If 8.65 g of a tin - fluorine compound contains 5.28 g of tin, what is its empirical formula?

First, we need to know the amounts of the elements in the compound.

Tin (Sn) = 5.28 g

Fluorine (F) = 8.65 - 5.28 = 3.37 g

Convert these to units of moles by dividing the molar masses.

Tin (Sn) = 5.28 g / 118.71 g/mol = 0.044 mol

Fluorine (F) = 3.37 g / 19.00 g/mol = 0.177 mol

Divide both by the least number of moles of the two.

Tin (Sn) = 0.044 mol / 0.044 mol = 1

Fluorine (F) = 0.177 mol / 0.044 mol = 4

Therefore, the empirical formula would be:

SnF4