Experiment-7: Analysis of a Hydrate
 To find the % of water in Alum
 To find the % of water in an unknown solid
 To become more familiar with hydrates and their properties
 To use mass relationships
Hydrates are ionic compounds (salts) that have a definite amount of water as part of their
chemical formula. The number of moles of water per mole of ionic substance is usually
an integer. This “water of hydration” is released as vapor when the hydrate is heated.
The water in a hydrate is bound loosely, and so is relatively easily removed by heating.
Most hydrates lose their water of hydration at temperatures slightly above 100 oC.
Sometimes the water is liberated in stages, with one or more lower hydrates being
observed during the heating process. The remaining solid is known as the anhydrous
salt. The general reaction for heating a hydrate is:
Hydrate (s) 
Anhydrous salt (s) +
water (g) 
The percent of water in a hydrate can be found experimentally by accurately determining
the mass of the hydrate and the mass of the anhydrous salt. The difference in mass is due
to the water lost by the hydrate. The percent of water in the original hydrate can be
calculated easily: % H2O = (Mass of water/ Mass of hydrate) x 100
Commonly encountered hydrates are the following:

Cupic Sulfate pentahydrate- CuSO4•5H2O , a blue solid

Magnesium Sulfate heptahydrate – MgSO4•7H2O - Epsom salt

Magnesium Chloride hexahydrate – MgCl2•6H2O

Calcium Chloride dihydrate – CaCl2•2H2O

Barium Chloride dihydrate – BaCl2•2H2O

Strontium Chloride hexahydrate – SrCl2•6H2O

Alum- KAl(SO4)2 •12H2O

Borax- Na2 B4O7•10H2O

Washing soda- Na2CO3•10H2O

Gypsum- CaSO4•2H2O

Plaster of Paris-(CaSO4 )2•H2O
Example
1.654 g of solid Alum sample was heated and which loses 0.706 g of water. Calculate the
a. Experimental percent of water in Alum sample
b. Theoretical percent of water in Alum, using its chemical formula
c. Percent error
Experimental % H2O in Alum = (Mass of water/ Mass of hydrate) x 100
= (0.706/1.654) x 100
= 42.7 %
Chemical formula of Alum is KAl(SO4)2 •12H2O which indicated that there are 12 mol
of water attached to 1 mol of KAl(SO4)2. First, let’s calculate the molar mass of Alum,
then the theoretical % of water in Alum, KAl(SO4)2 •12H2O can be calculated easily:
1xK
= 1 x 39.10 g = 39.10 g
1 x Al
= 1 x 26.98 g = 26.98 g
2xS
= 2 x 32.07
= 64.14 g
8xO
= 8 x 16.00
= 128.00 g
12 x H2O
= 12 x 18.02 = 216.24 g
KAl(SO4)2 •12H2O
= 474.46 g/mol
Theoretical percent of water in Alum = (Mass of water/ Mass of hydrate) x 100
= (12H2O / KAl(SO4)2 •12H2O ) x 100
= (216.24 g / 474.46 g ) x 100
= 45.58 %
% Error = (Difference between experimental & theoretical valuetheoretical value) x 100
= (45.58-42.7)  45.58 x 100
= 6.32 %
Procedure:
Part-A: Percent of water in Alum Hydrate
Do not round off the balance/scale readings.
Carryout entire experiment using one balance/scale.
Beakers, ring stand etc. will become very hot during
this experiment; so do not touch them directly. Use
crucible tongs to handle beakers. Do not weigh any
hot beakers! Review the figure before proceeding.
1. Find the mass of a clean, dry 250-mL beaker with a watchglass.
2. Using a spatula, transfer approximately 1.0 to 2.0 g of Alum hydrate into the 250mL beaker. Find the mass of 250-mL beaker with Alum and watchglass.
3. Place the beaker with watchglass on the wire gauze, which is placed on top of a
round ring stand connected to an iron stand. Your teacher will demonstrate the
assembly and the process of heating. Observe carefully!
4. Start heating the beaker gently. You will observe moisture on the sides of the
beaker and bottom of the watch glass. Crystals melt and become jellylike.
Continue heating until all moisture has been evaporated and dry powdered
substance is left in the beaker. Intensity of heat may be gradually and carefully
increased during this process. The appearance of a dry power form indicates that
process of decomposition is complete. At this point, turn-off the Bunsen burner
and allow the beaker to cool for 10 –12 minutes. Now, find the mass of the 250mL beaker with anhydrous Alum and watchglass. Dispose off the anhydrous
Alum into the designated container.
5. Do trial#2 using the Alum from the same bottle by repeating step-1 to step-4.
Part-B: Percent of water in Unknown Hydrate
6. Record the your Unknown Hydrate number __________________
7. Repeat step-1 thru step-5 just as in Part-A, except use your Unknown Hydrate in
place of Alum.
There are a total of four trails in this experiment. First two trails are in Part-A where you
will be using Alum, and other two trials in Par-B you will be using Unknown Hydrate. In
each trial you will use the balance thrice.
Laboratory Report#7: Analysis of a Hydrate
Last Name_________________________, first name________________
Date of Experiment___________ Instructor’s Initials_______
Data and Calculations
Part-A: Percent of water in Alum Hydrate
Observation
Trial#1
Trial#2
1. 250-mL beaker with a watchglass
2. 250-mL beaker with Alum and watchglass
3. 250-mL beaker with anhydrous Alum and watchglass
4. Mass of Alum hydrate used (step-2-step-1)
5. Mass of water lost (step-2- step-3)
6. % of water in Alum = (step-5/step-4) x 100
7.
Experimental % of water in Alum= Average of Trail#1 and Trail#2 =_________
8. Theoretical % of water in Alum= (Mass of water/ Mass of hydrate)x100=45.58 %
9.
% Error = (Difference between step-7 and step-8•step-8) x 100=_______%
Part-B: Percent of water in Unknown Hydrate
Observation
10. 250-mL beaker with a watchglass
11. 250-mL beaker with Unknown Hydrate and watchglass
12. 250-mL beaker with anhydrous solid and watchglass
13. Mass of Unknown Hydrate used (step-11-step-10)
14.Mass of water lost (step-11- step-12)
15 % of water in Unknown Hydrate=(step-14/step-13)x 100
Trial#1
Trial#2
Show all your work! Use proper units and significant figures!!
1.
What will be the mass of anhydrous Alum left over if one starts with 100. 00 g of
Hydrated Alum?
2. Calculate Theoretical percent of water in Cupic Sulfate pentahydrate- CuSO4•5H2O
3. Define

Hydrate:

Anhydrous compound:

Water of Hydration:

% Error: