Chemistry 12th Edition

Published by McGraw-Hill Education
ISBN 10: 0078021510
ISBN 13: 978-0-07802-151-0

Chapter 3 - Mass Relationships in Chemical Reactions - Questions & Problems - Page 108: 3.40


Please see the work below.

Work Step by Step

We know that: $molar\space mass\space of CHCl_3=12.01+1.008+3(35.45)=119.4\frac{g}{mol}$ Now, $\%C=\frac{12.01}{119.4}\times 100\%=10.06\%$ $\%H=\frac{1.008}{119.4}\times 100\%=0.8442\%$ $\%Cl=\frac{3(35.45)}{119.4}\times 100\%=89.07\%$
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