Question

A 0.04403 g sample of gas occupies 10.0-ml at 289.0 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?

Answer 1

We can first find the number of moles of gas in the container using the ideal gas law equation 
PV = nRT 
where P - pressure - 1.10 atm x 101 325 Pa/atm = 111 458 Pa
V - volume - 10.0 x 10⁻⁶ m³
n - number of moles
R - universal gas constant - 8.314Jmol⁻¹K⁻¹
T - temperature - 289.0 K
substituting these values in the equation 
111 458 Pa x 10.0 x 10⁻⁶ m³ = n x 8.314Jmol⁻¹K⁻¹ x  289.0 K
n = 0.463 x 10⁻³ mol
molar mass of the compound = 0.04403 g/ (0.463 x 10⁻³ mol)
molar mass = 95.1 g/mol

first we need to find the empirical formula of the compound, this is the simplest ratio of whole numbers of components in the compound.
assuming 100 g of the compound is present 
element                     C                            Cl
mass present        25.305 g                  74.695 g
number of moles   $\frac{25.305g}{12 g/mol} \frac{74.695g}{35.5 g/mol}$
                             = 2.11                       = 2.10
divide by the least number of moles 
                              2.11/2.10 = 1.00      = 2.10/2.10 = 1.00
therefore C - 1 
Cl - 1
empirical formula - CCl
molecular formula is actual composition of components in the compound 
mass of empirical unit - 12 + 35.5 = 47.5
mass of molecular formula = 95.1 g/mol
number of empirical units = 95.1 / 47.5 = 2.00
therefore molecular formula = 2(CCl)

molecular formula = C₂Cl₂