Hydrated Ionic Compounds
and how they fit into the
molecular formula thing
Many ionic compounds crystallize from
aqueous solution with one or more water
molecules incorporated into their crystal
structure.
These are called hydrated ionic compounds,
or “hydrates” to their close friends.
Hydrates have a specific number of water
molecules associated with each formula unit
(fu) of the compound.
eg. copper II sulfate·x hydrate—blue
The water molecules are weakly bonded, and
can be removed by
heating.
CuSO4·xH2O(s) + heat  CuSO4(s) + xH2O(g)
MgSO4·7H2O(s) + heat  MgSO4(s) + 7H2O(g)
magnesium sulfate
anhydrous
heptahydrate
magnesium sulfate
Note:
1. The dot in the formula of an ionic hydrate
does not mean multiplication. It denotes a
loose association of the water molecules.
2. eg. MgSO4—without any waters of hydration—
is known as anhydrous magnesium sulfate.
3. Usually the number of waters of hydration is a
whole number.
Some Ionic Hydrates
formula
name
CaSO4·2H2O
calcium sulfate dihydrate; gypsum
CaCl2·2H2O
calcium chloride dihydrate
LiCl·4H2O
lithium chloride tetrahydrate
MgSO4·7H2O
magnesium sulfate heptahydrate, epsom
salts
Ba(OH)2·8H2O
barium hydroxide octahydrate
Na2CO3·10H2O
sodium carbonate decahydrate
KAl(SO4)2·12H2O
potassium aluminum sulfate
dodecahydrate; alum
sample problem
Let’s say you want to determine the number
of waters of hydration in hydrated barium
hydroxide, Ba(OH)2·xH2O.
How would do this? What measurements
would you take?
• mass the hydrate (be more specific . . .)
• weigh empty test tube; add hydrate;
reweigh
• heat test tube over bunsen burner flame to
drive off water
• re-weigh test tube with anhydrous
Ba(OH)2
here are some data . . .
6.00 g of barium hydroxide hydrate is heated
over a bunsen burner flame to drive off the
waters of hydration.
After heating, 3.26 g of anhydrous Ba(OH)2
remains. Determine the number of waters
of hydration in hydrated barium hydroxide.
Calculate the formula of Ba(OH)2·xH2O.
set up a ratio:
mm Ba(OH)2 :
mm Ba(OH)2·xH2O
mass Ba(OH)2 obtained
mass hydrate heated
171.4 g
y
3.26 g
6.00 g
:
cross-multiply to solve for y:
y = 315.5 g  which is
the molar mass of the hydrate.
mass of water in hydrate =
315.5 g – 171.4 g = 144.1 g which is
mass of H2O/mol hydrate. Convert to mol H2O
144.1 g/18.02 g/mol = 8 mol H2O. And so...
Ba(OH)2·8H2O is the formula of hydrate.
Homework
p 277 13 – 18
p 278 52, 54, 57 – 60
We’ll do a version of Inv 6-C on p 286
Section Review and Chapter Review Questions
are all good. Knock yourself out . . .