Between the copper and the sulfate, CuSO4, there is an ionic bond.
Cu 2+
and
SO4 2-
---------------
The hydrated part, water, is a polar covalent molecule with the unequal sharing of electrons between the oxygen atom and the two hydrogen atoms.
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 is copper (II) sulfate. The balanced equation for CuSO4 with water is CuSO4 + H2O reacts to become Cu+2 + HSO4-2 + OH-.
CuSO4 * 5H2O ----> CuSO4 + 5H2O. This is true because CuSO4 * 5 H2O is a salt weakly bounded to water, that is why it is hydrous. When it decomposes, the weak bonds are broken making the products above. CuSO4*5H2O formula is [Cu(OH2)4]SO4*H2O CuSO4 + 5H2O --> [Cu(OH2)4]SO4*H2O
Copper sulfate dissolves in water, it does not react.
No. The % of Cu by mass in CuSO4 will be greater than the % of Cu by mass in the pentahydrate (5H2O) because in the hydrate there is added mass (5 H2O = 90 g) but no added Cu.
Yes. They do not contain the same proportion of CuSO4, however. The moles of CuSO4 . 5 H2O are more massive.
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
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
penta = 5 copper(II) sulphate pentahydrate = CuSO4*5 H2O CuSO4*5 H2O + heat --> CuSO4 + 5 H2O
CuSO4 is copper (II) sulfate. The balanced equation for CuSO4 with water is CuSO4 + H2O reacts to become Cu+2 + HSO4-2 + OH-.
H2O CuSo4 Ca2So4
These are some possible ionic equation for CuSO4 plus H2O: Cu2+ + 6 H2O --> Cu(H2O)6 Cu(H2O)62+ + H2O <--> Cu(OH)(H2O)5+ + H3O+ This makes a solution of copper sulfate weakly acidic.
CuSO4 * 5H2O ----> CuSO4 + 5H2O. This is true because CuSO4 * 5 H2O is a salt weakly bounded to water, that is why it is hydrous. When it decomposes, the weak bonds are broken making the products above. CuSO4*5H2O formula is [Cu(OH2)4]SO4*H2O CuSO4 + 5H2O --> [Cu(OH2)4]SO4*H2O
In its crystallic (solid) form there are 5 molecules H2O regularly build in the crystal lattice, not ionically but rather polar hydrogen bonds strong enough to keep'm in place: CuSO4.(H2O)5
CuCO3(s) + H2SO4(aq) → CuSO4(aq) + CO2(g) + H2O(l)
Copper sulfate dissolves in water, it does not react.
No. The % of Cu by mass in CuSO4 will be greater than the % of Cu by mass in the pentahydrate (5H2O) because in the hydrate there is added mass (5 H2O = 90 g) but no added Cu.