Complete combustion of 8.00g of a hydrocarbon produced 25.6g of CO2 and 9.15g of H2O. What is the emperical formula for the hydrocarbon?
Now I'm going to figure out how many moles of carbon are produced from burning the unknown item
\[25.6g CO_2 (\frac{1 mole CO_2}{44.01 g CO_2})\frac{1 mole C}{1 mole CO_2})=0.582 mole C\] And we will do the same thing for H
\[9.15 g H_2O (\frac{1 mole H_2O}{18.02 g H_2O})(\frac{2 mole H}{1 mole H_2O})=1.02 mole H\]
And now to do the O, \[9.15 g H_2O (\frac{1 mole H_2O}{18.02 g H_2O})(\frac{1 mole O}{1 mole H_20})=.508 mole O\] we also have to add the O from the CO2\[25.6 g CO_2 (\frac{1 mole CO_2}{44.01 g CO_2})(\frac{2 mole O}{1 mole CO_2})=1.16 mole O\] for a total of 1.67 mole of O, Now use the mole to mole comparasion thing you were showed in your previous problem by Jfrasher.
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