Compound in the vaporstate density of 3,27g/L 95 C 758 tor whatismolecular weight 24.2%C4.1%H71.7%Cl molformua

ok what is the certain organic compond in the vapor state has a denstity of 3.27 g/L at 95 degrees celcius and 758 torr. What is the molecular weight? if the compound contains 24.2% C, 4.1%H and 71.7% Cl what is the molecuar formula?

Answer:
We use the equation
d = M x p / RT
M is the molecular weight
p = 758 / 760 = 0.997 atm
T = 95 + 273 = 368 K

M = d RT / p = 3.27 x 0.0821 x 368 / 0.997 = 99 g/mol .
This is the molecular weight

we consider 100 g of gas
we have 24.2 g of C . 24.2 / 12.011 = 2.0
4.1 g / 1.00794 = 4.1
71.7 / 35.453 = 2
the empirical formula is
CH2Cl ( mass = 49.45 g/mol)
99 / 49 = 2.00
we mulpiply each number for 2
and we get molecular formula : C2H4Cl2 ( molecular weight = 99 g/mol)
It may be : CH3CHCl2 or CH2CLCH2Cl

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Compound in the vaporstate density of 3,27g/L 95 C 758 tor whatismolecular weight 24.2%C4.1%H71.7%Cl molformua

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