Answer:
Molecular formula = empirical formula = COCl2
Explanation:
Step 1: Data given
MAss of carbon = 0.305 grams
MAss of oxygen = 0.407 grams
Mass of chlorine = 1.805 grams
A different sample of the same gas having a mass of 8.84 g occupies 2 L at STP
Atomic mass of C = 12.01 g/mol
Molar mass of O2 = 16.0 g/mol
Molar mass of Cl2 = 35.45 g/mol
Step 2: Calculate moles
Moles = mass / molar mass
Moles C = 0.305 grams / 12.01 g/mol
Moles C = 0.0254 moles
Moles O = 0.407 grams / 16.0 g/mol
Moles O = 0.0254 moles
Moles Cl = 1.805 grams / 35.45 g/mol
Moles Cl = 0.0509 moles
Step 3: Calculate the mol ratio
We divide by the smallest amount of moles
C: 0.0254 moles / 0.0254 moles = 1
O: 0.0254 moles / 0.0254 moles = 1
Cl: 0.0509 / 0.0254 = 2
The empirical formula is COCl2
Step 4: Calculate moles
At STP we have 22.4 L for 1 mol
For 2 L we have 1 / 11.2 = 0.0893 moles
Step 5: Calculate the molar mass
Molar mass = mass / moles
Molar mass = 8.84 grams / 0.0893 moles
Molar mass = 98.99 g/mol
Step 6: Calculate the molar mass of the empirical formula
COCl2 = 12.01 + 16.0 + 70.9 ≠ 98.91 g/mol
Step 7: Calculate the molecular formula
Molecular formula = empirical formula = COCl2
