To determine the mass percent of oxygen in ethanol, we need to calculate the mass of oxygen in ethanol and then divide it by the total mass of ethanol.

Ethanol (C2H5OH) consists of carbon (C), hydrogen (H), and oxygen (O) atoms. The molecular formula of ethanol indicates that it contains two carbon atoms, six hydrogen atoms, and one oxygen atom.

The molar mass of ethanol can be calculated by adding the atomic masses of its constituent elements:
2(C) + 6(H) + 1(O) = 12.01 g/mol + 6(1.008 g/mol) + 16.00 g/mol = 46.07 g/mol

Now, let's determine the mass of oxygen in ethanol:
1(O) = 16.00 g/mol

To calculate the mass of oxygen in ethanol, we can multiply the molar mass of oxygen by the number of moles of oxygen in one mole of ethanol:
(16.00 g/mol) * (1 mol of O / 46.07 g/mol of ethanol) = 0.3473 mol of O

Finally, to find the mass percent of oxygen in ethanol, we divide the mass of oxygen by the total mass of ethanol and multiply by 100:
(0.3473 mol of O) / (46.07 g/mol of ethanol) * 100 = 0.3473 * 100 / 46.07 = 0.7534%

Therefore, the mass percent of oxygen in ethanol is approximately 0.7534%. When rounded to two decimal places, the correct answer is option 'D': 34.73%.