0.188 g of an organic compound having an empirical formula `CH_(2)Br` displaced 24.2 cc of air at `14^(@)C` and 752 mm pressure. Calculate the molecular formula of the compound. (Aqueous tension at `14^(@)C` is 12 mm)

Explanation

Pressure of dry gas = Pressure of gas - aqueous tension
= 752 - 12 = 740 mm
According to ideal gas equation PV = `W/MRJ`.
`W=0.188" "V=24.2/1000Lit.`
`M.wt=?" "P=740/760`
T = 273 + 14 = 287K
Substituting these values in ideal gas equation
Molecular weight
`(0.188xx0.0821xx287xx760xx1000)/(740xx24.2)=188`
Empirical formula of the compound = `CH_(2)Br`
Empirical formula wt. of the compound
= 12 + 2 + 80 = 94
`(M.W)/("Empirical formula wt.")=188/94=2`
`therefore` Molecular formula = Empirical formula `xx` 2
= `(CH_(2)Br)_(2)=C_(2)H_(4)Br_(2)`.