0.188 g of an organic compound having an empirical formula CH2Br displaced 24.2 cc of air at 14 degree Celsius and 752 mm pressure . Calculate the molecular formula of the compound ( Aqueous tension at 14 degree Celsius is 12 mm
Hello aspirant,
First we have to calculate the molar mass of organic compound having an empirical formula, CHBr
using ideal gas equation:
PV=nRT
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=0.0242L24.2cm3=24.2ml=0.0242L
conversion used :
(1cm^3=1ml)
(1L=1000ml)
(1cm3=1ml)
(1L=1000ml)
T = temperature of gas = 14^oC=273+14=287K14oC=273+14=287K
R = gas constant = 0.0821 L.atm/mol.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) * (0.0242L)=[{0.188g}/{M}] * (0.0821L.atm/mol.K)* (287K)
M=185.08g/mole
The Empirical formula = C H Br
The empirical formula weight = 12 + 1 + 80 = 93 gram/eq
Now we have to calculate the valency factor.
Formula used :
n= {Molecular formula}/{Empirical formula weight}}= {185.08}/{93}=1.99≈2
Molecular formula = (C H Br)n=(C H Br)2=C2H2Br2
Therefore, the molecular of the compound is, C2H2Br2.
Hope, this helps you.
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