Upon combustion, a 0.8009 g sample of a compound containing only carbon, hydrogen, and oxygen produces 1.6004 g co2 and 0.6551 g h2o. determine the empirical formula of the compound.

The compound contains Carbon, Hydroxide and Oxide
1 mole of carbon iv oxide contains 44 g, out of which 12 g are carbon.
Therefore, 1.6004 g of CO2 will contain;
    1.6004 ×12/44 = 0.4365 g of carbon
1 mole of water contains 18 g of which 2 g is hydrogen,
Therefore, 0.6551 g of H2O will hace ;
         0.6551 × 2/18 = 0.0728 g of hydrogen.
The total mass of the compound is 0.8009 g,
Thus the mass of oxygen = 0.8009 -(0.4365 +0.0728)
                                         = 0.2916 g
To get the empirical formula we first get the number of moles of each element;\
 Carbon = 0.4365/12= 0.036375 moles
Hydrogen = 0.0728/1 = 0.0728 moles
Oxygen = 0.2916/16 = 0.018225 moles
Then, to get the smallest ratio we divide each with the smallest value;
Carbon :                        Hydrogen :                    Oxygen
= (0.036375/0.018225) : (0.0728/0.018225) : ( 0.018225/0.018225)
= 1.996                         :   3.995                    : 1
≈ 2 : 4 : 1
Therefore, the empirical formula is C2H4O

D-no effect on the rate