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
