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Warm-up

Warm-up. How many grams are in 3.45 X 10 4 formula units of iron (II) oxide?. You may work with a partner. Answers are on page 996. In textbook: Pg 323 #1&3 Pg 324 #5 Pg 328 #15 Pg 329 #17 Pg 331 #19 Pg 335 #35 Pg 337 #41. Stoichiometry Chemistry Chapter 11. Flowchart.

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Warm-up

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  1. Warm-up • How many grams are in 3.45 X 104 formula units of iron (II) oxide?

  2. You may work with a partner.Answers are on page 996. • In textbook: • Pg 323 #1&3 • Pg 324 #5 • Pg 328 #15 • Pg 329 #17 • Pg 331 #19 • Pg 335 #35 • Pg 337 #41

  3. StoichiometryChemistry Chapter 11

  4. Flowchart Atoms or Molecules Divide by 6.02 X 1023 Multiply by 6.02 X 1023 Moles Multiply by atomic/molar mass from periodic table Divide by atomic/molar mass from periodic table Mass (grams)

  5. Review Practice • Calculate the Molar Mass of calcium phosphate • Formula = • Masses elements: • Ca: 3 Ca’s X 40.1 = • P: 2 P’s X 31.0 = • O: 8 O’s X 16.0 = • Molar Mass = Ca3(PO4)2 120.3 g 62.0 g 128.0 g 120.3g + 62.0g +128.0g 310.3 g/mol

  6. Calculations molar mass Avogadro’s numberGrams Moles particles Everything must go through Moles!!!

  7. Chocolate Chip Cookies!! 1 cup butter 1/2 cup white sugar 1 cup packed brown sugar 1 teaspoon vanilla extract 2 eggs 2 1/2 cups all-purpose flour 1 teaspoon baking soda 1 teaspoon salt 2 cups semisweet chocolate chips Makes 3 dozen How many eggs are needed to make 3 dozen cookies? How much butter is needed for the amount of chocolate chips used? How many eggs would we need to make 9 dozen cookies? How much brown sugar would I need if I had 1 ½ cups white sugar?

  8. Cookies and Chemistry…Huh!?!? • Just like chocolate chip cookies have recipes, chemists have recipes as well • Instead of calling them recipes, we call them reaction equations • Furthermore, instead of using cups and teaspoons, we use moles • Lastly, instead of eggs, butter, sugar, etc. we use chemical compounds as ingredients Ms. Ramsey’s

  9. Chemistry Recipes • Looking at a reaction tells us how much of something you need to react with something else to get a product (like the cookie recipe) • Be sure you have a balanced reaction before you start! • Example: 2 Na + Cl2  2 NaCl • This reaction tells us that by mixing 2 moles of sodium with 1 mole of chlorine we will get 2 moles of sodium chloride • What if we wanted 4 moles of NaCl? 10 moles? 50 moles?

  10. Practice • Write the balanced reaction for hydrogen gas reacting with oxygen gas. 2 H2 + O2 2 H2O • How many moles of reactants are needed? • What if we wanted 4 moles of water? • What if we had 3 moles of oxygen, how much hydrogen would we need to react and how much water would we get? • What if we had 50 moles of hydrogen, how much oxygen would we need and how much water produced? 2 mol H21 mol O2 4 mol H22 mol O2 6 mol H2, 6 mol H2O 25 mol O2, 50 mol H2O

  11. Mole Ratios • These mole ratios can be used to calculate the moles of one chemical from the given amount of a different chemical • Example: How many moles of chlorine is needed to react with 5 moles of sodium (without any sodium left over)? 2 Na + Cl2 2 NaCl 5 moles Na 1 mol Cl2 2 mol Na = 2.5 moles Cl2

  12. Mole-Mole Conversions • How many moles of sodium chloride will be produced if you react 2.6 moles of chlorine gas with an excess (more than you need) of sodium metal? 2 Na + Cl2 2 NaCl 2.6 moles Cl2 2 mol NaCl 1 mol Cl2 = 5.2 moles NaCl

