HYDRATES Unit 7 Lesson 3.

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Presentation transcript:

HYDRATES Unit 7 Lesson 3

HYDRATES Ionic compound that has water incorporated into its crystal structure

HYDRATES Written as MgSO4●7H2O Read as “Magnesium Sulfate Heptahydrate” Means for every one MgSO4 you have 7 water molecules. Notice the prefixes we learned for naming covalent compounds are used to tell how many H2O molecules you have If finding the molar mass, include the water!

HYDRATE PRACTICE Magnesium nitrate Tetrahydrate Name the following Hydrates and find their molar masses CuSO4 • 5H2O 63.55 + 32.07 + 4(16) + 5(18.016) = 249.70 g/mol ZnCl2 • 3H2O 65.39 + 2(35.45) + 3(18.016) = 190.34 g/mol Mg(NO3)2•4H2O 24.31 + 2(14.01) + 6(16) + 4(18.016) = 220.39 g/mol Copper(II)sulfate Pentahydrate Zinc chloride Trihydrate Magnesium nitrate Tetrahydrate

DETERMINING THE FORMULA OF A HYDRATE Example: Find the formula of a hydrate that is 48.8% MgSO4 and 51.2% H2O Step 1 – assume exactly 100 grams 48.8 g MgSO4 and 51.2 g H2O Step 2 – calculate moles 48.8 g MgSO4 (1 mol/120.38g) = .405 moles MgSO4 51.2 g H2O (1mol/18.02g) = 2.84 moles H2O Step 3 – calculate mole ratio by dividing by smallest number of moles .405/.405 = 1 mole MgSO4 2.84/.405 = 7.01 moles H2O So there are 7 moles of water per 1 mole of MgSO4 MgSO4●7H2O

EXAMPLE A hydrated Cobalt (II) Chloride sample has a mass of 11.75g. After being heated (to remove all water and become anhydrous) the mass is 9.25 g. What is the formula for the hydrate? 9.25 g CoCl2 and (11.75-9.25) = 2.50 g H2O CoCl2 = 58.93 + 2(35.45) = 129.83 g/mol H2O = 2(1.008) + 16 = 18.016 g/mol 9.25 g CoCl2 x 1mol/129.83 g = 0.0712 mol CoCl2 2.50 g H2O x 1mol/18.016 g = 0.139 mol H2O 0.0712/0.0712 = 1 mole CoCl2 0.139/0.0712 = 1.95 moles H2O CoCl2 •2H2O

EXAMPLE: FINDING MASS PERCENT FROM A HYDRATE FORMULA Determine the mass percent of the anhydrous portion and the water in the hydrate: Na3PO4 •10H2O Find the mass of the anhydrous Na3PO4 portion: 3(22.99) + 30.97 + 4(16) = 163.94g/mol Find the mass of the ten attached waters: 10(18.016) = 180.16g/mol Find the mass of the entire hydrate Na3PO4•10H2O: 163.94g/mol + 180.16g/mol = 344.10g/mol Mass percent of the anhydrous: (163.94g/mol/344.10g/mol)100 = 47.6% anhydrous Na3PO4 Mass percent of the water: (180.16g/mol/344.10g/mol)100 = 52.4% water