Question

0.246 gm of an organic compound containing `58.53%` carbon and `4.06%` hydrogen gave 22.4 ml of nitrogen at STP. What is the empirical formula of the compound?

Answer 1

calculation of composition:
(i). Percentage of carbon `=58.53` (given)
(ii). Percentage of hydrogen `=4.06` (given)
(iii) percentage of nitrogen
`=(28xxvol.of N_2" at "STPxx100)/(22400xx"Wt. Of compound")`
`=(28xx22.4xx100)/(22400xx0.246)=11.37%`
(iv) percentage of oxygen `=100-` ("percentage of "`C+`" percentage of" `H+`" percentage of N")
`=100-(58.53+4.06+11.38)=26.03`

Hence, the empirical formula of the compound is `C_(6)H_(5)NO_(2)`.