Question

In: Chemistry

when 4.027 grams hydrocarbon, CxHy, we're burned in a combustion analysis apparatus, 12.64 grams of CO2...

when 4.027 grams hydrocarbon, CxHy, we're burned in a combustion analysis apparatus, 12.64 grams of CO2 and 5.174 grams H2O we're produced.

I need empirical and molecular formulas

Solutions

Expert Solution

C%   = 12*wt of CO2*100/44*wt of hydrocarbon

       = 12*12.64*100/44*4.027   = 85.6%

H%   = 2*wt of H2O*100/18*wt of hydrocarbon

          = 2*5.174*100/18*4.027 = 14.4%

element         %                     A.wt             relative number   simple ratio

C                85.6                    12                  85.6/12 = 7.13             7.13/7.13 = 1

H               14.4                      1                   14.4/1   = 14.4            14.4/7.13    = 2

               Empirical formula = CH2

queen_honey_blossom answered 2 years ago
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