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 CuSOCuSO44(s)(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 Hydratesformula 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 problemLet’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 : mass Ba(OH)2 obtained

mm Ba(OH)2·xH2O mass hydrate heated

171.4 g : 3.26 g

y 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 . . .