Question
If 3.943 g of anhydrous salt remains after heating 5.00 g of CuCℓ2·xH2O, determine the number of molecules of water of hydration in the original hydrate.
Answers
DrBob222
5.00 g = initial mass
-3.943 g after heating
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1.057 g = mass H2O driven off in heating
mols H2O = 1.057/18 = about 0.0587
mols CuCl2 = 3.943/134.5 = about 0.0293
Find the ratio and round to whole numbers with the smaller number being 1.00
0.0293/0.0293 = 1.00
0.0587/0.0293 = 2.00
So formula is (CuCl2)1Cl2.2H2O which would be written as CuCl2.2H2O
-3.943 g after heating
-------
1.057 g = mass H2O driven off in heating
mols H2O = 1.057/18 = about 0.0587
mols CuCl2 = 3.943/134.5 = about 0.0293
Find the ratio and round to whole numbers with the smaller number being 1.00
0.0293/0.0293 = 1.00
0.0587/0.0293 = 2.00
So formula is (CuCl2)1Cl2.2H2O which would be written as CuCl2.2H2O