CH 12 GAS LAWS PRACTICE QUESTIONS

CH 12 GAS LAWS PRACTICE QUESTIONS - Gas Law questions for...

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Gas Law questions for Class II 5.94 An anesthetic gas contains 64.81% carbon, 13.60% hydrogen, and 21.59% oxygen by mass. If 2.00 L of the gas at 25 o C and 0.420 atm weights 2.57 grams, what is the molecular formula of the anesthetic? 5.94 It is necessary to determine both the empirical formula and the molar mass of the gas. Empirical formula: Assume 100.0 grams of sample to make the percentages of each element equal to the grams of that element. Thus, 64.81% C = 64.81 g C, 13.60% H = 13.60 g H, and 21.59% O = 21.59 g O. Moles C = (64.81 g C) (1 mol C / 12.01 g C) = 5.39634 mol C (unrounded) Moles H = (13.60 g H) (1 mol H / 1.008 g H) = 13.49206 mol H (unrounded) Moles O = (21.59 g O) (1 mol O / 16.00 g O) = 1.349375 mol O (unrounded) Divide by the smallest number of moles (O): C = (5.39634 mol) / (1.349375 mol) = 4 H = (13.49206 mol) / (1.349375 mol) = 10 O = (1.349375 mol) / (1.349375 mol) = 1 Empirical formula = C 4 H 10 O (empirical formula mass = 74.12 g/mol) Molar Mass: M = mRT / PV = () ( ) L•atm 2.57 g 0.0821 273 25 K mol • K 0.420 atm 2.00 L ⎛⎞ + ⎜⎟ ⎝⎠ = 74.85 g/mol (unrounded) Molecular formula: Since the molar mass and the empirical formula mass are similar, the empirical and molecular formulas must both be: C 4 H 10 O
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This note was uploaded on 05/31/2008 for the course CHM 2045 taught by Professor Mitchell during the Spring '07 term at University of Florida.

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CH 12 GAS LAWS PRACTICE QUESTIONS - Gas Law questions for...

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