The empirical formula of a gas is CH₃O.If 2.77 g of the gas occupies 1.00 L at exactly 0 $\circ$ C at a pressure of 760 Torr,what is the molecular formula of the gas? A)C₄H₁₂O₄ B)C₂H₆O₂ C)C₅H₁₅O₅ D)CH₃O E)C₃H₉O₃

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1) Calculate the number of moles of gas using the ideal gas law: PV = nRT P = 760 Torr = 1 atm V = 1 L T = (0+273) K = 273 K R = 0.08206 L atm/mol K n = (1 atm * 1 L)/(0.08206 L atm/mol K * 273 K) = 0.0448 mol 2) Calculate the molar mass of the empirical formula: C = 1*12.01 = 12.01 H = 1*1.01 = 1.01 S = 1*32.07 = 32.07 U = 1*238.07 = 238.07 B = 13*10.81 = 140.53 O = 1*15.99 = 15.99 Total mass = 439.68 g/mol 3) Calculate the empirical formula mass: C = 1*12.01 = 12.01 H = 1*1.01 = 1.01 S = 1*32.07 = 32.07 U = 1*238.07 = 238.07 B = 13*10.81 = 140.53 O = 1*15.99 = 15.99 Total mass = 439.68 g/mol Empirical formula mass = 12.01 + 1.01 + 32.07 + 238.07 + 140.53 + 15.99 = 440.68 g/mol 4) Calculate the molecular formula mass: Molecular formula mass = empirical formula mass * n = 440.68 g/mol * n 5) Divide the given molar mass by the empirical formula mass to determine n: n = 2.77 g / 439.68 g/mol = 0.0063 mol Molecular formula mass = 440.68 g/mol * 0.0063 mol = 2.78 g/mol 6) Calculate the molecular formula: C = 1*12.01 = 12.01 H = 16*1.01 = 16.16 S = 1*32.07 = 32.07 U = 1*238.07 = 238.07 B = 12*10.81 = 129.72 O = 1*15.99 = 15.99 Total mass = 454.02 g/mol The molecular formula must be a multiple of the empirical formula, so divide the molecular formula mass by the empirical formula mass to determine the multiple: 454.02 g/mol / 440.68 g/mol = 1.030 Round to the nearest whole number to get the multiple: 1 The molecular formula is therefore the same as the empirical formula: C₂H₆O₂

The Empirical Formula of a Gas Is CH3O ∘\circ∘ C at a Pressure of 760 Torr,what Is the Molecular

The Empirical Formula of a Gas Is CH₃O $\circ$ C at a Pressure of 760 Torr,what Is the Molecular