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Entry task: Feb 13 th -14 th Block #2 NOT AN ENTRY TASK! Agenda: Sign off on Post Lab question- Discuss Notes on Molecular Formulas and Hydrates HW: Molecular formula and hydrates #1
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Fast Flash of Molecular formula Empirical formula is CH But it was experimentally determined to have a molecular mass of 26 grams How many CH masses are there in 26 g? What is the Empirical mass of CH? C= 12.01 + H = 1.0079= 13g is the empirical mass
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The Set up: Experimental molar mass (given in problem) Empirical mass 26g 13g Divide the two! 2 is the amount that the empirical formula is off by
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The Fix: 2 (CH) Multiply 2 through the empirical formula The molecular formula is (C 2 H 2 )
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Empirical to Molecular Formula 49.98 g carbon 1 mole of C 12.01 of C --------- = 4.16 Moles of Carbon ------------- 10.47 g hydrogen 1 mole of H 1.007g of H --------- = 10.39 Moles of Hydrogen ------------- A compound was found to contain 49.98g carbon and 10.47g hydrogen. The molar mass is found to be 58.12 g/mol. Determine the molecular formula. Start with empirical formula
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Empirical to Molecular Formula Divide by the smallest ratio. 4.16 Moles of Carbon = 1 Moles of Carbon 4.16 Moles of Carbon 10.39 Moles of Hydrogen = 2.5 Moles of hydrogen 4.16 Moles of Carbon C2H5C2H5 CAN’T HAVE ½ a mole X 2
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Empirical to Molecular Formula 2 Moles of Carbon C 2 H 5 empirical formula- get its mass 1 mole of C 12.01 of C --------- = 24.02 g of Carbon ------------- 5 Moles of hydrogen 1 mole of H 1.007 of H --------- = 5.04 g of Hydrogen ------------------- Empirical mass is 29.055g
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Empirical mass to the fix 58.12 g/mole 29.055 g/mole A compound was found to contain 49.98g carbon and 10.47g hydrogen. The molar mass is found to be 58.12 g/mol. Determine the molecular formula. 2.00 Multiply the empirical formula by 2. Experimental molar mass (given in problem) Mass of empirical mass 2.00(C 2 H 5 ) C 4 H 10 is the molecular formula
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You Try! 46.68 g nitrogen 1 mole of N 14.006 of N --------- = 3.33 Moles of Nitrogen ------------- 53.32 g oxygen 1 mole of O 15.999 g of O --------- = 3.33 Moles of Oxygen ------------- A colorless liquid composed of 46.68% nitrogen and 53.32% oxygen has a molar mass of 60.01 g/mol. What is the molecular formula?
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Empirical Formula Divide by the smallest ratio. 3.33 Moles of nitrogen = 1 Moles of nitrogen 3.33Moles of nitrogen 3.33 Moles of oxygen = 1 Moles of oxygen 3.33 Moles of oxygen NO
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Empirical mass 1 Moles of nitrogen NO empirical formula 1 mole of N 14.00 of N --------- = 14.00 g of Nitrogen ------------- 1 Moles of oxygen 1 mole of O 15.999 of O --------- = 15.999 g of Oxygen ------------------- Empirical mass is 30.00 g
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Empirical mass to the fix 60.01 g/mole 30.00 g/mole 2.00 Multiply the empirical formula by 2. Experimental molar mass (given in problem) Empirical mass 2.00(NO) N 2 O 2 is the molecular formula A colorless liquid composed of 46.68% nitrogen and 53.32% oxygen has a molar mass of 60.01 g/mol. What is the molecular formula?
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What are hydrates? Have you heard of something being hydrated? Something to do with water? Yes! You are right.
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Hydrates Hydrates are compounds that has a specific number of water molecules bound to its atoms. This is methane surrounded by water molecules. Opals are hydrates, the trapped water molecules give its unusual color
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Hydrates It’s a RATIO of the compound and its water companion.
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Naming Hydrates Na 2 CO 3 10H 2 O Sodium carbonate decahydrate FePO 4 4H 2 O Iron III phosphate tetrahydrate
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Analyzing a Hydrates You can drive off the water by heating it. When this is done the substance is called anhydrous. Hydrated Cobalt II chloride Anhydrous Cobalt II chloride
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How much water? Suppose you have 5.0 grams of hydrated Barium chloride. BaCl 2 XH 2 O Before heating = 5.0g After heating = 4.26g 0.74 g of H 2 O But how many moles is this? 0.74 g of H 2 O ------ 1 mole of H 2 O 18.0 g of H 2 O --------- = 0.041 moles of H 2 O How much water was driven off?
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What is the relationship to its compound? BaCl 2 XH 2 O Use the mass of the anhydrous BaCl 2 = 4.26 g 4.26 g of BaCl 2 ------ 1 mole of BaCl 2 208.23 of BaCl 2 --------- = 0.0205 moles of BaCl 2 How many moles are in this mass?
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What is the relationship to its compound? 0.0205 moles of BaCl 2 = 2 mole of H 2 O = 0.0205 moles of BaCl 2 = 0.041 moles of H 2 O Get the ratios divide smallest mole into others 0.041 moles of H 2 O BaCl 2 2H 2 O
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You try! Suppose you have 2.50 grams of hydrated Copper II sulfate. CuSO 4 XH 2 O Before heating = 2.50g After heating = 1.59g 0.91 g of H 2 O But how many moles is this? 0.91g of H 2 O ------ 1 mole of H 2 O 18.0 g of H 2 O --------- = 0.050 moles of H 2 O How much water was driven off?
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What is the relationship to its compound? CuSO 4 XH 2 O Use the mass of the anhydrous CuSO 4 = 1.59 g 1.59 g of CuSO 4 ------ 1 mole of CuSO 4 159.6 of CuSO 4 --------- = 0.00996 moles of CuSO 4 How many moles are in this mass?
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What is the relationship to its compound? 0.00996 moles of CuSO 4 = 5 mole of H 2 O = 0.00996 moles of CuSO 4 = 0.050 moles of H 2 O Get the ratios divide smallest mole into others 0.050 moles of H 2 O CuSO 4 5H 2 O Copper II sulfate pentahydrate
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