Question

Example 62
An organic compound on analysis gave the following percentage composition : C = 57.8%,
H = 3.6% and the rest is oxygen. The vapour density of the compound was found to be 83. Find out the
molecular formula of the compound.​

Answer 1

$ANSWER</p><p></p><p>We have,</p><p></p><p></p><p>An organic compound C=57.8%</p><p></p><p></p><p>H=3.6%</p><p></p><p></p><p>Thus the oxygen is 38.6% Since the total %is100</p><p></p><p></p><p>So, </p><p></p><p></p><p>Molecular weight i=2×vapordensity</p><p></p><p></p><p>=2×83=166</p><p></p><p></p><p>So, Mass of C=\dfrac{57.8}{100}\times166 </p><p></p><p></p><p></p><p>=95.948g</p><p></p><p></p><p>No pof carbon atoms =1295.948=7.99∼8atoms Since </p><p></p><p></p><p>12=mass of the carbon</p><p></p><p></p><p>Mass of H=1003.6×166=5.976∼6 atoms</p><p></p><p></p><p>Mass of O=10038.6×166=64.076g</p><p></p><p></p><p>No of atoms =1664.076=4atoms</p><p></p><p></p><p>Thus. The formulaa will be C8H6O4</p><p></p><p>$