Combustion analysis of a 0.2608 g sample of a compound containing carbon, oxygen, and hydrogen only...

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Chemistry

Combustion analysis of a 0.2608 g sample of a compoundcontaining carbon, oxygen, and hydrogen only produced 0.5501 g ofcarbon dioxide, and 0.2703 g of water.
Determine the empirical formula of this compound. If the molarmass for this compound is 104.1 g/mole, find the molecular formula.Write a complete balanced equation for the combustion reaction ofthis compound.

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4.1 Ratings (804 Votes)

c%Â Â Â Â Â Â = 12* weight of CO2 *100/44*weigh of organic compound

Â Â Â Â Â Â Â Â = 12*0.5501*100/44*0.2608Â Â = 57.53%

H%Â Â = 2* weight of H2O *100/18*weight of organic compound

Â Â Â Â Â Â = 2*0.2703*100/18*0.2608 = 11.52%

O%Â Â = 100-(C%+ H%)

Â Â Â Â Â Â Â Â = 100-(57.53+11.52) = 30.95%

elementÂ Â Â Â Â Â Â Â Â Â Â Â Â Â %Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â A.WtÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Relative numberÂ Â Â Â Â Â Â Â Â Â Â simple ratioÂ Â Â Â Â Â simplest ratio

CÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 57.53Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 12Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 57.53/12 = 4.79Â Â Â Â Â Â Â Â Â Â 4.79/1.934Â Â =2.5Â Â Â Â Â 2.5*2 = 5

HÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 11.52Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 1Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 11.52/1Â Â Â Â = 11.52Â Â Â Â Â Â Â Â Â Â Â 11.52/1.934 = 6Â Â Â Â Â Â 6*2Â Â = 12

OÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 30.95Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 16Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 30.95/16Â Â Â Â = 1.934Â Â Â Â Â Â Â Â Â Â Â 1.934/1.934 = 1Â Â Â Â Â Â 1*2 = 2

Â Â Â Â Â Â Empirical formula = C5H12O2

Â Â Â Â molecular formula = ( empirical formula)nÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â { E.F.Wt Of C5H12O2 = 12*5 + 1*12 + 16*2 = 104}

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â nÂ Â Â Â Â Â Â Â Â = E.F.Wt/M.wt

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 104/104 = 1

Â Â Â Â Â Â Â Â Â Â molecular formula = ( C5H12O2)1 = C5H12O2

C5H12O2 + 7O2 ----------> 5CO2 + 6H2O

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