  13. Mole-Mass Conversions • Most of the time in chemistry, the amounts are given in grams instead of moles • We still go through moles and use the mole ratio, but now we also use molar mass to get to grams • Example: How many grams of chlorine are required to react completely with 5.00 moles of sodium to produce sodium chloride? 2 Na + Cl2 2 NaCl 5.00 moles Na 1 mol Cl2 70.90g Cl2 2 mol Na 1 mol Cl2 = 177g Cl2

  14. Practice • Calculate the mass in grams of Iodine required to react completely with 0.50 moles of aluminum. 2 Al + 3 I2  2 AlI3 0.50 moles Al 3 moles I2 253.8g I2 2 moles Al 1 mole I2 = 190 g I2

  15. Mass-Mole • We can also start with mass and convert to moles of product or another reactant • We use molar mass and the mole ratio to get to moles of the compound of interest • Calculate the number of moles of ethane (C2H6) needed to produce 10.0 g of water • 2 C2H6 + 7 O2 4 CO2 + 6 H20 10.0 g H2O 1 mol H2O 2 mol C2H6 18.0 g H2O 6 mol H20 = 0.185 mol C2H6

  16. Practice • Calculate how many moles of oxygen are required to make 10.0 g of aluminum oxide • 4 Al + 3 O2  2 Al2O3 10.0g Al2O3 1 mol Al2O3 3 mol O2 101.96g Al2O3 2 mol Al2O3 = 0.147 mol O2

  17. Mass-Mass Conversions • Most often we are given a starting mass and want to find out the mass of a product we will get (called theoretical yield) or how much of another reactant we need to completely react with it (no leftover ingredients!) • Now we must go from grams to moles, mole ratio, and back to grams of compound we are interested in

  18. Mass-Mass Conversion • Ex. Calculate how many grams of ammonia (NH3) are produced when you react 2.00g of nitrogen with excess hydrogen. • N2 + 3 H2  2 NH3 2.00g N2 1 mol N2 2 mol NH3 17.031g NH3 28.014g N2 1 mol N2 1 mol NH3 = 2.43 g NH3

  19. Practice • How many grams of calcium nitride are produced when 2.00 g of calcium reacts with an excess of nitrogen? • 3 Ca + N2 Ca3N2 2.00g Ca 1 mol Ca 1 mol Ca3N2 148.248g Ca3N2 40.078g Ca 3 mol Ca 1 mol Ca3N2 = 2.47 g Ca3N2

  20. Limiting Reactant: Cookies 1 cup butter 1/2 cup white sugar 1 cup packed brown sugar 1 teaspoon vanilla extract 2 eggs 2 1/2 cups all-purpose flour 1 teaspoon baking soda 1 teaspoon salt 2 cups semisweet chocolate chips Makes 3 dozen If we had the specified amount of all ingredients listed, could we make 4 dozen cookies? What if we had 6 eggs and twice as much of everything else, could we make 9 dozen cookies? What if we only had one egg, could we make 3 dozen cookies?

  21. Limiting Reactant • Most of the time in chemistry we have more of one reactant than we need to completely use up other reactant. • That reactant is said to be in excess (there is too much). • The other reactant limits how much product we get. Once it runs out, the reaction s. • This is called the limiting reactant.

  22. Limiting Reactant • To find the correct answer, we have to try all of the reactants. We have to calculate how much of a product we can get from eachof the reactants to determine which reactant is the limiting one. • The lower amount of a product is the correct answer. • The reactant that makes the least amount of product is the limiting reactant. Once you determine the limiting reactant, you should ALWAYS start with it! • Be sure to pick a product! You can’t compare to see which is greater and which is lower unless the product is the same!

  23. LimitingReactant Limiting Reactant: Example • 10.0g of aluminum reacts with 35.0 grams of chlorine gas to produce aluminum chloride. Which reactant is limiting, which is in excess, and how much product is produced? 2 Al + 3 Cl2 2 AlCl3 • Start with Al: • Now Cl2: 10.0 g Al 1 mol Al 2 mol AlCl3 133.341 g AlCl3 26.982 g Al 2 mol Al 1 mol AlCl3 = 49.4g AlCl3 35.0g Cl2 1 mol Cl2 2 mol AlCl3 133.341 g AlCl3 70.906 g Cl2 3 mol Cl2 1 mol AlCl3 = 43.9g AlCl3

  24. LR Example Continued • We get 49.4g of aluminum chloride from the given amount of aluminum, but only 43.9g of aluminum chloride from the given amount of chlorine. Therefore, chlorine is the limiting reactant. Once the 35.0g of chlorine is used up, the reaction comes to a complete .

