A0.3870-g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in oxygen to yield 0.7191â g of co2 and 0.1472â g of h2o. what is the empirical formula of the compound?

The chemical formula for the compound can be written as,

    CxHyOz 

where x is the number of C atoms, y is the number of H atoms, and z is the number of O atoms. The combustion reaction for this compound is,
   
    CxHyOz + O2 --> CO2 + H2O 

number of moles of C:
     (0.7191 g)(1 mol CO2/44 g of CO2) = 0.0163 mol CO2
 This signifies that 0.0163 mole of C and the mass of carbon in the compound,
        (0.0163 mols C)(12 g C/ 1 mol C) = 0.196 g C

number of moles H:
      (0.1472 g H2O)(1 mol H2O/18 g H2O) = 0.00818 mol H2O

This signifies that there are 0.01635 atoms of H in the compound.
      mass of H in the compound = (0.01635 mols H)(1 g of H) = 0.01635 g H

Mass of oxygen in the compound,
   0.3870 - (0.196 g C + 0.01635 g H) = 0.1746 g

Moles O in the compound = (0.1746 g O)(1 mol O/16 g O) = 0.0109 mols O

The formula of the compound is,
      C0.0163H0.01635O0.0109

Dividing the numbers by the least number,
    C3/2H3/2O

The empirical formula of the compound is therefore,
      C₃H₃O₂