1. Hydrated Ionic Compounds and how they fit into the molecular formula thing

2. 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.

3. 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. CuSO 4 (s) CuSO 4 ·xH 2 O(s) + heat  CuSO 4 (s) + xH 2 O(g)

4. MgSO 4 ·7H 2 O(s) + heat  MgSO 4 (s) + 7H 2 O(g) magnesium sulfateanhydrous 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. MgSO 4 —without any waters of hydration— is known as anhydrous magnesium sulfate. 3.Usually the number of waters of hydration is a whole number.

5. Some Ionic Hydrates formulaname CaSO 4 ·2H 2 Ocalcium sulfate dihydrate; gypsum CaCl 2 ·2H 2 Ocalcium chloride dihydrate LiCl·4H 2 Olithium chloride tetrahydrate MgSO 4 ·7H 2 Omagnesium sulfate heptahydrate, epsom salts Ba(OH) 2 ·8H 2 Obarium hydroxide octahydrate Na 2 CO 3 ·10H 2 Osodium carbonate decahydrate KAl(SO 4 ) 2 ·12H 2 Opotassium aluminum sulfate dodecahydrate; alum

6. sample problem Let’s say you want to determine the number of waters of hydration in hydrated barium hydroxide, Ba(OH) 2 ·xH 2 O. How would do this? What measurements would you take?

7. 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

8. 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 ·xH 2 O.

9. set up a ratio: mm Ba(OH) 2 :mass Ba(OH) 2 obtained mm Ba(OH) 2 ·xH 2 O mass hydrateheated 171.4 g:3.26 g y6.00 g

10. 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 H 2 O/mol hydrate. Convert to mol H 2 O 144.1 g/18.02 g/mol = 8 mol H 2 O. And so... Ba(OH) 2 ·8H 2 O is the formula of hydrate.

11. 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...