If 1.6045g of anhydrous salt remains after heating 1.9045 g of CuC§¤2¡¤xH2O, determine the number of molecules of water of hydration in the original hydrate

1 answer

I guess that is CuC2 hydrated with x H2O

CuC2 = 63.5 + 24 = 87.5 grams/mol
so
1.6045 grams * 1 mol/87.5 grams
= .0183 mol of CuC2

H2O = 18 grams/mol
we have how many mols of H2O?
1.9045 - 1.6045 = 0.3 grams H2O
so
0.3 / 18 = .01667 mol H2O
looks like about as many mols of H2O as of CuC2 like 183/167 = 1.01
and x = 1