Exp 3B - Hydrates and Thermal Decomposition
• Background
– In chemistry, a neutral ionic compound is typically referred to as a “ salt ”
– Hydrates are inorganic salts that contain a specific number of water molecules crystallized with the salt (“water of crystallization” or “water of hydration“)
– Examples are
• CaCl
2
.
4 H
2
O
• CuSO
4
.
5 H
2
O
• BaCl
2
.
2 H
2
O
• Na
2
HPO
4
.
12 H
2
O
4 mol H
5 mol H
2
2 mol H
2
2
O/mol CaCl
O/mol CuSO
O/mol BaCl
2
2
4
(blue)
12 mol H
2
O/mol Na
2
HPO
4
– Most hydrates have simple formulas with a ratio of
1 salt: x H
2 not always the case
O,but it is
• cadmium sulfate hydrate is best represented as (CdSO
4
)
3
.
8 H
2
O
• When determining the formula of a hydrate you must not assume that it is one with a simple formula.

Exp 3B Hydrates and Thermal Decomposition
• Hydrates can normally be decomposed into the anhydrous
( without water ) salt and water by gentle heating .
• From mass data collected the number of molecules of hydrated water can be determined per molecule of anhydrous salt.
• Heating removes H
2
O gradually
– yields anhydrous form
– CoCl
2
.6H
2
O
– CoCl
2
.2H
– CoCl
2
2
O
6 mol H
2
O/mol CoCl
2
2 mol H
2
O/mol CoCl
2
( red ) (hexahydrate)
( violet ) (dihydrate) no H
2
O, anhydrous ( blue )
CoCl
2
.6H
2
O red
D
-4 H
2
O
CoCl
2
.2H
2
O violet
D
-2 H
2
O
CoCl
2 blue

Exp 3B Hydrates and Thermal Decomposition
• Thermal decomposition of BaCl
2
.
2H
2
O (barium chloride dihydrate)
– BaCl
2
.
2H
D
2
O (s)

BaCl
2
.
H
2
O (s) + H
2
O (g)
– BaCl
2
.
2H
D
2
O (s)  BaCl
2
(s) + 2 H
2
O (g)
– BaCl
2
.
2H
D
2
O (s)

BaO (s) + H
2
O (g) + 2HCl (g)
“ D ”= “Heat”

Exp 3B Hydrates and Thermal Decomposition
Part A: Preparation
• Heat a crucible + lid on a clay triangle in a the flame of a lab burner for about 3 minutes (as you did for Exp. 3A)
• Let the crucible+lid cool to warm-to-thetouch and weigh. Record in 4 decimal points
• Reheat the crucible+lid to make sure the crucible is really dry and repeat the weighing as above
• Repeat until two consecutive weightings are within 0.001
g of each other
• Report the weights on the lab report sheet and determine the average of your weightings

Exp 3B Hydrates and Thermal Decomposition
Part A: Thermal decomposition product of BaCl
2
.2H
2
O Mass(g)
1.
Average mass of crucible+lid _14.3087_
2.
Weigh about 1.4 –1.5 g BaCl
2
.2H
2
O on an electronic balance
(exact weight is not that important!!)
3.
Transfer the barium chloride to the crucible
4.
Weigh crucible + lid + BaCl
2
.2H
2
O ( 0.0001
g precision) _15.7175_
5.
Mass of BaCl
2
.2H
2
O = ( crucible + lid + BaCl
2
.2H
2
O ) – ( crucible+lid )
_1.4088__
6.
Heat the crucible gently for 2-3 min
6.
Continue heating at full flame for 15 min
7.
Cool crucible to “warm to touch” and weigh
8.
Repeat heating for 3 min, and cool crucible (“warm to touch”)
_15.5059__
9.
Weigh crucible: difference not more than 0.001 g
10. Weigh crucible + lid + residue ( 0.0001
g precision) _15.5050_

