860\ mol}{0.7625\ mol}\approx9$ $\frac{n_H}{n_N}=\frac{6.87\ mol}{0.7625\ mol}\approx9$ The formula of this compound should follow these proportions, so the formula is: $C_9H_9N$ b) Calculate the molar mass of this compound, given the formula obtained in part (a): $M=9\cdot 12.011+9\cdot1.008+14.007=131.178\ g/mol$
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