Acompound is found to contain 58.80 % xenon, 7.166 % oxygen, and 34.04 % fluorine by mass. what is the empirical formula for this compound?

To make it easier, assume that we have a total of 100 g of a compound. Hence, we have 58.80g of xenon, 7.166g of oxygen, and 34.04g of fluorine. 
Know we will convert each of these masses to moles by using the atomic masses:

58.8/131.3 = 0.45 mole of Xe
7.166/16 = 0.45 mole of O
34.04/19 = 1.79 mole of F

Now, we will divide all the mole numbers by the smallest among them and get the number of atoms in the compound:

Xe = 0.45/0.45 = 1
O = 045/0.45 = 1
F = 1.79/0.45 = 3.98 = 4

So, the empirical formula of the compound XeOF₄

Sorry i dont understand the question can you rephase it?