0.220g of a sample of a volatile compound, containing carbon, hydrogen and chlorine yielded on combustion in oxygen 0.195g of CO_(2), 0.0804g of H_(2)O. 0.120g of the compound occupied a volume of 37.24mL at 105^(@)C and 768mm Hg pressure. Calculate the molecular formula of the compound.

0.220g of a sample of a volatile compound, containing carbon, hydrogen

1.	0.220g of a sample of a volatile compound, containing carbon, hydrogen and chlorine yielded on combustion in oxygen 0.195g of CO_(2), 0.0804g of H_(2)O. 0.120g of the compound occupied a volume of 37.24mL at 105^(@)C and 768mm Hg pressure. Calculate the molecular formula of the compound.
Answer» Solution :Moles of C in `CO_(2)=1 xx` moles of `CO_(2)= (0.195)/(44)=0.00443` Weight of `C= 0.00443 xx 12=0.05316g` Moles of H in `H_(2)O =2xx` moles of `H_(2)O= 2 xx (0.0804)/(18)= 0.00893` Weight of `H= 0.00893 xx1= 0.00893g` `therefore` weight of Cl= 0.22 `-(0.05316 + 0.00893)g= 0.15791g` Moles of `Cl= (0.15791)/(35.5)= 0.00445` Moles of `C: H: Cl= 0.00443: 0.00893: 0.00445` `=443: 893: 445=1:2:1` i.e., the empirical formula is `CH_(2)Cl` Now, volume of the vapour of 0.12g of compound at NTP `=(37.24 xx 768)/(378)xx (273)/(760)=27.18mL` Moles of the compound `=(0.12)/(M)` (M = mol .wt) `=(27.18)/(22400)` `thereforeM = 99` `therefore ("molecular formula weight")/("empirical formula weight")= (99)/(49.5)=2` Hence, the empirical formula is `(CH_(2)Cl)_(2)`, i.e. `C_(2)H_(40Cl_(2)`