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Topic: percent compostion  (Read 3994 times)

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mrstrh

  • Guest
percent compostion
« on: September 28, 2004, 04:33:14 PM »
combustion of 6.51grams of and organic compound gave 20.47grams of carbon dioxide and 8.36grams of water. How would you calculate the percent composition and empirical formula.

I know how to find just the percent comp if it was just like how much percent of carbon is in 50grams of CO but i dont know how to do multiple things like this really. Any help would be greatly appreciated.i think i could find the empirical formula i think you have to get the molecular and divide by the empirical to get your answer

Tetrahedrite

  • Guest
Re:percent compostion
« Reply #1 on: September 28, 2004, 11:44:07 PM »
Finding the empirical formula is a simple matter of finding the ratio between the number of moles of carbon and the number of moles of hydrogen. In this case the number of moles of C = moles CO2 = 0.465 and,
moles of H = 2xmoles H20 = 0.928
As you can see there is almost exactly a 1:2 ratio between C:H so the empirical formula is CH2

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