Answer: The empirical formula is C_2H_6O.

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 52.14 g

Mass of H = 13.13 g

Mass of O = 34.73 g

Step 1 : convert given masses into moles.

Moles of C =frac{text{ given mass of C}}{text{ molar mass of C}}= frac{52.14g}{12g/mole}=4.35moles

Moles of H =frac{text{ given mass of H}}{text{ molar mass of H}}= frac{13.13g}{1g/mole}=13.13moles

Moles of O =frac{text{ given mass of O}}{text{ molar mass of O}}= frac{34.73g}{16g/mole}=2.17moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C =frac{4.35}{2.17}=2

For H =frac{13.13}{2.17}=6

For O =frac{2.17}{2.17}=1

The ratio of  C: H: O = 2: 6: 1

Hence the empirical formula is C_2H_6O.