A substance was known by its mode of synthesis to contain `10` atoms of carbon per molecule along with unknown number of atoms of chlorine, hydrogen and oxygen. Analysis showed 60.5% carbon 5.5% hydrogen, 16.10% oxygen and 17.9% chlorine. The molecular formula of compound is

Options

A. `C_(10)H_(8)Cl_(2)`
B. `C_(10)H_(11)O_(2)Cl`
C. `C_(10)H_(10)Ocl`
D. `C_(10)H_(12)O_(2)Cl`

Answer

B

Explanation

`C:H:O:Cl=60.5/12:5.55/10:16.10/16:17.9/935.5`
`=5.04:5.55:1.006:0.0504`
`=10:11:2:1`
`EF=C_(10)H_(11)O_(2)Cl`
Since th e compound contains 10 carbons atoms per molecule and M.f. is the same as E.F.