Determine the empirical formula of the analysis of a compound being 9.2% b and 90.8% cl

Let's begin by assuming that the mass of the molecule is 100 grams. This means that per 100 grams of molecule there are 90.8 grams of Chlorine, and 9.2 grams of Boron. Using the molar mass of each element we can find the number of moles of each element there are per 100 grams of molecule.

90.8 / 35.453 = ~ 2.56 moles of Chlorine 
9.2 / 10.812 = ~ 0.851 moles of Boron
Now divide by the smallest mole number to find how many moles of each element there are in a 100 gram sample. 
0.851/0.851 = 1
and 2.56/0.851 = ~3
Meaning that for every one mole of Boron there are 3 moles of Chlorine atoms
so the empirical formula is BCl3

Another way you can do this is by knowing that Boron has 3 valence electrons, and Chlorine has 1 meaning Boron can form 3 bonds with 3 different Chlorines, and that BCl3 is the simplest Boron Chlorine molecule. 

1. o2(g)as a product