1. 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

4. 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

5. 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

6. The Fix: 2 (CH) Multiply 2 through the empirical formula The molecular formula is (C 2 H 2 )

7. 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

8. 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

9. 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

10. 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

11. 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?

12. 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

13. 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

14. 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?

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

16. 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

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

18. Naming Hydrates Na 2 CO 3  10H 2 O Sodium carbonate decahydrate FePO 4  4H 2 O Iron III phosphate tetrahydrate

19. 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

20. 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?

21. 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?

22. 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

23. 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?

24. 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?

25. 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