the combustion of 1.83 grams of a compound which only contains C,H, and O yields 4.88 grams of CO2, and 1.83 grams of H2O. What is the empirical formula of this compound?

Question

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4.88 g CO2 x 1 mol CO2/44.01 g CO2 x 2 mol O2= .2217 mol O2

1.38 g of H20 x 1mol H20/ 18.02 g H20 x 2 mol H2/ 1 mol H20= .15316 mol H2

How do I get find mols of C? What is the 1.83 grams used for? thanks!

Answer 1

Convert 4.88 g CO2 to grams C.
Convert 1.83 g H2O to grams hydrogen (H, not H2).Your answer looks ok for this part.
Then 1.83 - g C - g H = g O (not O2)

Then g C/12 = mols C
g H/1 = mols H
g O/16 = mols O

Then find the ratio.

Answer 2

Thanks!