How do you find the empirical formula of antimony and two chlorides?

The first compound is 46.8% chlorine and 53.2 % antimony.  These are mass-type numbers.  Divide each by the corresponding atomic mass to get mole type numbers.  Divide these two answers by the smallest to get a ratio where one of them is 1.  If the other is a whole number, write down the formula.  If it is something .5, double them both to get whole numbers first, if it ends in .33 or .67 triple them, etc.  
   46.8/35.453 = 1.320 for the Cl
   53.2/121.75 = .4370 for Sb
Dividing each by .4370 gives 3.0 and 1.  The empirical formula is SbCl3

The second compound is 59.3% chlorine and 40.7% antimony.
    59.3/35.453 = 1.673 for the chlorine
    40.7/121.75 = .334 for the antimony
Dividing each by .334 gives 5.00 and 1.  The empirical formula is SbCl5

The first SbClX is 46.8 w/o (weight percent) Cl and therefor 53.2 w/o Sb.
Now let 1 mole of Sb (121.75g) represent 53.2 w/o of the entire compound.
Thus, 1 mole of the compound = 121.75g/0.532 = 228.85g
Next find the grams of Cl in the compound. (228.85g)(0.468) = 107.103g
Moles of Cl present is: (107.103g) / (35.453g/moleCl) = 3.021 moles Cl
The compound is SbCl3

Use the same method for the second compound.

You will find it to be SbCl5