Chapter 8
Hydrates
Hydrates
• A hydrate is a solid crystalline substance that has a certain
number of water molecules as part of its structure.
– Example: CoCl2 • 6H2O
Hydrates
• A hydrate is a solid crystalline substance that has a certain
number of water molecules as part of its structure.
– Example: CoCl2 • 6H2O
• An anhydrous salt is the hydrate minus the water.
– Example: CoCl2
Determination of a hydrate formula:
CoCl2 • 6H2O
•
•
Although all problems for hydrates
may sound a little different they all
require that you use the information in
the problem to find the:
Grams of water
Grams of anhydrous substance.
Determination of a hydrate formula
1. Calculate the mass of water and convert
it moles.
2. Calculate the mass of the anhydrous salt
(dried salt) and convert it to moles.
3. Divide your moles of water by your moles
of anhydrous salt. Round your answer to
the nearest whole number to determine
the number of water molecules in the
hydrate formula.
4.32g of the hydrate FeSO4  xH2O is heated to drive
off the water. After heating we find that we have 2.36g
of anhydrous salt. What is the hydrate formula?
Example
problems are
in the note
outline.
4.32g of the hydrate FeSO4  xH2O is heated to drive
off the water. After heating we find that we have 2.36g
of anhydrous salt. What is the hydrate formula?
4.32g FeSO4  xH2O – 2.36g FeSO4 = 1.96g H2O
4.32g of the hydrate FeSO4  xH2O is heated to drive
off the water. After heating we find that we have 2.36g
of anhydrous salt. What is the hydrate formula?
4.32g FeSO4  xH2O – 2.36g FeSO4 = 1.96g H2O
1.96g H2O mol
= 0.109 mol H2O
18.0 g
2.36g FeSO4
mol
= 0.0155 mol FeSO4
151.9g
4.32g of the hydrate FeSO4  xH2O is heated to drive
off the water. After heating we find that we have 2.36g
of anhydrous salt. What is the hydrate formula?
4.32g FeSO4  xH2O – 2.36g FeSO4 = 1.96g H2O
1.96g H2O mol
= 0.109 mol H2O
18.0 g
2.36g FeSO4
mol
= 0.0155 mol FeSO4
151.9g
0.109 mol H2O
0.0155 mol FeSO4
= 7.03
4.32g of the hydrate FeSO4  xH2O is heated to drive
off the water. After heating we find that we have 2.36g
of anhydrous salt. What is the hydrate formula?
4.32g FeSO4  xH2O – 2.36g FeSO4 = 1.96g H2O
1.96g H2O mol
= 0.109 mol H2O
18.0 g
2.36g FeSO4
mol
= 0.0155 mol FeSO4
151.9g
0.109 mol H2O
0.0155 mol FeSO4
= 7.03 ≈ 7
4.32g of the hydrate FeSO4  xH2O is heated to drive
off the water. After heating we find that we have 2.36g
of anhydrous salt. What is the hydrate formula?
4.32g FeSO4  xH2O – 2.36g FeSO4 = 1.96g H2O
1.96g H2O mol
= 0.109 mol H2O
18.0 g
2.36g FeSO4
mol
= 0.0155 mol FeSO4
151.9g
0.109 mol H2O
0.0155 mol FeSO4
= 7.03 ≈ 7
FeSO4 • 7H2O
• A student uses a balance to measure the mass of an
evaporating dish to be 19.82g. She then pours some
of the hydrate CuSO4  xH2O into the evaporating dish
and measures the mass again to obtain a reading of
21.54g. She heats the hydrate gradually at first and
then more rapidly to vaporize the water from the
hydrate. After vaporizing the water she removes the
evaporating dish from the hotplate and allows it to
cool. She then measures the mass for the last time to
be 20.94g. What is the formula of the hydrate?
• A student uses a balance to measure the mass of an
evaporating dish to be 19.82g. She then pours some
of the hydrate CuSO4  xH2O into the evaporating dish
and measures the mass again to obtain a reading of
21.54g. She heats the hydrate gradually at first and
then more rapidly to vaporize the water from the
hydrate. After vaporizing the water she removes the
evaporating dish from the hotplate and allows it to
cool. She then measures the mass for the last time to
be 20.94g. What is the formula of the hydrate?
•Find the grams of water and the grams of anhydrous
salt before attempting to solve the problem.
• A student uses a balance to measure the mass of an
evaporating dish to be 19.82g. She then pours some
of the hydrate CuSO4  xH2O into the evaporating dish
and measures the mass again to obtain a reading of
21.54g. She heats the hydrate gradually at first and
then more rapidly to vaporize the water from the
hydrate. After vaporizing the water she removes the
evaporating dish from the hotplate and allows it to
cool. She then measures the mass for the last time to
be 20.94g. What is the formula of the hydrate?
21.54g – 19.82g = 1.72g hydrate
21.54g – 20.94g = 0.60g H2O
1.72g – 0.60g = 1.12g CuSO4
The mass of an evaporating dish is 19.82g. When the hydrate
CuSO4  xH2O is added to the evaporating dish the mass is 21.54g.
After heating for the last time the mass is 20.94g. What is the
formula of the hydrate?
0.60g H2O
mol
18.0 g
1.12g CuSO4 mol
159.6g
0.033 mol H2O
= 0.033 mol H2O
= 0.00702 mol CuSO4
= 4.7
≈5
0.00702 mol CuSO4
CuSO4 • 5H2O
Homework
• Hydrate Worksheet
• Do the lab summary for the Lab:
“Determination of a Hydrate Formula”
• Study Guide Chapter 8