1. Entry Task: Feb 8 th 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 #1 In-class Practice HW: 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? 92.25 g of Carbon 1 mole of C 12.011 g of C --------- = 7.68 Moles of Carbon ------------- 7.75 g of Hydrogen 1 mole of H 1.0079 g of H --------- = 7.68 Moles of Hydrogen ------------- 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. 56.0 g of vanadium 1 mole of V 50.9 g of V --------- = 1.1 Moles of Vanadium ------------- 44.0 g of oxygen 1 mole of O 15.999 g of O --------- = 2.75 Moles of Oxygen -------------

6. Empirical Formula 1.1 moles of Vanadium = 1 Moles of Vanadium Take the smallest mole amount and divided it into all others. 1.1 Moles of Vanadium 2.75 Moles of Oxygen = 2.5 Moles of Oxygen 1.1 Moles of Vanadium CAN’T HAVE ½ a mole V2O5V2O5 X 2

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 52.9 g of Aluminum 1 mole of Al 26.982 g of Al --------- = 1.96 Moles of Aluminum ------------- 47.1 g of oxygen 1 mole of O 15.999 g of O --------- = 2.94 Moles of Oxygen -------------

8. Empirical Formula 2.94 moles of Oxygen = 1.5 Moles of Oxygen Take the smallest mole amount and divided it into all others. 1.96 Moles of Aluminum = 1 Moles of Aluminum 1.96 Moles of Aluminum CAN’T HAVE ½ a mole Al 2 O 3 Aluminum Oxide X 2

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? 59.37 g of lead 1 mole of Pb 207.2 g of Pb --------- = 0.287 Moles of Lead ------------- 40.63 g of Chlorine 1 mole of ClO 35.45 g of Cl --------- = 1.14 Moles of Chlorine -------------

10. Empirical Formula 1.146 moles of Chlorine = 3.99 Moles of Chlorine Take the smallest mole amount and divided it into all others. 0.287 Moles of Lead = 1 Moles of Lead 0.287 Moles of Lead PbCl 4 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. 35.5 g of Carbon 1 mole of C 12.011 g of C --------- = 2.95 Moles of Carbon ------------- 4.77 g of hydrogen 1 mole of H 1.0079 g of H --------- = 4.7 Moles of Hydrogen -------------

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. 8.29 g of Nitrogen 1 mole of N 14.007 g of N --------- = 0.59 Moles of Nitrogen ------------- 13.6 g of sodium 1 mole of Na 22.99 g of Na --------- = 0.59 Moles of Sodium ------------- 37.9 g of oxygen 1 mole of O 15.999 g of O --------- = 2.36 Moles of Oxygen -------------

14. Empirical Formula 2.95 Moles of Carbon = 5.0 Moles of Carbon Take the smallest mole amount and divided it into all others. 0.59 Moles of N or Na 4.7 Moles of Hydrogen = 8.0 Moles of Hydrogen 0.59 Moles of N or Na C 5 H 8 NO 4 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 = 1 Moles of Sodium 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? 13.5 g of Calcium 1 mole of Ca 40.078 g of Ca --------- = 0.337 Moles of Calcium ------------- 10.8 g of oxygen 1 mole of O 15.999 g of O --------- = 0.675 Moles of Oxygen ------------- 0.675 g of hydrogen 1 mole of H 1.0079 g of H --------- = 0.669 Moles of hydrogen -------------

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

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

18. Empirical Formula 0.967 Moles of Chlorine = 2 Moles of Chlorine Second: Take the smallest mole amount and divided it into all others. 0.48 Moles of Calcium 2.89 Moles of Oxygen = 6 Moles of Oxygen 0.48 Moles of Calcium CaO 6 Cl 2 Ca(ClO 3 ) 2 0.48 Moles of Calcium = 1 Moles of Calcium 0.48 Moles of Calcium

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

20. Empirical Formula 5.4 Moles of Carbon = 4 Moles of Carbon Second: Take the smallest mole amount and divided it into all others. 1.35 Moles of Oxygen 13.4 Moles of Hydrogen = 9.9 Moles of Hydrogen 1.35 Moles of Oxygen C 4 H 10 O 1.35 Moles of Oxygen = 1 Moles of Oxygen 1.35 Moles of Oxygen

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