saplinglearning.com The percent yield for the reacti... When lead(l) nitrate sol

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saplinglearning.com The percent yield for the reacti... When lead(l) nitrate solution i... eous solution containin... Jump to E166-spring 17- MORRIS I Activities and Due Dates l Topic 15 Extra Credit 5/9/2017 11:55 PM 1/5 Print Calculator Periodic Table Question 1 of 6 Sapling Learning complete combustion of 6.7o g of a hydrocarbon produced 20.5 g of co2 and 9.80 g of H20. What is the empirical formula for the hydrocarbon? CH Insert subscripts as necessary AO Previous ® Give Up & View Solution 2 Check Answer Next Hint

Explanation / Answer

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% by mass of C=12*mass of CO2*100/(44*mass of hydrocarbon)=12*20.5*100/(44*6.70)=83.4%

%by mass of H=2*mass of H2O*100/(18*mass of hydrocarbon) = 2 * 9.80 * 100 / (18*6.70) = 16.2 %

Element Mass     Number of moles               SImplest ratio

C             83.4      83.4 /12 = 6.95                  6.95 / 6.95 = 1          *3                   3

H            16.2      16.2 / 1 = 16.2                    16.2 / 6.95 = 2.33     *3                   7

Thereofre, the emperical formula is C3H7

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