  25. Limiting Reactant Practice • 15.0 g of potassium reacts with 15.0 g of iodine. Calculate which reactant is limiting and how much product is made.

  26. Limiting Reactant Practice 2 K + I2 2 KI • Potassium: • Iodine: • Iodine is the limiting reactant and we get 19.6 g of potassium iodide 15.0g K 1 mol K 2 mol KI 166.002 g KI 39.098 g K 2 mol K 1 mol KI = 63.7g KI 15.0 g I2 1 mol I2 2 mol KI 166.002 g KI 253.808 g I2 1 mol I2 1 mol KI = 19.6 g KI

  27. Finding the Amount of Excess • By calculating the amount of the excess reactant needed to completely react with the limiting reactant, we can subtract that amount from the given amount to find the amount of excess. • Can we find the amount of excess potassium in the previous problem?

  28. Finding Excess Practice • 15.0 g of potassium reacts with 15.0 g of iodine. 2 K + I2 2 KI • We found that Iodine is the limiting reactant, and 19.6 g of potassium iodide are produced. 15.0 g I2 1 mol I2 2 mol K 39.1 g K 254 g I2 1 mol I2 1 mol K = 4.62 g K USED! 15.0 g K – 4.62 g K = 10.38 g K EXCESS Given amount of excess reactant Amount of excess reactant actually used Note that we started with the limiting reactant! Once you determine the LR, you should only start with it!

  29. Limiting Reactant: Recap • You can recognize a limiting reactant problem because there is MORE THAN ONE GIVEN AMOUNT. • Convert ALL of the reactants to the SAME product (pick any product you choose.) • The lowest answer is the correct answer. • The reactant that gave you the lowest answer is the LIMITING REACTANT. • The other reactant(s) are in EXCESS. • To find the amount of excess, subtract the amount used from the given amount. • If you have to find more than one product, be sure to start with the limiting reactant. You don’t have to determine which is the LR over and over again!

  30. Percent Yield and Stoichiometry

  31. Percent Yielddescribes how much product was actually made in the lab versus the amount that theoretically could be made. Actual Yield  100 = Percent Yield Theoretical Yield • Percent yield tells you how close you were to the 100% mark. • Reactions do not always work perfectly. Experimental error (spills, contamination) often means that the amount of product made in the lab does not match the ideal amount that could have been made.

  32. Theoretical Yield = The maximum amount of product that could be formed from given amounts of reactants. • Actual Yield = The amount of product actually formed or recovered when the reaction is carried out in the laboratory.

  33. Example: • A chemist was supposed to produce 75.0 g of aspirin. However, he squandered some of the reactants for his own personal use (he was later fired), and so only actually made 50.0 g. What was his percent yield? 50.0 g = 66.7 % X 100 75.0 g

  34. More Examples! • When I was a sophomore in college, we had a lab where we were supposed to isolate caffeine from tea leaves. Most of my caffeine was washed down the drain in a freak accident. Although I should have had 5.0 g of caffeine, I only ended up with 0.040 g of caffeine and a bad grade on the lab. What was my percent yield? 0.040 g X 100 = 0.80 % 5.0 g

  35. 2. A very sloppy student did not wait until his NaCl was dry before he weighed it. As a result, his product weighed 2.25 g when it should have been 1.75 g. What was his percent yield? 2.25 g = 129% X 100 1.75 g

  36. 3. What is the theoretical yield if 5.50 grams of hydrogen react with nitrogen to form ammonia? H2 + N2 NH3 3 2 5.50 g H2 1 mole H2 17.031 g NH3 2 moles NH3 1 mol NH3 2.016 g H2 3 moles H2 = 31.0 grams NH3 = Theoretical Yield!!!! Only 20.4 grams of ammonia is actually produced in the lab. What is the percent yield? 20.4 g = 65.8 % yield X 100 31.0 g

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