The complete combustion of a sample of propane produced 2.641 grams of carbon dioxide and 1.442 grams of water as the only products. find the empirical formula of propane.

thnx!
20pts on this one.

A general equation for a combustion reaction would be expressed as follows:

CxHy + (x+y/2)O2 = xCO2 + y/2H2O

Propane would obviously would only have carbon and hydrogen in its structure. Assuming a complete combustion, all of the carbon atoms would go to carbon dioxide and all of the hydrogen atoms to water. To determine the empirical, we determine the number of carbon and hydrogen atoms present.

moles C = 2.461 g CO2 ( 1 mol / 44.01 g ) ( 1 mol C / 1 mol CO2 ) = 0.06 mol C

moles H = 1.442 g H2O ( 1 mol / 18.02 g ) ( 2 mol H / 1 mol H ) = 0.16 mol H

Then, we divide the smallest amount to the each mole of the atoms. We do as follows:

C = 0.06 / 0.06 = 1
H = 0.16 / 0.06 = 2.67

Then we multiply a number in order to obtain a whole number ratio between the atoms.

1        CH2.67
2       C2H5.34
3       C3H8   < empirical formula