0.246 gm of an organic compound gave 0.198 gm of carbon dioxide and 0.1014 gm of water on complete combustion. 0.37 gm of the compound gave 0.638 gm of silver bromide. What is the molecular formula of the compound if its vapour density is `54.4?`

Explanation

calculation of percentage composition:
(i) percentage of carbon
`=(12)/(44)xx("mass of " CO_2" produced")/("mass of substance taken")xx100`
`=(12)/(44)xx(0.198)/(0.246)xx100=21.95%`
(ii). Percentage of hydrogen
`=(2)/(18)xx("mass of "H_2O" produced")/("mass of substance taken")xx100`
`=(2)/(18)xx(0.1014)/(0.246)xx100=4.57%`
(iii). percentage of bromine
`=(80)/(180)xx("mass of AgBr formed")/("mass of substance taken")xx100`
`=(80)/(188)xx(0.638)/(0.37)xx100=73.37%`
(iv). The given compound doen not contain oxygen since the sum of the percentages of carbon, hydrogen, and bromine is approximately 100, i.e., `21.95+4.58+73.37=99.90`
b.
(c). Determination of molecular formula,
Vapour density of the compound `=54.4`
molecular mass of the compound `=2xx`vapour density
`=2xx54.4=108.8`
But the empirical formula mass of the compound `C_2H_5Br=2xx12+5xx1+1xx80=109`
`n=("molecular mass")/("empirical formula mass")=(108.8)/(109)=1` (approx)
Thus the molecular formula of the compound `=nxx`(Empirical)`=1xx(C_2H_5Br)=C_2H_5Br`.