1. Laura Conrad Teaching Stoichiometry Using Recipes: Quick! Quick! Easy! Easy! Student-Friendly! Student-Friendly!

2.   Chemistry 8E: Perform stoichiometric calculations, including determination of mass relationships between reactants and products, calculation of limiting reagents, and percent yield.  Chemistry 9B: Perform stoichiometric calculations, including determination of mass and volume relationships between reactants and products for reactions involving gases. TEKS We’ll Address

3.   A recipe for Thanksgiving cranberry sauce calls for 1 pound of cranberries, 2 cups of water, and 2 cups of sugar to produce 6 servings of cranberry sauce.  What ingredients (and how much) do I need to buy if there will be 18 people at my family’s Thanksgiving?  Solve using ANY method, but you must be able to explain your thought process! A Recipe Scenario

4.   Logic: “I need three times as many servings, so I’ll just multiply everything by three!”  Algebraic charts (increase by 6 servings each time, count up till you have 18 servings)  Proportions: These are the most popular and the ones that will transfer to Stoichiometry problems. Methods Students Use

5.   Logic: “I need three times as many servings, so I’ll just multiply everything by three!”  Algebraic charts (increase by 6 servings each time, count up till you have 18 servings)  Proportions: These are the most popular and the ones that will transfer to Stoichiometry problems. Methods Students Use

6.  1 C + 2 W + 2 S  6 CS This means that 1 pound of Cranberries, 2 cups of Water, and 2 cups of Sugar will make 6 servings of Cranberry Sauce. The Recipe as a Chemical Reaction

7.  x C 18 CS 1 C + 2 W + 2 S  6 CS Inserting the “Given” numbers

8.  x C 18 CS 1 C + 2 W + 2 S  6 CS 1 C 6 CS Inserting the “Recipe” numbers

9.  x C18 CS 1 C + 2 W + 2 S  6 CS 1 C 6 CS x C 18 CS 1 C 6 CSx = 3 C (3 pounds of cranberries) Using Proportions to solve =

10.  y W z S 18 CS 1 C + 2 W + 2 S  6 CS 2 W 2 S 6 CS Solving for y W and z S in the same way gives us 6 cups of water and 6 cups of sugar in order to make the 18 servings of cranberry sauce. Writing Proportions to solve

11. Chemical Reactions are JUST LIKE Recipes!!

12.  Balanced Chemical Equations  Ingredients look just a little different  Units are all Moles

13.  Balanced Chemical Equations  Ingredients look just a little different  Units are all Moles 2 H 2 + O 2  2 H 2 O This “recipe” in the equation itself says that 2 moles of hydrogen and 1 mole of oxygen (invisible coefficient = 1) make 2 moles of water.

14.  A Stoichiometry Question How many moles of H 2 are required to make 8 moles of water??? 2 H 2 + O 2  2 H 2 O Traditional Method: 8 mol H 2 O x 2 moles H 2 = 8 mol H 2 2 moles H 2 O

15.  A Stoichiometry Question How many moles of H 2 are required to make 8 moles of water??? x mol H 2 8 mol H 2 O 2 H 2 + O 2  2 H 2 O mol H 2 mol H 2 O

16.  A Stoichiometry Question How many moles of H 2 are required to make 8 moles of water??? x mol H 2 8 mol H 2 O 2 H 2 + O 2  2 H 2 O 2 mol H 2 2 mol H 2 O x mol H 2 8 mol H 2 O 2 mol H 2 2 mol H 2 O x = 8 mol H 2 =

17.  Another Stoichiometry Question How many moles of H 2 O could be produced from 0.60 moles of oxygen? 2 H 2 + O 2  2 H 2 O

18.  Another Stoichiometry Question How many moles of H 2 O could be produced from 0.60 moles of oxygen? 0.60 mol O 2 x mol H 2 O 2 H 2 + O 2  2 H 2 O 1 mol O 2 2 mol H 2 O 0.60 mol O 2 x mol H 2 O 1 mol O 2 2 mol H 2 O x = 1.2 mol H 2 O =

