How to find the mass of hydrate, anhydrous salt, evaporated water, and % water in hydrated salt?

Question

anonymous · Answer

mass of empty crucible = 39.43 g
mass of crucible and hydrated salt = 42.43 g
mass of hydrate = 3 g
mass of crucible and dehydrated salt (after 1st heating) = 41. 49 g
mass of crucible and dehydrated salt (after 2nd heating) = 41.39 g
mass of anhydrous salt =?
mass of water released = ?
% water in hydrated salt =?

Type of salt:

I believe the mass of anhydrous salt is 1.96 g (41.39 g - 39.43 g)
mass of water released is 1.04 g (42.43 g - 41.39 g)? Or is it (42.43 g - 3g)?
% water in hydrated salt = 34. 67% (1.04 g / 3.00 g x 100% = .346667 x 100 = 34.6667.

HELP!

anonymous · Answer

Also I have to calculate the % of water of my unknown salt and compare it to the % water in the salts: CuSO4 . 5H2O (which is %5H2O = 36.08%), MgSO4 . 7H2O (which is %7H2O = 51.2%), CoCl2 . 6H2O (which is %6H2O = 45.42%), SnCl2 . 2H2O (which is %2H2O = 15.97%), and FeSO4 . 7H2O (which is %7H2O = 45.4%)

Please... help.