posted by digits on .
how do I solve for:
How many grams of water would be lost form 100.0grams of CuCl2 x 5H2O when the hydrated salt was heated at 1000degreesC for one hour in a crucible?
What is the percent composition of water in the compound:
Percentwater= 5*(2+16)/(5*18 + atomic mass Cu + 2*atomicmassCl) * 100
That will tell you how much water is there.
Bob Pursley is right. Essentially all of the hydrated water would be lost at that extreme temperature, and you would be left with the anhydrous form, CuCl2.