Your alcohol's empirical formula is #C_2H_6O#.
Once again, your dealing with an organic compound that only contains carbon, hydrogen, and oxygen. You will use the mass of water and the volume of #CO_2# absorbed following the combustion of your alcohol to determine how much hydrogen and carbon your initial compound contained.
The percent composition of hydrogen is water is
#(2 * 10.01cancel("g/mol"))/(18.02cancel("g/mol")) * 100 = "11.21%"#
This means that you get 11.21 g of hydrogen for every 100 g of water. The amount of hydrogen your alcohol contained will be
#1.174cancel("g"H_2O) * "11.21 g H"/(100cancel("g"H_2O)) = "0.1316 g H"#
At STP, one mole of any ideal gas occupies exactly 22.4 L - this is known as the molar volume of a gas at STP. In your case, the reaction produced #"974 cm"^3# of carbon dioxide under these conditions, which is equivalent to
#974cancel("cm"^3) * (1cancel("L"))/(1000cancel("cm"^3)) * "1 mole"/(22.4cancel("L")) = "0.04348 moles C"#
The alcohol contained
#0.04348cancel("g") * "12.0 g"/(1cancel("mole")) = "0.5218 g C"#
The mass of oxygen will be
#m_"alcohol" = m_"C" + m_"O" + m_"H"#
#m_"O" = 1 - 0.5218 - 0.1316 = "0.3466 g O"#
Since 1 mole of #CO_2# contains 1 mole of carbon, you already know how many moles of carbon your sample contained. Use the molar masses of hydrogen and oxygen to determine how many moles of each you started with
#"For H": (0.1316cancel("g"))/(1.01cancel("g")/"mol") = "0.1303 moles H"#
#"For O": (0.3466cancel("g"))/(16.0cancel("g")/"mol") = "0.02176 moles O"#
Divide these numbers by the smallest one to get the mole ratios the elements have in the alcohol
#"For C": (0.04348cancel("moles"))/(0.02176cancel("moles")) = 2#
#"For H": (0.1313cancel("moles"))/(0.02176cancel("moles")) = 6#
#"For O": (0.02176cancel("moles"))/(0.02176cancel("moles")) = 1#
The empirical formula for your alcohol will be
#C_2H_6O#
SIDE NOTE Once again, I assumed that your problem intended for you to use the old value of the molar volume of a gas as STP, which is 22.4 L.
The current value of 22.7 L will produce the same results, though.
