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

Given Emperical formula `CH_(2)Br`
Wt. of compound =0.188 gms
Volume of displaced air = 24.2 CC
Temperature = `14^(@)C=287 K `
Pressure = 752 mm.
STP Conditions
`P_(1)=760` " mm"
`V_(1)=?`
`T_(1)=273` "K"
Given conditions
`P_(2)` = Pressure - aqueous tension
= 752 -12 mm=740 mm
`V_(2)=24.2 K`
`T_(2)=287 K`
Formula
`(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))implies(760x x V_(1))/(273)=(740 xx 24.2)/(287)`
`V_(1)=(740 xx 24.2)/(287)xx(273)/(760)=22.41 ml (or)22.414 " CC"`
:. 0.188 gms of organic compound displaced 22.414 CC of air
gms of organic compound displaced 22400 CC of air
`(22400)/(22.414)xx0.188 =188 :.` Gram molecular wt of compound= 188
Molecular formula = n (Emperical formula)
=`n(CH_(2)Br)`
n= `("Molecular wt")/("Emperical wt")=(188)/(94)=2 :." Molecular fornula"=C_(2)H_(4)Br_(2)`