What is the empirical formula of a compound that contains 15.9% boron and 84.1% fluorine?

To find the empirical formula of a compound, follow these steps:

1. Assume that you have 100g of the substance. This means you have 15.9g of boron and 84.1g of fluorine.

2. Convert the mass of each element to moles. Boron has an atomic mass of about 10.81g/mol and fluorine has an atomic mass of about 18.998g/mol.

   Moles of boron = 15.9g / 10.81g/mol = 1.47 mol Moles of fluorine = 84.1g / 18.998g/mol = 4.43 mol

3. Divide each mole quantity by the smallest number of moles calculated. In this case, boron has the smallest number of moles.

   Boron = 1.47 mol / 1.47 = 1 Fluorine = 4.43 mol / 1.47 = 3

4. If the numbers obtained are very close to whole numbers, these are the subscripts for each element in the empirical formula. If they are not close to whole numbers, multiply them by integers (such as 2, 3, etc.) until you get whole numbers.

In this case, the empirical formula of the compound is BF3.