Chemistry 12th Edition

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

Chapter 5 - Gases - Questions & Problems - Page 220: 5.98

Answer

$C_{6}H_{6}$

Work Step by Step

Number of moles of the compound $n=\frac{PV}{RT}=\frac{(0.74\,atm)(97.2\times10^{-3}\,L)}{(0.0821\,L\,atm\,K^{-1}mol^{-1})(200+273)K}$ $=0.0018522\,mol$ Molar mass= $\frac{\text{mass in grams}}{n}=\frac{0.145\,g}{0.0018522\,mol}=78.285\,g/mol$ As the molecular formula is a multiple of empirical formula, the molar mass is also a multiple of empirical formula mass. To know by what number the empirical formula is multiplied, we divide molar mass by empirical formula mass. $\frac{\text{molar mass}}{\text{empirical formula mass}}=\frac{78.285\,g/mol}{(12+1)g/mol}=6.0$ Then, Molecular formula= $C_{6}H_{6}$
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