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: 3.53


Please see the work below.

Work Step by Step

We know that $molar\space mass \space of C_4H_5N_2O=4\times 12.0+5\times 1.0+2\times 14+1\times 16.0=97\frac{g}{mol}of C_4H_5N_2O$ Now, we can find the ratio as: $\frac{194.19}{97}=2$ It shows that the ratio of the empirical formula to the molecular formula is 2. Hence, we multiply the subscripts of the empirical formula by 2, and we obtain the molecular formula as: $C_8H_{10}N_4O_2.$
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