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anonymous
 4 years ago
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?
anonymous
 4 years ago
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?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Now I'm going to figure out how many moles of carbon are produced from burning the unknown item

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[25.6g CO_2 (\frac{1 mole CO_2}{44.01 g CO_2})\frac{1 mole C}{1 mole CO_2})=0.582 mole C\] And we will do the same thing for H

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[9.15 g H_2O (\frac{1 mole H_2O}{18.02 g H_2O})(\frac{2 mole H}{1 mole H_2O})=1.02 mole H\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0And 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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