Lab Report: Water Within a Hydrate
Name of Course:
Instructor’s Name:
Date:
Student Name:
Honors Chemistry
Mrs. Thompson
08/31/2012
Awesome Blossom
Introduction
Many salts are associated with water in their crystalline structure. Such compounds are hydrates. The
bonds holding the water molecules to the salt are not very strong and are due to the ionic nature of the
salt and the polarity of water. Simple heating can drive off the water and leave an anhydrous salt behind.
Since the mass of water is considered part of the hydrate's mass, the molar ratio between water and the
salt corresponds to n, the number of water molecules associated to a salt group or formula unit.
The formula of a hydrate can be written in the form XYŸnH2O
Hypothesis
After heating a hydrate, the mass of the compound will change. The compound will have a lower molar
mass, making it possible to determine the number of water molecules in each formula unit of the
anhydrous salt.
Materials and Methods
1. After recording the name and formula of the unknown hydrate, an evaporating dish was placed on
the center of the wire gauze atop a stand. A Bunsen burner was used to heat the dish and drive
off any impurities such as oil or water that may have collected on it.
2. After about 2 minutes of the initial heating, the evaporating dish was masses using a balance.
3. The evaporating dish was placed on the table and about two spoons of the unknown were placed
on it.
4. The mass of the evaporating dish with the unknown was measured and recorded using a
balance.
5. The evaporating dish and the unknown were returned to the stand and heated for 3 minutes at a
low flame.
6. The evaporating dish was kept in place and heated for another 5 minutes on high, after which it
was removed and allowed to cool before measuring its mass on the balance. This mass was
recorded
7. The evaporating dish and anhydrous salt were returned to the stand and heated for another 2
minutes. The mass was determined again with the balance. This action was repeated until 2
successive masses were the same. This was done to ensure that all of the water had been
removed from the compound. This value was recorded as mass of evaporating dish and
anhydrous salt.
Observations/Data
Water in a Hydrate Data Table
Unknown Name
Unknown Formula
Mass of empty evaporating dish
Mass of dish with unknown
Mass of unknown compound
Mass of dish with anhydrous salt
Mass of anhydrous salt
Mass of water lost
Moles of anhydrous salt
Moles of water
Number of water molecules per formula
Magnesium sulfate
MgSO4 * ?H2O
168.386 g
181.368 g
12.982 g
174.725 g
6.339 g
6.643 g
0.05266 mol
0.3691 mol
7 mol
Analysis
Mass of unknown: 181.386 – 168.386 = 12.982 g
Mass of water lost: 181.386 – 174.725 = 6.643 g
Mass of anhydrous salt: 12.982 – 6.643 g = 6.339 g
Moles of water: 6.643 g (1mol/18g H2O) = 0.3691mol
Moles of anhydrous salt: 6.339 g (1mol/120.38g) = 0.05266mol
Molar ratio between water and anhydrate: (0.3691/0.05266) = 7:1
Empirical formula of the salt hydrate: MgSO4
Ÿ
7H2O
Discussion and Conclusion
1. What are three potential sources of error specific to this lab and how may they have affected your
results?
Three potential sources of error are:
1. Not heating the evaporating dish before adding the sample
If this step is skipped, any water in the evaporating dish prior to adding the sample will be
accounted as being part of the compound. This will give results of a higher mass of
water being lost, which be will analyzed as a greater number of water molecules per
formula unit.
2. Starting to heat the sample on high flame instead of gradually
This could cause the compound to foam and spill over the evaporating dish. If the lost
mass is not accounted for the calculations of molar mass will be messed up.
3. Taking the mass of the anhydrous salt too soon before having two consecutive masses being
the same
This will yield a lower number of water molecules per formula unit.
2. The dehydration and re-hydration is a reversible process. How can you re-hydrate the anhydrous salt
so it will become a hydrate again?
Water can be added to the anhydrous salt to make a solution. Then the solution would be allowed
evaporate the excess water.
3. What was the percentage of water present in the hydrate?
MgSO4
Ÿ
7H2O = 246.38g/mol
Moles of water: 7(18.00) = 126.00g
%Mass: (126.00/246.38)100 = 51.14%
4. Use the internet to list 5 examples of salt hydrates and their empirical formulas.
MgSO4 Ÿ 7H2O: magnesium sulfate heptahydrate
CaSO4 Ÿ 2H2O: calcium sulfate dihydrate
CuSO4 Ÿ 5H2O: copper(II) sulfate pentahydrate
ZnSO4 Ÿ 7H2O: zinc sulfate heptahydrate
Ba(OH)2 Ÿ 8H2O: barium hydroxide octahydrate
5. Discuss if the hypotheses for the experiment was supported or refuted. Be sure to reference all of the
results from the lab in your discussion.
The hypothesis was supported. The data collected in this lab supports that water molecules are loosely
bound to a salt under normal conditions. When these conditions were changed, like applying direct heat
to the salt, water evaporated leaving an anhydrous salt behind. The molar mass ratio of the lost water
and anhydrous salt was calculated using stoichiometry. This ratio was the number of moles of water
molecules per formula unit. With all of this data the empirical formula of the hydrate compound was
found.