# empirical formula of propane after complete combustion

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Many homes in rural America are heated with propane gas, a compund that contains only carbon and hydrogen. COmplete combustion of a sample of propane produced 2.641g of carbon dioxide and 1.442g of water as teh only products. Find the empirical formula of propane.

### If C=12, H=1 and O=16 then 1

If C=12, H=1 and O=16 then 1 mole CO2 weighs 44g and contains 12g of Carbon. So find the mass of C in 2.641g CO2 ( =0.7203g) . 1mole H2O has a mass of 18g and contains 2 gram hydrogen. So find the mass of hydrogen in 1.442g of water (=0.1602 g). Divide the masses of these elements by 12 for carbon and 1 for hydrogen. (=0.7203/12 = 0.060025 moles C and 0.1602/1 = 0.1602 moles H. This will change the masses into moles. Divide each of the mole values by the smallest value (should be moles of C) and this will tell you the ratio of H to each atom of C.

0.060025/0.060025 = 1 mole C and 0.1602/0.060025 = 2.6689 moles H. This works out at  C3H8 if the calculation is correct.

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