1. Chapter #9 Stoichiometry

2. Chapter 9.1 Composition stoichiometry deals with the mass relationships of elements in compounds. Reaction stoichiometry involves the mass relationships between reactants and products in a chemical reaction. Always starts with a balanced equation!

3. Mole ratio is a conversion factor that relates the amounts in moles of any two substances involved in a chemical reaction. 2Al + 3H 2 SO 4 Al 2 (SO 4 ) 3 + 3H 2 2 mol Al 2 mol Al 2 mol Al 3 mol H 2 SO 4 Al 2 (SO 4 ) 3 3H 2 Then they can all be flipped over. Then do the second compound and the 3 rd & 4 th.

4. Molar Mass is the mass, in grams, of one mole of a substance. Al 2 O 3 Al (26.981) x 2=53.962 g 0 (15.999) x 3= 47.997 g 101.959 g 1 mol or 101.959 g 101.959 g 1 mol

5. Chapter 9.2 1.Mole to Mole conversions Given # of x what you want mols what you have Example: 2H 2 + O 2 2H 2 O How many moles of oxygen are required to react with 45.2 moles of hydrogen? 45.2 mol H 2 X 1 mol O 2 2 mol H 2 Mol to mol ratio from your balanced equation = 22.6 mol of O 2 = Mol of what you want

6. 2.Mol to Mass (grams) Given # x mol want x molar mass of mols mol have of what you want Example: 2H 2 + O 2 2H 2 O How many grams of oxygen are required to react with 45.2 moles of hydrogen? 45.2 mol H 2 X 1 mol O 2 X 31.998 g 2 mol H 2 1 mol O 2 = Grams of what you want Mol to mol ratio from balanced equation = 723.1548 g of O 2

7. 3. Mass (grams) to Moles Given # x 1 x mol want of grams molar mass mol have of given Example: 2H 2 + O 2 2H 2 O How many moles of oxygen are required to react with 63.56 grams of hydrogen? 63.56 g H 2 X 1 mol H 2 X 1 mol O 2 2.014 g H 2 2 mol H 2 = Moles of what you want Mol to mol ratio from balanced equation = 15.78 mol of O 2

8. 4. Mass (g) to Mass (g) Given # x 1 x mol want x molar mass of grams molar mass mol have of what you want of given Example: 2H 2 + O 2 2H 2 O How many grams of oxygen are required to react with 63.56 grams of hydrogen? 63.56 g H 2 X 1 mol H 2 X 1 mol O 2 X 31.998g = 2.014 g H 2 2 mol H 2 = Mol to mol ratio 504.91 g of O 2

9. Chapter 9.3 Limiting reactant is the reactant that limits the amounts of the other reactants that can combine and the amount of product that can form in a chemical reaction. The substance that is not used up completely in a reaction is called the excess reactant.

10. If you are given: 12 legs, 2 heads, 2 bodies, 3 eyes, 4 antenna, and 5 mouth parts. Which part is the limiting reactant and which is the excess? Cootie example: It takes 6 legs, 1 head, 1 body, 2 eyes, 2 antenna, and 1 mouth part to make a cootie. 6 legs + 1 head + 1body+ 2 eyes + 2 antenna + 1 mouth part 1cootie

11. Silicon dioxide is usually unreactive but reacts with hydrogen fluoride according to the following reaction: SiO 2 + 4HF SiF 4 + 2H 2 0 If 2 moles of HF react with 4.5 moles of SiO 2 which is the limiting reactant? How much excess reactant? 2 moles HF X 1 mol SiO 2 4 mol HF 4.5 mol SiO 2 X 4 mol HF 1 mol SiO 2 0.50 mol SiO 2 = = 18 mol HF Given is what you have Answers are what you Need

12. Percent Yield Percent Yield = Theoretical yield is the amount in grams that you should have gotten math problem (moles to grams or grams to grams). Actual yield is the amount that you did get in lab. Actual yield Theoretical yield X 100

13. Aluminum reacts with an excess of copper (II) sulfate. If 1.85 g of Al react with an excess of copper (II) sulfate, the actual yield of copper is 3.70 g. Calculate the percent yield of Cu. 2Al + 3CuSO 4 Al 2 (SO 4 ) 3 + 3Cu 1.85 g Al X 1 mol X 3 mol Cu X 63.546g 26.98g 2 mol Al 1 mol Cu Percent yield = actual 3.70 theoretical 6.536 = 6.536 g Cu X 100 = 56.6 %

14. “Cartoon”. January 9, 2007. http://www.washbac.org/images/farside.gif http://www.washbac.org/images/farside.gif “Cootie Toys”. January 10, 2007. http://static.flickr.com/48/124090469_d6d0 95d9f9_t.jpg http://static.flickr.com/48/124090469_d6d0 95d9f9_t.jpg “Cootie parts”. January 10, 2007. http://shadeaux.hypermart.net/PICS/cootie 2.jpg http://shadeaux.hypermart.net/PICS/cootie 2.jpg