0.246 g of an organic compound gave 0.198 g of carbon dioxide and 0.1014 g of water on complete combustion. 0.37 g of the compound gave 0.638 g of silver bromide. What is the molecular formula of the compound if its vapour density is 54.4?

Options

A. `C_(2)H_(5)Br`
B. `CH_(3)Br`
C. `C_(2)H_(7)Br`
D. `C_(3)H_(8)Br`

Answer

B

Explanation

Calculation of percentage composition :
(i) Percentage of carbon ` = (12)/(44) xx ("Mass of " CO_(2)"produced")/("Mass of substance taken") xx 100`
` = (12)/(44) xx (0.198)/(0.246) xx 100 = 21.95%`
(ii) Percentage of hydrogen
` = (2)/(18) xx ("Mass of " H_(2)O "produced ")/("Mass of substance take") xx 100`
`= (2)/(18) xx (0.1014)/(0.246) xx 100 = 4.58%`
(iii) Percentage of bromine
` = (80)/(180) xx ("Mass of AgBr formed")/("Mass of substance taken") xx 100`
` = (80)/(180) xx (0.368)/(0.37) xx 100= 73.37%`
(iv) The given compound does not contain oxygen since the sum of the percentage of carbon hydrogen and bromine is approximately 100, i.e., 21.95 + 4.58 + 73.37 = 99.90
Calculation of empirical formula :

Thus the empirical formula of the compound is `C_(2)H_(5)Br`
(B) Determination of molecular formula :
Vapour density of the compound = 54.5
`therefore ` molecular mass of the compound
` 2 xx ` vapour density ` = 2 xx 54.4 = 108.8`
But the empirical formula mass of the compound `C_(2)H_(5)Br = 2 xx 12 + 5 xx 1+1xx80 = 109`
`therefore n = ("Molecular mass ")/("Empirical formula maass") = (108)/(109) = 1` (approx)
Thus , the molecular formula of the compound
` = n xx ` (Empirical formula)
`= 1 xx (C_(2)H_(5)Br) = C_(2)H_(5)Br`