An organic compound on analysis was found to contain `16.27%` carbon. `0.67%` Hydrogen, `72.2%` chlorine. The V.D. of the compound is equal to 73.75. Calculate the empirical formula and molecular formula of the compound. 

Explanation

Empirical formula of the compound `=C_(2)HCl_(3)O`
Empirical formula weight
`=(2xx12)+(1xx1)+(3xx35.5)+(1xx16)=147.5`
`"Molecular weight "=2xxV.D=2xx73.75=147.50`
`n("Molecular weight")/("Empirical formula weight")=(147.50)/(147.50)=1`
`therefore " Molecular formula "=C_(2)HCl_(3)O`