Chemistry HW Help...Empirical Formula?

Combustion of a 1.031 g sample of an unknown compound produced 2.265 g of carbon dioxide gas and 1.236 g of water vapor. What is the empirical formula of the compound?

Please show your work so I know how you got there.
THANKS!!!!!!!!!

Answer:
Moles C = 2.265 g / 44.010 = 0.05146
g C = 0.05146 x 12.011 = 0.618 g

Moles H = 1.236 g / 18.02 x 2 = 0.137
g of H = 0.137 x 1.008 g/mol = 0.138 g

grams of O = 1.031 - 0.618 - 0.138 = 0.275 g
moles O = 0.275 / 15.999 = 0.01719

44.010 is the molecular weight of CO2.
18.02 is the molecular weight of H2O . For each mole of H2O there are 2 moles of H so to get moles of H we multiply for 2.

So we have :
0.05146 moles of C , 0.137 moles of H and 0.01719 moles of O.
We divide each number for the smallest :
0.05146 / 0.01719 = 3 ( C)
0.137 / 0.01719 = 8 ( H)
0.01719 / 0.01719 = 1 ( O )
the empirical formula is :
C3H8O

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Chemistry HW Help...Empirical Formula?

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