Question

0.188 g of an organic compound having an empirical
formula, CH2Br 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.)​

Answer 1

Answer : C2H2Br2

Explanation:

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

using ideal gas equation:

PV=nRT  ==  PV=w/M  *RT

where,

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

conversion used : (1 atm = 760 mmHg)

V = volume of gas =  0.0242 L

conversion used :  1 cm³= 1 ml & 1 L= 1000 ml

T = temperature of gas =  287 K

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:

M= 185.05 g/mole

The Empirical formula =  CHBr

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

Now we have to calculate the valency factor.  n = 2

Formula used : n =  w/M

Molecular formula =  ( C H Br )₂ == C₂ H₂ Br₂

Therefore, the molecular of the compound is,  ""  C₂ H₂ Br₂""