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Nitrobenzene 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 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 benzene. Calculate the molar mass of the minor product and determine its molecular and structural formula. (Molal boiling point elevation constant of benzene is 2.53 K kg mol^-1.)

Shane Macguire

12 Years agoGrade upto college level

1 Answer

Deepak Patra

12 Years ago

Hello Student,

Please find the answer to your question

Element	%	Relative no. of atoms	Simplest ratio
C	42.86	42.86/12 = 3.57	3.57/1.19 = 3
H	2.40	2.40/1 = 2.40	2.40/1.19 = 2
N	16.67	16.67/14 = 1.19	1.19/1.19 = 1
O	38.07	38.07/16 = 2.38	2.38/1.19 = 2

∴ Empirical formula of the minor product is C₃H₂NO₂ Molar empirical formula mass of the minor product = 3 * 12 + 2 * 1 * 14 + 2 * 16 = 84 g mol^-2

Let M be the molar mass of the minor product. For 5.5 g of the minor product dissolved in 45 g benzene, the molality

(m) of the solution = 5.5 g/M/0.045 kg

Substituting this in the expression of elevation of boiling point,

∆T_b = k_bm⇒ 1.84 K = (2.53 K kg mol^-1) (5.5 g/M/0.045 kg)

Or M = 168 g mol^-1

No. of unit of empirical formula in molecular formula

= 168g mol^-1/84 g F = 2

Hence the molecular formula of the minor product is

2 (C₃H₂NO₂), i.e., C₆H₄(NO₂)₂.

The product is m – dinitrobenzene

Thanks
Deepak patra
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