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Calculations of Empirical Formula for Unknown Compound

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Combustion analysis of 1.0g of an unknown compound was found to give the following data: 2.97 g CO2 and 0.73 g of H2O. Dtermine the empirical and molecular formula for the unknown compound.

So far, I took:

2.97/44.01 = .067 moles of CO2
0.73/18.01 = .0405 moles of H2O

then I took 2.97 + 0.73 g to = 3.7g - 1 g of unknown to give us 2.7 grams of O2. Since C?H?O? + O2 = CO2 + H2O

2.7/32g = 0.0804 g

.0804:.067: .0405 divide all by .0405 so o2=2.07 CO2 = 1.65 H2O=1 times all by 2 to give whole numbers

so C?H?O? + 4O2 = 3CO2 + 2H2O but I can't figure out the empirical formula. Am I on the right path? and I don't think I am doing this right.

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Hi,

There are 0.0675mol of CO2, and since there are 2 molecules of oxygen, there are 2x0.0675 = 0.135mol of ...

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