Upon combustion, a compound containing only carbon and hydrogen produces 1.34 gCO2 and 0.547 gH2O. Find the empiri. formula of the compound.?

Note: Oxygen is involved whenever there is a flame.

C(_a)H(_b) + __ O2 ----> __ CO2 + __ H2O

_a for how much carbon, _b for how much hydrogen in the molecule.

CO2 is the only source of carbon:

1.34 g CO2 * (1 mole CO2 / 44.009 g CO2) * (1 mole carbon / 1 mole CO2) = ~ 0.0304 moles carbon.

0.547 g H2O * (1 mole H2O / 18.015 g H2O) * (2 moles H / 1 mole H2O) = ~ 0.0607 moles H

Technically these are the moles of Carbon and Hydrogen in the products. But no other sources of "C" or "H" reacted with this hydrocarbon so by the conservation of mass all those moles were in this compound.

Divide by least moles so that of carbon:

_a = 0.0304 moles C / 0.0304 moles C = 1

_b = 0.0607 moles H / 0.0304 moles C = ~ 2

CH2 is the empirical formula.

The molecular formula is some multiple of this so it is probably C2H4 (ethene) as methylene is unstable by the rules of organic chemistry.