Anhydrous salt problem!

I am given the following info (i'm not sure if these are the exact numbers but it goes about like this; i just want to know how to do the following type of question for the future):

You are given a hydrate of Barium Chloride
The crucible and hydrate together weigh 23.4 g
The crucible alone weighs 18.82 g
The crucible and anhydrous salt produced weigh 20.94 g

Determine the formula of the salt, balance the equation, and find the experimental % of H2O

the part i'm unsure of is the determining the formula of the anhydrous salt

like i said, these probably aren't exact numbers, but if someone could explain the general procedure it would help a lot

Weight of hydrate:
23.4 g - 18.82 g = 4.58 g

Weight of water lost:
23.4 g - 20.94 g = 2.46 g

Weight of anhydrous salt:
20.94 g - 18.82 g = 2.12 g

Convert the anhydrous salt and water to moles:
(2.12 g BaCl2) / (208.2341 g BaCl2/mol) = 0.010181 mol BaCl2
(2.46 g H2O) / (18.01532 g H2O/mol) = 0.13655 mol H2O

Divide the number of moles of water by the number of moles of anhydrous salt:
(0.13655 mol H2O) / (0.010181 mol BaCl2) = 13.412
This number should be near an integer (and it probably should be a lot less than 13). (This is where having inexact numbers messes you up.) It is the number that replaces "x" in this formula:
BaCl2 · x H2O

There's no equation to balance -- there's been no chemical reaction.

The % of H2O is the weight of water divided by the weight of the hydrated salt:
(2.46 g) / (4.58 g) = 0.537 = 53.7% water