This is a question that requires you to first figure out the empirical formula of the compound. You can think of the empirical formula of a compound as the formula that shows the ratio of atoms in a molecule to one another but not the actual numbers. The molecular formula shows the actual number of elements in a molecule.
If we have 38.8% Cl and 61.2% O, it is easiest to act as if we have a 100g sample. This is because 38.8% and 61.2% will directly correspond to the number of grams of each element in the sample:
Use the molar mass of both of these elements from the periodic table in order to determine their numerical ratio to one another in terms of mols:
38.8g Cl x (1 mol Cl/35.45g Cl) ~ 1.09 mol Cl
61.2g O x (1 mol O/16g O) ~ 3.8 mol O.
You cannot have decimals in the empirical formula. But all we are really interested in here is the ratio of Cl atoms (in terms of mols) to O atoms (in terms of mols). Just divide the mols of O by the mols of Cl to figure out what our ratio is:
3.8/1.09 = 3.49 ~ 3.5; that is, we have 3.5 mol of O for every 1 mol of Cl. Again, we cannot have decimals in our empirical formula, so simply multiply these number by 2 to get rid of the decimal:
3.5(2) = 7; 1(2) = 2.
Thus, our empirical formula is Cl2O7.
To calculate the molecular formula, simply divide the molecular weight of the empirical formula by the real molecular weight that was given to you at the start of the problem (182.9g).
Empirical formula molecular weight: (molar mass Cl x 2) + (molar mass O x 7) = (35.45 x 2) + (16 x 7) = (70.9) + (112) = 182.9.
Thus, the empirical formula weight is 182.9. To calculate the ratio of the empirical formula to the molecular formula (although at this point it should be obvious), divide the empirical formula weight by the molecular formula weight:
182.9/182.9 = 1.
Thus, the empirical formula weight to molecular formula weight occur in the same ratio. Therefore, they are the same formula.
The answer is Cl2O7.