Relevant but random chemistry questions?

ex5)

3)Give the correct name for the following:
a) CaCl2 x 2H2O
b) Cu(NO3)2 x 3H2O

what are fromulas for the following hydrated salts:
a) Cobalt (II) cholride hexahydrate
b) Iron (II) sulfate heptahydrate

2) A 2.815 g sample of hydrated copper sulfate was heated untill all of the water was removed. Calculate the percentage of water of hydration and the formula of the hydrate if the residue after heating weighted 1.802 g.

6) A 4.00 g sample of hydrate of nickel (III) bromid looses 0.739 g of water when heated. Determine mass percentage of water in the hydrate and the formula of the hydrate.

8) The masses of hydrates listed below were measured, heated to drive off the water of hydration, and cooled. Then the masses of the residues were measured. Find the formulas of the following hydrates.
8a) 1.04 g NiSO4 x n H2O gave a residue of 0.61 g
8b) 1.26 g CaSO4 x n H2O gave a residue of 0.99 g

Answer:
Hello-

3)
a) Calcium Chloride Dihydrate
There is only one type of calcium ions, so the number is not needed. Otherwise this is self-explanatory.

b) Copper (II) Nitrate Trihydrate (no expl. needed)

**

a) CoCl2 * 6H2O
Just read the name: You have Co2+ so you need 2 Cl- to cancel it, and you have 6 H2Os.

b) FeSO4 * 7H20 (no expl. needed)

**

2) You assume that all mass lost was water. Therefore you have 1.802 g copper sulfate and 1.013 g H2O.

1.013g/2.815g = 35.99%
That is the percent water of hydration.

Copper sulfate is probably CuSO4 (Copper (II) Sulfate). The MW is 159.62 g/mol (63.55+32.07+4*16.00).

Water has an MW of 18.02 g/mol (16.00+2*1.01).

Therefore you have:
1.013 g H2O * mol/18.02 g = 0.05622 mol H2O
1.802 g CuSO4 * mol/159.62 g = 0.01129 mol CuSO4

The mole ratio of H2O to CuSO4 is 0.05622/0.01129 = 5
Therefore the formula is CuSO4 * 5H2O

**

2) You assume that all mass lost was water. Therefore you have 3.26 g copper sulfate and 0.739 g H2O.

0.739g/4.00g = 18.5%
That is the percent water of hydration.

Nickel (III) Bromide is NiBr3. The MW is 298.39 g/mol (58.69+3*79.9).

Water has an MW of 18.02 g/mol (16.00+2*1.01).

Therefore you have:
0.739 g H2O * mol/18.02 g = 0.0410 mol H2O
3.26 g CuSO4 * mol/298.39 g = 0.0109 mol CuSO4

The mole ratio of H2O to NiBr3 is 0.0410/0.0109 = 4
Therefore the formula is NiBr3 * 4H2O

**

8)**As this is the same process as above I will show less work to save myself/the reader time.

a) 0.61g NiSO4
0.43g H2O

0.61g * mol/154.76g = .0039 mol NiSO4
0.43g * mol/18.02g = .024 mol NiSO4

ratio = .024/.0039 = 6

NiSO4 * 6H2O

b) 0.99g CaSO4
0.27 g H2O

0.99g * mol/136.16g = .0073 mol CaSO4
0.27g * mol/18.02g = .015 mol NiSO4

ratio = .015/.0073 = 2

CaSO4 * 2H2O

Hope that helps. Have a good day.
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