A0.04336 g sample of gas occupies 10.0-ml at 293.5 k and 1.10 atm. upon further analysis, the compound is found to be 25.305% c and 74.695% cl. what is the molecular formula of the compound?

The molecular formula  of the compound  is C₂Cl₂

 calculation

Step 1 : calculate the  empirical  formula  as below

 find the moles of each  C and  cl

 moles =  % composition / molar mass

from periodic table the molar  mass of C = 12  g/mol , Cl = 35.5 g/mol

moles of C = 25.305 / 12 = 2.11   moles

moles for Cl = 74.695/35.5 =2.10 moles

find   the mole ratio   by diving each mole  by smallest number  of mole

that  is for C = 2.11 moles/ 2.10 moles = 1

            for  Cl = 2.10 moles / 2.10 moles = 1

Therefore the empirical formula = CCl

Step 2:  find the  molar weight  of the compound  

 molar weight  =  mass/ moles

 moles  of the compound is calculated using the ideal gas equation

 = PV=nRT

where;

       P(pressure) = 1.10 atm

       V(volume) =10.0 ml in  to liters = 10/1000 = 0.01 L

   n( number  of moles) = ?

 R( gas constant)  = 0.0821 L.atm/mol.K

T( temperature) = 293.5 k

make n the subject  of the formula by diving both side of equation by RT

n=Pv/RT

n= {(1.10 atm x 0.01 L)/ ( 0.0821 L.atm/mol.K  x  293.5 K) =0.00046 moles

molar weight  is therefore = 0.04336 g /0.00046 mol =  94 g/mol

Step 3: find the molecular formula of the compound

  [CCl]n = 94 g/mol

[12 + 35.5]n = 94 g/mol

47.5 n = 94 g/mol

divide both side by n

n =2

Therefore  the  molecular formula = [CCl]2 = C₂Cl₂