Chemistry: A Molecular Approach (3rd Edition)

Published by Prentice Hall
ISBN 10: 0321809246
ISBN 13: 978-0-32180-924-7

Chapter 2 - Sections 2.1-2.9 - Exercises - Cumulative Problems: 95


$1.50\ g\ C$

Work Step by Step

1. Using the first sample, subtract the remaining elements from the mass of HCN in order to determine the mass of carbon 2. \begin{align} m_{HCN}=m_{H}+m_{C}+m_{N}\\ m_{C}=m_{HCN}-m_{H}-m_{N}\\ (7.83-0.29-4.06)g= 3.48\ g \end{align} 3. Then use the ratio $m_{C}\ :m_{HCN}$ from the first sample as a conversion factor in order to calculate the mass of carbon from the second sample. 4. $3.37\ g\ HCN\times \frac{3.48\ g\ C}{7.83\ g\ HCN}=1.50\ g\ C$
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