(4.951 / molar mass CO2) = moles C = x
(2.027 / molar mass H2O) / 2 = moles H = y
k[x(molar mass C) + y(molar mass H)] = 56.11
molecular formula is ... C(kx)H(ky)
divide kx and ky by GCF to find empirical formula
When 1.578 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 4.951 grams of CO2 and 2.027 grams of H2O were produced. In a separate experiment, the molar mass of the compound was found to be 56.11 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon.
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