H Y D R A T E S
1. What is the difference between two types of compounds hydrides and hydrates ? Give one example of each.
Hydrides are compounds containing hydride ions (H – ). Example: LiH or CaH
2
.
Hydrates are compounds that have in their structures chemically bonded water molecules.
Example: BaCl
2
∙2H
2
O.
2. Fill the blanks in the following table.
Formula Common Name Systematic Name
CuSO
4
∙
5H
2
O blue vitriol copper (II) sulfate pentahydrate
CaSO
4
∙
2H
2
O gypsum calcium sulfate dihydrate
CaSO
4
∙
½H
2
O plaster of Paris calcium sulfate hemihydrate
MgSO
4
∙
7H
2
O Epsom salt magnesium sulfate heptahydrate
KAl(SO
4
)
2
∙
12H
2
O potash alum potassium aluminum sulfate dodecahydrate
KCr(SO
4
)
2
∙
12H
2
O chrome alum potassium chromium (III) sulfate dodecahydrate
(NH
4
)
2
Fe(SO
4
)
2
∙
6H
2
O Mohr’s salt ammonium iron (II) sulfate hexahydrate
KMgCl
3
∙
6H
2
O carnallite potassium magnesium chloride hexahydrate
3. What is the difference between the following chemical notations?
CuSO
4
∙
5H
2
O copper (II) sulfate pentahydrate ionic compound blue crystalline (solid) substance
CuSO
4
( aq ) copper (II) sulfate dissolved in water aqueous solution; mixture of two substances clear blue liquid
4. Calculate the theoretical percentage of water for the following hydrates
(a) iron(II) sulfate heptahydrate
FeSO
4
∙7H
2
O
7 × 18.02 g
%H
2
O = × 100% = 45.37 %
278.01 g
(b) aluminum nitrate nonahydrate
Al(NO
3
)
3
∙9H
2
O
9 × 18.02 g
%H
2
O = × 100% = 43.23 %
375.13 g

5. A sample of an unknown hydrate had a mass of 1.550 g before heating. After heating the mass of the remaining solid was found to be 0.994 g.
(a) What is the experimental percentage of water in the hydrate? mass of H
2
O = 1.550 g – 0.994 g = 0.556 g
0.556 g
%H
2
O = × 100% = 35.87 %
1.550 g
(b) What is the percent error if the theoretical percentage of water in the hydrate is 36.08%
│ theoretical value – experimental value
│
%error = × 100% = theoretical value
│36.08%
–
35.87 %│
= × 100% = 0.6%
36.08%
6. A 2.62-gram sample of sodium sulfate hydrate lost 1.02 grams of water when heated to constant mass. Give the formula of the hydrate and name it.
2.62 g Na
2
SO
4
∙ xH
2
O mass H
2
O = 1.02 g mass Na
2
SO
4
= 2.62 g – 1.02 g = 1.60 g
1 mol 0.0113 mol
Na
2
SO
4
1.60 g × = 0.0113 mol = 1.00 mol
142.04 g 0.0113
H
2
O 1.02 g ×
1 mol
= 0.0566 mol
0.0566 mol
= 5.01 mol
18.02 g 0.0113
Formula of the hydrate: Na
2
SO
4
∙
5H
2
O
7. A hydrate of strontium chloride contains 18.52% water. Determine the formula of the hydrate and name it.
1 mol 0.5140 mol
SrCl
2
81.48 g × = 0.5140 mol = 1.000 mol
158.53 g 0.5140
1 mol 1.0277 mol
H
2
O 18.52 g × = 1.0277 mol = 2.000 mol
18.02 g 0.5140
Formula of the hydrate: SrCl
2
∙
2H
2
O
8. Calculate the formula of the compound, which exists as a hydrate, and contains 44.6% ytterbium and
27.5% chlorine.
1 mol 0.2577 mol
Yb 44.6 g × = 0.2577 mol = 1.00 mol
173.04 g 0.2577
1 mol 0.7757 mol
Cl 27.5 g × = 0.7757 mol = 3.01 mol
35.45 g 0.2577
1 mol
H
2
O 27.9 g × = 1.548 mol
18.02 g
Formula of the hydrate: YbCl
3
∙
6H
2
O
1.548 mol
0.2577
= 6.01 mol