1. Assume a 100-gram sample:

* This makes the calculations easier. If 71.4% is bromine, then there are 71.4 grams of bromine in the sample.

* The remaining mass is oxygen: 100 g - 71.4 g = 28.6 g of oxygen.

2. Convert grams to moles:

* Moles of Bromine (Br): 71.4 g / 79.90 g/mol (molar mass of Br) = 0.894 moles

* Moles of Oxygen (O): 28.6 g / 16.00 g/mol (molar mass of O) = 1.788 moles

3. Find the simplest whole-number ratio:

* Divide both mole values by the smallest number of moles (0.894 in this case):

* Br: 0.894 / 0.894 = 1

* O: 1.788 / 0.894 ≈ 2

4. Write the empirical formula:

* The ratio of bromine to oxygen is 1:2.

* The empirical formula is BrO₂.