anonymous
  • anonymous
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?
Chemistry
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
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anonymous
  • anonymous
Now I'm going to figure out how many moles of carbon are produced from burning the unknown item
anonymous
  • anonymous
\[25.6g CO_2 (\frac{1 mole CO_2}{44.01 g CO_2})\frac{1 mole C}{1 mole CO_2})=0.582 mole C\] And we will do the same thing for H
anonymous
  • anonymous
\[9.15 g H_2O (\frac{1 mole H_2O}{18.02 g H_2O})(\frac{2 mole H}{1 mole H_2O})=1.02 mole H\]

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anonymous
  • anonymous
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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