Question

A 0.271-g sample of a gaseous hydrocarbon (C_xH_y) occupies 226 mL at 22.0 ºC at 735 torr. Calculate the molar mass of this hydrocarbon in g/mol. Express your answer as a number (not scientific notation) without units; for full credit, the last digit of your answer should be within ±1 of the correct value, with an appropriate number of significant figures.

Molar mass: Answer          g/mol.

Complete combustion of 1.532 g of this gaseous hydrocarbon (C_xH_y) yields 4.484 g of carbon dioxide and 2.754 g of water. Calculate the moles of C and H in this amount of the hydrocarbon. Express your answers as a number (not scientific notation) without units; for full credit, the last digit of your answer should be within ±2 of the correct value, with an appropriate number of significant figures.

Answer moles C in C_xH_y

Answer moles H in C_xH_y

Determine the empirical and molecular formulas of this hydrocarbon. If a subscript in the formula is "1", enter it explicitly.

Empirical formula: CAnswerHAnswer.

Molecular formula: CAnswerHAnswer.

Answer 1

i 1 me = 103 226 mL = 2268103 L ó, Oe = 273 k. 22.0°e = 273+ 2205 295.0 .'. I torr = 0.00 13 atm 735 tort = 0.967 atm using Ideal gas equation: Pr = nRT 0,967 atm X 226X163 L=nX 0.0821 Latmmork X 295.0k (mole) = 9.02 x 103 mol

:. mote = mass molar mass molar mass = mass = 0,271 g = 30g/mol 9,02 x 103g 1mol . Cuty + puty > xCO2 + y the moles of co2 = 4.484 g - o. lol mal 44 g/mol moles of tho= mass = 2.7549 - TO = 0.153 mol malar mass 18g/mol moles of Cuty = mass = 1.532g molar mass -= OOSmol 30 g/mol from stoichiometric relation; moles of cutly = moles of coa x = moles of co -opolo moles of Cutty AISO, moles of Cutty = то] у но 7/2 moles of the X2 _ 0.153x2 6 0.05 y = Trnoles of Cutty) In Cutly; moles at c = xx moles of Cretty = 2 x 0.05 mol = oil mol moles of H = y x moles of Cutty = 6x 0.05 mol = 0.3 mol .. Cutty & Cats so molecular formula = cho Empirical formula = CH₂