A compound contains `4.07% H, 24.27% C`, and `71.65% Cl`. If its molar mass is `98.96`, the molecular formula will be

Options

A. `C_(2) H_(4) Cl_(2)`
B. `CH_(2) Cl`
C. `C_(3) H_(6) H_(3)`
D. `C_(4) H_(8) Cl_(4)`

Answer

A

Explanation

`n_(H) = (4.07g)/(1.008g mol^(-1)) = 4.04`
`n_(C) = (24.27g)/(2.01g mol^(-1)) = 2.02`
`n_(Cl) = (71.65g)/(31.45g mol^(-1)) = 2.02`
Dividing by the smallest number, we get
`(4.04)/(2.02) : (2.02)/(2.02) : (2.02)/(2.02)` `= 2 : 1 : 1`
`:. EF` is `CH_(2) Cl` and `EFM` is `(12.01 + 2 xx 1.008 + 35.45 )`
`= 49.48 u`
`n = (MM)/(EFM) = (98.96u)/(49.48u) = 2`
`:. MF = 2xx FE = 2(CH_(2) Cl) = C_(2) H_(4) Cl_(2)`