Step 1: Find the number of moles of the gas
Use PV = nRT
P = pressure = 1.10 atm
V = volume = 10 ml = 0.01 L
T = temperature = 293.5 K
R = gas constant = 0.0821 L,atm/mol-K
n = PV/RT = 1.10 * 0.0l/0.0821*293.5 = 0.000456 moles
Step 2: Find the moles of C and Cl in the given mass of the sample
Mass of sample = 0.04336 g
% C = 25.305% and % Cl = 74.695%
Therefore mass of C = 0.25305 * 0.04336 = 0.01097 g
mass of Cl = 0.74695*0.04336 = 0.03239 g
# moles of C = 0.01097 g/12 g.mol-1 = 0.000914 moles
# moles of Cl = 0.03239 g/ 35 g.mol-1 = 0.000925 moles
Step 3: Find the empirical formula and the empirical formula mass
C = 0.00914/0.00914 = 1
Cl = 0.00925/0.00914 = 1.01 = 1
Empirical formula = CCl
Empirical formula mass = 12 + 35 = 47 g.mol-1
Step 4: Find the molecular mass
# moles of the gas from step 1 = 0.000456 moles
Mass of the gas = 0.04336 g
Molecular mass = 0.04336/0.000456 = 95.087 g.mol-1
Step 5: Find the molecular formula
n = molecular mass/empirical mass = 95.087/47 = 2.02 = 2
Molecular formula = n(Empirical formula) = 2(CCl) = C₂Cl₂
Ans: the molecular formula of the compound is C₂Cl₂