Exp 3B Hydrates and Thermal Decomposition
• Part A: Thermal decomposition of BaCl
2
.2H
2
O - Analysis
– Average mass of crucible+lid
_14.3087_ (g)
– Mass of crucible + lid + barium chloride hydrate
– Mass of BaCl
2
.2H
2
O
_15.7175_ (g)
_1.4088 (g)
– Mass of crucible + lid + anhydrous salt
– Mass of anhydrous product
_15.5050_ (g)
_1.1963 (g)
– Ratio of anhydrous product/hydrate _0.84916__
Share your ratio with other groups/write your results on the white board

Exp 3B Hydrates and Thermal Decomposition
Part A: Calculate correct decomposition products
Substance
Formula
Weight
244.27
Moles Mass (g)
0.0057674
1.4088
Ratio
BaCl
2
.2H
2
O
BaCl
2
. H
2
O
226.25
0.0057674
1.3048
0.926
BaCl
2
208.24
0.0057674
1.2010
0.852
BaO
153.33
0.0057674
0.8843
0.628
Because the observed mass of the anhydrous product is _ 1.1963
__ g, ratio = 1.1963/1.4088 = 0.849
 the product is _ BaCl
2
_.

Exp 3B Hydrates and Thermal Decomposition
Part B: Identifying hydrates
1.
Put in 4 test tubes ~ 0.5 g each (small amount, pea sized) a) Cobalt(II) chloride hexahydrate, CoCl
2
.
6H
2
O b) c) d)
Nickel(II) chloride hexahydrate, NiCl
2
.
6H
2
O
Iron(II) sulfate heptahydrate, FeSO
4
.7
H
2
O
Copper(II) sulfate pentahydrate, CuSO
4
.5H
2
O
1.
Heat gently and observe H
2
O vapor on side of tube
2.
Record color changes of hydrate
3.
Test condensation with blue litmus paper; record response
If the color of the litmus paper changes, an acid is formed and released
5.
Record color and appearance of residue
6.
Heat 1 min in full flame; retest with moistened litmus paper
7.
Record color observations

Exp 3B Hydrates and Thermal Decomposition
Part B: Identifying hydrates
Observations
Example: CrCl
3
.6H
2
O a) Gentle heating: green solid changes to dark liquid; water vapor
(condensation on side of tube) and acid (change in color of litmus paper) evolve b) Intense heating: liquid changes to violet solid that increases in size; evolution of water and acid continues
Tested Samples a) Cobalt(II) chloride hexahydrate, CoCl
2
.
6H
2
O b) Copper(II) sulfate pentahydrate, CuSO
4
.5H
2
O c) Nickel chloride hexahydrate, NiCl
2
.
6H
2
O d) Iron(II) sulfate heptahydrate, FeSO
4
.
7H
2
O

Exp 3B Hydrates and Thermal Decomposition
Part B: Identifying hydrates
Reaction equations for
• 2CrCl
3
.6H
2
O (s)  Cr
2
O
3
(s) + 6HCl (g) + 9H
2
O (g)
•
CoCl
2
.
6H
2
O (s)

?
•
NiSO
4
.
6H
2
O (s)

?
• FeCl
3
.
6H
2
O (s)  ?
• Fe(NH
4
)
2
(SO
4
)
2
.6H
2
O (s)

?

Next week
Lab report for Exp 3B
• Summary of experiment
• Results sheets (p. 73-74)
•
Questions 1a – d, 2 (p. 75-76)
• Conclusion & Discussion of your results
Exp 4A: Conductivity of Aqueous Solutions
• Prelab assignments: 1a-d, 2a-b, 3, 4a-b, 6
•
Lab preparations
–
Goal of Experiment
–
Experimental Procedures
– Physical, chemical & and toxicological properties of:
• hydrochloric acid (HCl)
• potassium hydroxide (KOH)
• Potassium nitrate (KNO
3
)
• ethanol (C
2
H
5
OH)
• acetic acid (C
2
H
4
O
2
)