When 5.454 grams of a hydrocarbon, CxHy, were burned in a combustion analysis ap

Question

When 5.454 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 17.11 grams of CO2 and 7.006 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.

Explanation / Answer

CxHy+(x+y/4)/2o2-->Xco2+y/2h20 weight of CO2 = 17.11 so moles of CO2 = 17.11 / 44 = 0.388 =x' weight of H2O = 7.006or 7.006 /18 moles of H2O = 0.389 moles==>y'=0.777 for 56.11 grams of Cxhy= 12x+y=56.11 ==> molar ratio of C : H = 0.388:0.777 = 1 : 2 CH2 12+2=14 = 1/4 Molecular weight so the molecular formula is C4H8 is the ans

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