29.8g H2O = 1.66 mol H2O
Molar Mass CuSO4 * 5H2O = 249.6 g
mol CuSO4 * 5H2O --> 5 mol H2O
249.6 g CuSO4 * 5H2O/1 mol CuSO4 * 5H2O Times *
1mol CuSO4 * 5H2O/5mol H2O Times*
1.66 mol H2O = 82.6 g CuSO4 * 5H2O
cuso4 - 5h2o= cuso4 + 5h20 + heat
The chemical formula for blue vitriol (copper sulphate pentahydrate) is CuSO4.
penta = 5 copper(II) sulphate pentahydrate = CuSO4*5 H2O CuSO4*5 H2O + heat --> CuSO4 + 5 H2O
CuSO4•5H2O
Copper sulfate in its solid crystal from present as pentahydrate (CuSO4 5 H2O). When heated strongly it loses water molecules and becomes CuSO4. Note that CuSO4 5 H2O is blue in color whereas CuSO4 is colourless since absorption region shifted from visible region. On further heating it is dissociated to CuO.
formula : cuso4 chemical name: copper sulphate
cuso4 - 5h2o= cuso4 + 5h20 + heat
Copper sulfate does not produce crystals. Copper (II) sulfate pentahydrate (CuSO4•5H2O) does. The formula units are attracted to each other and form a repeating lattice.
The chemical formula for blue vitriol (copper sulphate pentahydrate) is CuSO4.
penta = 5 copper(II) sulphate pentahydrate = CuSO4*5 H2O CuSO4*5 H2O + heat --> CuSO4 + 5 H2O
CuSO4•5H2O + heat ---> CuSO4 + 5H2O
This equation is CuSO4.5 H2O -> CuSO4 + 5 H2O.
The blue copper sulfate pentahydrate loss by heating water and become an anhydrous white sulfate.
Anhydrous Copper Sulfate will be white. The original crystals (CuSO4 pentahydrate) will be blue.
There is a balanced equation to use for decomposition of copper II sulfate pentahydrate. It is the following: CuSO4.5H2O+heat -->CuSO4(anhydrous)+5H2O.
CuSO4 x 5 H2O is the formula for Copper Sulfate Pentahydrate.
CuSO4•5H2O