Acertain alcohol contains only three elements, carbon, hydrogen, and oxygen. combustion of a 30.00 gram sample of the alcohol produced 57.30 grams of co2 and 35.22 grams of h2o. what is the empirical formula of the alcohol?

first we calculate the mass percent composition of three elements carbon hydrogen and oxygen.
Mass of CO₂ = 44.0096 g/mol

Mass of H₂O = 8.0153 g/mol

percent composition of carbon = (57.30 g CO2) / (44.0096 g CO2/mol) x (1/1) x (12.0108 g C/mol) / (30.00 g) = 0.521264 = 52.1264% C 

Percent composition of hydrogen = (35.22 g H2O) / (18.0153 g H2O/mol) x (2/1) x (1.0079 g H/mol) / (30.00 g) = 0.131363 = 13.1363% H 

percent composition of oxygen = 100% - 52.1264% C 13.1363% H = 34.7373% O 

Now calculate the number of moles in the compound; 
(52.1264 g C) / (12.0108 g C/mol) = 4.33996 mol C 
(13.1363 g H) / (1.0079 g H/mol) = 13.0333 mol H 
(34.7373 g O) / (15.9994 g O/mol) = 2.17116 mol O 

Now divide by the smallest number of moles: 
(4.33996 mol C) / (2.17116 mol) = 1.999 = 2
(13.0333 mol H) / (2.17116 mol) = 6.003 = 6
(2.17116 mol O)/ (2.17116 mol) = 1.000  = 1

So,  the empirical formula is : 
C₂H₆O

your answer is a. the enthalpy of formation for a pure element in its standard state is always positive.