Nirtobenzene is formed as the major product along with a minor product in the reaction of benzene with a hot mixture of nitric acid and sulphuric acid. The minor product consists of carbon `42.86%`, hydrogen `2.40%`, nitrogen `16.67%` and oxygen `38.07%`.
a. Calculate the empirical formula of the minor product.
b. When `5.5 g` of the minor product is dissolved in `45 g` of benzene, the boiling point of the solution is `1.84 C` higher than that of pure benzen. Calculate the molar mass of the minor product and determine its molecular and structural formulae. (Molar elevation constant of benzene is `2.53 K kg mol^(-1))`

Explanation

The rations of atoms in the minor products are :
`C:H:N:O` `{:( ::(42.86)/(12):(2.40)/(1):(16.67)/(14):(38.07)/(16)),( ::3.57 : 2.40 : 1.19 : 2.38), ( :: 3 : 2 : 1 : 2):}`
Empirical formula of the minor product: `C_(3)H_(2)NO_(2)` Molar empirical formula mass of the minor product is `(3xx12+2xx1+1xx14+2xx16)` gm `mol6(-1)` `= 84 gm mol^(-1)`
Let M be the molar mass of the minor product. For `5.5` gm of the minor product dissloved in 45 gm benzene, the molality of solution is given by
`m = (55gm//M)/0.045 kg)`
Subsituting this in the expression of elevation of boiling point, we get `Delta T_(a) = K_(b)m`
`1.84 K = (2.53K kg mol^(-1))` `((55gm//M)/(0.045kg))`
`M = ((2.53xx55)/(1.84xx0.045))gm mol^(-1)`
`=168gm mol^(-1)`
Number of unit of empirical formula in the moleculat formula `= (168gm mol^(-1))/(84gm mol^(-1))=2`
Hence, the molecular formula of the minor product is `2(C_(3)H_(2)NO_(2))`, i.e., `C_(6)H_(4)(NO_(2))` . The product is m-dinitrobenzene.