When 0.72 g of a liquid is vaporized at 110°C and 0.967 atm,the gas occupies a volume of 0.559 L.The empirical formula of the gas is CH₂.What is the molecular formula of the gas? A)CH₂ B)C₂H₄ C)C₃H₆ D)C₄H₈ E)none of these

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1. Calculate the moles of the gas: PV = nRT n = (PV)/(RT) n = (0.967 atm * 0.559 L) / (0.08206 L.atm/mol.K * 383 K) n = 0.0156 mol 2. Calculate the empirical formula mass: CH₂ = 12.01 + 1.01 + 32.07 = 45.09 g/mol 3. Calculate the empirical formula molar mass: (0.72 g) / (45.09 g/mol) = 0.016 mol 4. Divide the molecular mass by the empirical formula mass to find the ratio of the molecular formula to the empirical formula: Molecular ratio = Molecular mass / Empirical formula mass We don't have the molecular mass yet, so let's call it "x": x / 45.09 = Molecular ratio 5. Use the ratio of moles from step 1 and the empirical formula from the question to set up a second equation: 0.0156 mol = k(CH₂) We don't know what "k" is yet (it's the number of empirical formula units in the molecular formula), so let's call it "y": 0.0156 mol = y(CH₂) 6. Solve for "y" by dividing the molecular formula by the empirical formula: y(CH₂) = Molecular ratio y(CH₂) = x / 45.09 y = (x / 45.09) / (CH₂) 7. Substitute y into the second equation: 0.0156 mol = y(CH₂) 0.0156 mol = (x / 45.09) / (CH₂) * (CH₃H₆) 0.0156 mol = x(CH₃H₆) / 45.09 x = 1.38 g/mol 8. Calculate the molecular formula mass: C₃H₆ = 12.01 + 1.01 + 32.07 + 1.01 + 32.07 + 1.01 + 32.07 = 111.25 g/mol 9. Divide the experimental molecular mass by the calculated molecular mass to find the ratio of the experimental molecular formula to the true molecular formula: (1.38 g/mol) / (111.25 g/mol) = 0.0124 10. Multiply the empirical formula by this ratio to get the molecular formula: (CH₃H₆) * 0.0124 = C₃H₆

When 072 G of a Liquid Is Vaporized at 110°C and and 0.967