Final answer:
The empirical formula of the compound with 52.2% C, 13.0% H, and 34.8% O is C2H6O.
Step-by-step explanation:
To determine the empirical formula of the compound, we need to convert the percentage composition into grams. Assuming we have 100g of the compound, we have 52.2g of C, 13.0g of H, and 34.8g of O. Next, we convert the grams of each element to moles by dividing by their molar mass. The molar mass of C is 12.01 g/mol, H is 1.01 g/mol, and O is 16.00 g/mol. This gives us approximately 4.34 moles of C, 12.87 moles of H, and 2.18 moles of O.
We then divide each of the mole ratios by the smallest number of moles, which is 2.18. This gives us a ratio of approximately 2 moles of C, 6 moles of H, and 1 mole of O. To find the empirical formula, we write the elements and their respective subscripts, which are C2H6O. Therefore, the empirical formula of the compound is C2H6O.