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Entry task: Feb 13 th -14 th Block #2

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. Fast Flash of Molecular formula. Empirical formula is CH.

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Entry task: Feb 13 th -14 th Block #2

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  1. Entry task: Feb 13th-14thBlock #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

  2. Fast Flash of Molecular formula Empirical formula is CH But it was experimentally determined to have a molecular mass of 26 grams What is the Empirical mass of CH? How many CH masses are there in 26 g? C= 12.01 + H = 1.0079= 13g is the empirical mass

  3. 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

  4. The Fix: (CH) 2 Multiply 2 through the empirical formula The molecular formula is (C2H2)

  5. Empirical to Molecular Formula 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 1 mole of C ------------- 49.98 g carbon = 4.16 Moles of Carbon --------- 12.01 of C 1 mole of H ------------- 10.47 g hydrogen = 10.39 Moles of Hydrogen --------- 1.007g of H

  6. Empirical to Molecular Formula Divide by the smallest ratio. 4.16 Moles of Carbon = 1 Moles of Carbon 4.16 Moles of Carbon CAN’T HAVE ½ a mole 10.39 Moles of Hydrogen = 2.5 Moles of hydrogen X 2 4.16 Moles of Carbon C2H5

  7. Empirical to Molecular Formula C2H5 empirical formula- get its mass ------------- 12.01 of C 2 Moles of Carbon = 24.02 g of Carbon --------- 1 mole of C ------------------- 1.007 of H 5 Moles of hydrogen = 5.04 g of Hydrogen 1 mole of H --------- Empirical mass is 29.055g

  8. Empirical mass to the fix 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. 58.12 g/mole 29.055 g/mole Experimental molar mass (given in problem) Mass of empirical mass 2.00 Multiply the empirical formula by 2. 2.00(C2H5) C4H10 is the molecular formula

  9. You Try! 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? 1 mole of N ------------- 46.68 g nitrogen = 3.33 Moles of Nitrogen --------- 14.006 of N 1 mole of O ------------- 53.32 g oxygen = 3.33 Moles of Oxygen --------- 15.999 g of O

  10. 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

  11. Empirical mass NO empirical formula ------------- 14.00 of N 1 Moles of nitrogen = 14.00 g of Nitrogen 1 mole of N --------- ------------------- 15.999 of O 1 Moles of oxygen = 15.999 g of Oxygen 1 mole of O --------- Empirical mass is 30.00 g

  12. Empirical mass to the fix 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? 60.01 g/mole 30.00 g/mole Experimental molar mass (given in problem) Empirical mass 2.00 Multiply the empirical formula by 2. 2.00(NO) N2O2 is the molecular formula

  13. What are hydrates? • Have you heard of something being hydrated? • Something to do with water? • Yes! You are right.

  14. 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

  15. Hydrates It’s a RATIO of the compound and its water companion.

  16. Naming Hydrates Na2CO3 10H2O Sodium carbonate decahydrate Iron III phosphate tetrahydrate FePO4 4H2O

  17. 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

  18. How much water? Suppose you have 5.0 grams of hydrated Barium chloride. BaCl2  XH2O Before heating = 5.0g After heating = 4.26g How much water was driven off? 0.74 g of H2O But how many moles is this? ------ 0.74 g of H2O 1 mole of H2O = 0.041 moles of H2O --------- 18.0 g of H2O

  19. What is the relationship to its compound? BaCl2  XH2O Use the mass of the anhydrous BaCl2 = 4.26 g How many moles are in this mass? ------ 4.26 g of BaCl2 1 mole of BaCl2 = 0.0205 moles of BaCl2 --------- 208.23 of BaCl2

  20. What is the relationship to its compound? = 0.0205 moles of BaCl2 = 0.041 moles of H2O Get the ratios  divide smallest mole into others 0.041 moles of H2O = 2 mole of H2O 0.0205 moles of BaCl2 BaCl2  2H2O

  21. You try! Suppose you have 2.50 grams of hydrated Copper II sulfate. CuSO4  XH2O Before heating = 2.50g After heating = 1.59g How much water was driven off? 0.91 g of H2O But how many moles is this? ------ 0.91g of H2O 1 mole of H2O = 0.050 moles of H2O --------- 18.0 g of H2O

  22. What is the relationship to its compound? CuSO4  XH2O Use the mass of the anhydrous CuSO4 = 1.59 g How many moles are in this mass? ------ 1.59 g of CuSO4 1 mole of CuSO4 = 0.00996 moles of CuSO4 --------- 159.6 of CuSO4

  23. What is the relationship to its compound? = 0.00996 moles of CuSO4 = 0.050 moles of H2O Get the ratios  divide smallest mole into others 0.050 moles of H2O = 5 mole of H2O 0.00996 moles of CuSO4 CuSO4  5H2O Copper II sulfate pentahydrate

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