Question

Nitrobezene 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%` (i) Calculate the empirical formula of the minor product, (ii) when `5.5` gm of the minor product is dissolved in 45 gm of benzene, the boioling point of the solution is `1.84^(@)C` higher than that of pure benzene. Calculate the molar mass of the minor product and determine its molecular its molecular and structural formula. (Molal boiling point elevation constant of benzene is `2.53 K kg mol^(-1)`.)

Answer 1

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.