Entry Task: Feb 8 th Friday

1. Entry Task: Feb 8th Friday Write the question down: A compound contained 92.25% carbon and 7.75% hydrogen. What is the empirical formula of the compound? You have ~10 minutes!

2. Agenda:Go over Empirical Formula ws #1In-class PracticeHW: Empirical Formula ws #2

4. Empirical Formula A compound contained 92.25% carbon and 7.75% hydrogen. What is the empirical formula of the compound? 1 mole of C ------------- 92.25 g of Carbon = 7.68 Moles of Carbon --------- 12.011 g of C 1 mole of H ------------- 7.75 g of Hydrogen = 7.68 Moles ofHydrogen --------- 1.0079 g of H CH

5. Empirical Formula Vanadium oxide is used as an industrial catalyst. The percent composition of this oxide is 56.0% vanadium and 44.0% oxygen. Determine the empirical formula for vanadium oxide. 1 mole of V ------------- 56.0 g of vanadium = 1.1 Moles of Vanadium --------- 50.9 g of V 1 mole of O ------------- 44.0 g of oxygen = 2.75 Moles of Oxygen --------- 15.999 g of O

6. Empirical Formula Take the smallest mole amount and divided it into all others. 1.1 moles of Vanadium = 1 Moles of Vanadium 1.1 Moles of Vanadium CAN’T HAVE ½ a mole 2.75 Moles of Oxygen = 2.5 Moles of Oxygen X 2 1.1 Moles of Vanadium V2O5

7. Empirical Formula While trace impurities of iron and chromium in natural corundum form the gemstones ruby and sapphire, they are basically a binary compound of aluminum and oxygen, with 52.9% Al. Find the empirical formula and give the chemical name for corundum 1 mole of Al ------------- 52.9 g of Aluminum = 1.96 Moles of Aluminum --------- 26.982 g of Al 1 mole of O ------------- 47.1 g of oxygen = 2.94 Moles of Oxygen --------- 15.999 g of O

8. Empirical Formula Take the smallest mole amount and divided it into all others. 2.94 moles of Oxygen = 1.5 Moles of Oxygen 1.96 Moles of Aluminum CAN’T HAVE ½ a mole 1.96 Moles of Aluminum = 1 Moles of Aluminum X 2 1.96 Moles of Aluminum Al2O3 Aluminum Oxide

9. Empirical Formula Analysis of a compound containing chlorine and lead reveals that the compound is 59.37% lead. What is the empirical formula for the chloride? 1 mole of Pb ------------- 59.37 g of lead = 0.287 Moles of Lead --------- 207.2 g of Pb 1 mole of ClO ------------- 40.63 g of Chlorine = 1.14 Moles of Chlorine --------- 35.45 g of Cl

10. Empirical Formula Take the smallest mole amount and divided it into all others. 1.146 moles of Chlorine = 3.99 Moles of Chlorine 0.287 Moles of Lead 0.287 Moles of Lead = 1 Moles of Lead 0.287 Moles of Lead PbCl4 Lead IV chloride

12. Empirical Formula Provide the empirical formula for a substance that consists of 35.5% carbon, 4.77% hydrogen, 8.29% nitrogen, 13.6% sodium, and 37.9% oxygen. 1 mole of C ------------- 35.5 g of Carbon = 2.95 Moles of Carbon --------- 12.011 g of C 1 mole of H ------------- 4.77 g of hydrogen = 4.7 Moles of Hydrogen --------- 1.0079 g of H

13. Empirical Formula Provide the empirical formula for a substance that consists of 35.5% carbon, 4.77% hydrogen, 8.29% nitrogen, 13.6% sodium, and 37.9% oxygen. 1 mole of N ------------- 8.29 g of Nitrogen = 0.59 Moles of Nitrogen --------- 14.007 g of N 1 mole of Na ------------- 13.6 g of sodium = 0.59 Moles of Sodium --------- 22.99 g of Na 1 mole of O ------------- 37.9 g of oxygen = 2.36 Moles of Oxygen --------- 15.999 g of O

14. Empirical Formula Take the smallest mole amount and divided it into all others. 2.95 Moles of Carbon = 5.0 Moles of Carbon 0.59 Moles of N or Na 4.7 Moles of Hydrogen = 8.0 Moles of Hydrogen 0.59 Moles of N or Na 2.36 Moles of Oxygen = 4.00 Moles of Oxygen 0.59 Moles of N or Na 0.59 Moles of N and Na = 1 Moles of Nitrogen 0.59 Moles of N and Na 0.59 Moles of N and Na = 1 Moles of Sodium C5H8NO4Na 0.59 Moles of N and Na

15. Empirical Formula A compound was analyzed and found to contain 13.5 g calcium, 10.8 g oxygen, and 0.675 g hydrogen. What is the empirical formula of the compound? 1 mole of Ca ------------- 13.5 g of Calcium = 0.337 Moles of Calcium --------- 40.078 g of Ca 1 mole of O ------------- 10.8 g of oxygen = 0.675 Moles of Oxygen --------- 15.999 g of O 1 mole of H ------------- 0.675 g of hydrogen = 0.669 Moles of hydrogen --------- 1.0079 g of H

16. Empirical Formula Second: Take the smallest mole amount and divided it into all others. 0.675 Moles of oxygen = 2 Moles of Oxygen 0.337 Moles of Calcium 0.669 Moles of Hydrogen = 1.98 or 2 Moles of hydrogen 0.337 Moles of Calcium 0.337 Moles of Calcium = 1 Moles of Calcium 0.337 Moles of Calcium CaO2H2 Ca(OH)2

17. Empirical Formula Find the empirical formula of a compound containing: 19.32 % Calcium, 34.30 % Chlorine, and 46.38 % Oxygen. 1 mole of Ca ------------- 19.32 g of Calcium = 0.48 Moles of Calcium --------- 40.078 g of Ca 1 mole of Cl ------------- 34.30 g of chlorine = 0.967 Moles of chlorine --------- 35.45 g of Cl 1 mole of O ------------- 46.38 g of oxygen = 2.89 Moles of Oxygen --------- 15.999 g of O

18. Empirical Formula Second: Take the smallest mole amount and divided it into all others. 0.967 Moles of Chlorine = 2 Moles of Chlorine 0.48 Moles of Calcium 2.89 Moles of Oxygen = 6 Moles of Oxygen 0.48 Moles of Calcium 0.48 Moles of Calcium = 1 Moles of Calcium 0.48 Moles of Calcium CaO6Cl2 Ca(ClO3)2

19. Empirical Formula Find the empirical formula of a compound containing: 64.86 % Carbon, 13.52 % Hydrogen, and 21.62 % Oxygen. 1 mole of C ------------- 64.86 g of Carbon = 5.4 Moles of Carbon --------- 12.011 g of C 1 mole of H ------------- 13.52 g of hydrogen = 13.4 Moles of Hydrogen --------- 1.0079 g of H 1 mole of O ------------- 21.62 g of oxygen = 1.35 Moles of Oxygen --------- 15.999 g of O

20. Empirical Formula Second: Take the smallest mole amount and divided it into all others. 5.4 Moles of Carbon = 4 Moles of Carbon 1.35 Moles of Oxygen 13.4 Moles of Hydrogen = 9.9 Moles of Hydrogen 1.35 Moles of Oxygen 1.35 Moles of Oxygen = 1 Moles of Oxygen 1.35 Moles of Oxygen C4H10O

21. Your Turn- Flip paper over to complete your homework