Determination Molecular Formula by Combustion Analysis and Gas Laws Complete combustion of 1.110 g of a gaseous hydrocarbon yields 3 ... empirical for...

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It is somewhat important to be sure that the hydrocarbon is only carbon and hydrogen. This can be easily accomplished by converting these molar quantities of C and H into mass and adding them together. If the sum of the masses adds to 1.110 g, then there is no other atom in the hydrocarbon. Without further proof, suffice it to say that the hydrocarbon is only C and H. Now, normalize to the least common multiple (i.e., divide by the smallest n).... x = 1 y = 1.5 so the empirical formula is C 2 H 3 Determine the molar mass of the compound: PV m = R and n = nT M

so

PVM mTR = R and M = mT PV

V = 0.131 L ⎛ ⎞ P = ⎜ 753 mmHg mmHg ⎟ = 0.9908 atm 760 atm ⎠ ⎝ T = 24.8° + 273.2 K = 298.0 K M=

L ⋅ atm mTR ( 0.288 g )( 298.0 K ) ( 0.082059 mol g ⋅K ) = = 54.26 mol PV ( 0.9908 atm )( 0.131 L )

Finally, determine the factor and molecular formula: f =

g 54.26 mol =2 g 27.044 mol

Molecular formula is C4 H 6