Question

0.188 g of an organic compound having an empirical
formula, CH Br displaced 24.2 cc. of air at 14°C and
752 mm pressure. Calculate the molecular formula of the
compound. (Aqueous tension at 14°C is 12 mm.)
(a) CH,Br
(b) C2H2Br2
(c) C_H_Br4
(d) C H Br4​

Answer 1

Answer : The correct option is, (b) $C_2H_2Br_2$

Explanation:

First we have to calculate the molar mass of organic compound having an empirical  formula, $CHBr$

using ideal gas equation:

$PV=nRT\\\\PV=\frac{w}{M}RT$

where,

P = pressure of gas = 752 mm Hg = 0.989 atm

conversion used : (1 atm = 760 mmHg)

V = volume of gas = $24.2cm^3=24.2ml=0.0242L$

conversion used : $(1cm^3=1ml)\text{ and }(1L=1000ml)$

T = temperature of gas = $14^oC=273+14=287K$

R = gas constant = 0.0821 L.atm/mole.K

w = mass of an organic compound = 0.188 g

M = molar mass of an organic compound = ?

Now put all the given values in the ideal gas equation, we get:

$(0.989atm)\times (0.0242L)=\frac{0.188g}{M}\times (0.0821L.atm/mole.K)\times (287K)$

$M=185.08g/mole$

The Empirical formula = $CHBr$

The empirical formula weight = 12 + 1 + 80 = 93 gram/eq

Now we have to calculate the valency factor.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}=\frac{185.08}{93}=1.99\approx 2$

Molecular formula = $(CHBr)_n=(CHBr)_2=C_2H_2Br_2$

Therefore, the molecular of the compound is, $C_2H_2Br_2$

Answer 2

Explanation:

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