Hydrates (11.5) Hydrate- compound with a specific amount of water molecules bound to it

Hydrates are often used as drying agents or desiccators because they can absorb moisture Silica gel- used in electronic equipment packaging, cat litter

Chemicals that do not contain water in their structures are called anhydrous (without water) Hydrates are easy to spot because their formulas show the water molecules Anhydrous cobalt (ii) chloride Cobalt (II) chloride hexahydrate CoCl2 CoCl2 . 6H2O

What is a Hydrate? Any salt that has water chemically bonded to the ions in the crystal structure is a hydrate or hydrated crystal. Copper(II) sulfate is a hydrate. Hydrated copper(II) sulfate is deep blue in color.

What does the Chemical Formula of A Hydrate Look Like? BaCl2•2H2O FeSO4•6H2O Na2CO3•10H2O CuSO4•5H2O

How are Names of Hydrates Written? BaCl2•2H2O barium chloride dihydrate FeSO4•6H2O iron(II) sulfate hexahydrate Na2CO3•10H2O sodium carbonate decahydrate CuSO4•5H2O copper(II) sulfate pentahydrate BaCl2•2H2O FeSO4•6H2O Na2CO3•10H2O CuSO4•5H2O

What prefixes are used? 0.5 hemi 7 hepta 1 mono 8 octa 2 di 9 nona 3 tri 10 deca 4 tetra 11 undeca 5 penta 12 dodeca 6 hexa 13 triskaideca

To determine hydrate formula Determine moles of water and of anhydrous compound Find x = moles of water moles of anhydrous compound Plug in x Formula for compound X H2O

The water molecule is highly polar

Water’s polar ends are attracted to the + and – elements of the crystal

The # of water molecules associated with the compound depends on the size of the crystal

The chemical formula for a hydrate includes the formula unit for the ionic compound and the number of complexed water molecules separated by a dot: a) CaSO4 • 2 H2O calcium sulfate dihydrate b) CoCl2 • 6 H2O cobalt II chloride hexahydrate

Analysis of an unknown hydrate: There is no simple way to predict the number of water molecules found in a hydrate compound. It must be determined empirically by heating the compound and evaporating the water

How can the water be removed? Heat the crystal. The water is loosely bound, and will come away as water vapor. Put the crystal in contact with or near a desiccant, maybe in a desiccator.

What is the compound called after the water has been removed? Anhydride (noun) The light blue powder is the anhydride. Anhydrous (adjective) Anhydrous copper(II) sulfate is left in the test tube after heating.

Example: A 14 gram sample of hydrated cobalt II chloride is heated thoroughly to remove all the water. The resulting anhydrous cobalt II chloride weighs 7.65 g. What is the chemical formula of the hydrated cobalt II chloride To know the chemical formula, you must determine the number of moles of water present per mole of cobalt II chloride Approach: a) the 7.65 g is only anhydrous cobalt II chloride b) the difference between 7.65 g and the original 14 g is the amount of water present in the original sample

Approach: a) the 7.65 g is only anhydrous cobalt II chloride b) the difference between 7.65 g and the original 14 g is the amount of water present in the original sample 7.65 g 1 mole = 0.0588 mol CoCl2 130 g 14 - 7.65 = 6.35 g H2O 1 mole = 0.353 mol H2O 18 g 0.353 mol H2O = 6 mol H2O  CoCl2 • 6 H2O 0.0588 mol CoCl2 1 mol CoCl2

CuSO4 x 5H2O 2.50 = CuSO4 + H2O CuSO4= 1.59 g A mass of 2.50 g blue hydrated copper sulfate (CuSO4 x XH2O) is place in a crucible and heated. After heating, 1.59 g of copper sulfate remains. What is the formula for the hydrate? 2.50 = CuSO4 + H2O CuSO4= 1.59 g 2.50 = 1.59 + H2O H2O = 0.91 g CuSO4 = 63.5 + 32.1 + 4 x 16 = 159.6 g/mol CuSO4 = 1.59 / 159.6 = 0.010 mol H2O = 2 x 1 + 16 = 18 g/mol H2O = 0.91 / 18 = 0.050 mol X = 0.050 / 0.010 = 5 CuSO4 x 5H2O

What is the mole ratio of the hydrate CuSO4 x ___H2O, if the anhydrous mass of CuSO4 is 25.42g and mass of water is 14.3 g? CuSO4 = 63.5 + 32.1 + 4 x 16 = 159.6 g/mol CuSO4 = 25.42 / 159.6 = 0.159 mol H2O = 2 x 1 + 16 = 18 g/mol H2O = 14.3 / 18 = 0.79 mol X = 0.79 / 0.159 = 4.96 ~ 5 CuSO4 x 5H2O 1:5

What is the mole ratio of hydrate Na2CO3 x ____ H2O, if the original mass of the hydrated Na2CO3 was 43.69 g and the mass of the anhydrous Na2CO3 is 16.19g? Hydrate = anhydrous + H2O 43.69 = 16.19 + H2O H2O = 27.50 g Na2CO3 = 16.19 g Na2CO3 = 2 x 24.3 + 12 + 3 x 16 = 108.6 g/mol Na2CO3 = 16.19 / 108.6 = 0.15 mol H2O = 2 x 1 + 16 = 18 g/mol H2O = 27.50 / 18 = 1.5 mol X = 1.5 / 0.15 = 10 Na2CO3 x 10 H2O