19.  Mole to Mass calculations How many grams of hydrogen are produced from 6.2 mol of ammonia? 2 NH 3  N 2 + 3 H 2

20.  Mole to Mass calculations  Units on top and bottom must match!  Instead of writing the coefficient (# of moles) under H 2, write the formula mass of 3 H 2. Student memory trick: g = “Go to the Periodic Table!” 6.2 mol x g 2 NH 3  N 2 + 3 H 2 mol g

21.  Mole to Mass calculations  Units on top and bottom must match!  Instead of writing the coefficient (# of moles) under H 2, write the formula mass of 3 H 2. Student memory trick: g = “Go to the Periodic Table!” 6.2 mol x g 2 NH 3  N 2 + 3 H 2 2 mol 6.048 g 6.2 molx g 2 mol 6.048 g x = 19 g H 2 =

22.  Mass to Mole calculation How many moles of nitrogen are produced from 6.35 grams of ammonia? 6.35 g x mol 2 NH 3  N 2 + 3 H 2 Follow the same rules to solve! The answer is: 0.186 moles

23.  Mass to Mass calculation How many grams of H 2 O must react with 0.025 grams of Na in the reaction below? 2 Na + 2 H 2 O  2 NaOH + H 2 This will still be only a ONE-STEP Problem!

24.  Mass to Mass calculation How many grams of H 2 O must react with 0.025 grams of Na in the reaction below? 0.025 g x g 2 Na + 2 H 2 O  2 NaOH + H 2 45.98 g 36.03 g 0.025 gx grams 45.98 g 36.03 g x = 0.020 g H 2 O =

25.  Gases: Liters calculation How many liters of H 2 gas (at STP) will be produced if you start with 0.15 g of Na? 0.15 g x L 2 Na + 2 H 2 O  2 NaOH + H 2 45.98 g 22.4 L  For the number on the bottom, multiply the coefficient x 22.4.  This only works at STP. Otherwise you have to find moles through stoichiometry and then use PV = nRT to find the volume.

26. Let’s Talk… ** Then we’ll go back to the math again.

27.   Chemical reaction terminology of “reactants” and “products” is tied to real-world experience. (“ingredients” and “servings”)  Utilizes logic and reasoning in calculations  More familiarity with proportions than dimensional analysis from math classes  It reduces all stoichiometry calculations to ONE STEP! Why does this process help students?

28.   Yes, we do!!!! (it’s in the TEKS: Chemistry 2G)  However, dimensional analysis is used for Unit Conversions instead of Stoichiometry.  In dimensional analysis, the answer is equal to the starting amount, but the units are changed.  For example: 0.15 g Na x 1 mole Na = 0.0065 mol Na 22.99 g Na 0.15 g Na = 0.0065 mol Na Wait a second – don’t we HAVE to teach Dimensional Analysis?

29.   Stoichiometry calculations tell us that, for this reaction, at STP: 0.15 g Na produces 0.073 L H 2  How people use Dimensional Analysis for this: 0.15 g Na x 1 mole Na x 1 mol H 2 x 22.4 L H 2 22.99 g Na 2 mol Na 1 mol H 2  Stoichiometry does NOT say 0.15 g Na = 0.073 L H 2 So why not Dimensional Analysis for Stoichiometry??

30.   A cookie recipe calls for 2 cups of flour, 1 cup of sugar, 2 eggs, and ½ cup of milk. This recipe claims that it makes 30 cookies.  How many cookies can I make if I have 8 cups of flour, 3 cups of sugar, a dozen eggs, and 1 cup of milk?  Solve using any methods you want, but you must be able to explain your reasoning.  Pre-AP: How much is left over of the other ingredients? Another Recipe Scenario

31.  Write this recipe as a chemical equation: 2 F + 1 S + 2 E + ½ M  30 C We have: 8F, 3S, 12E, 1M. Another Recipe Scenario

32.  8F 3S 12E 1 M 2 F + 1 S + 2 E + ½ M  30 C Fractions available: 8 F3 S12 E1 M 2 F1 S 2 E ½ M 4 batches3 batches6 batches2 batches Another Recipe Scenario

33.  8F 3S 12E 1 M 2 F + 1 S + 2 E + ½ M  30 C Fractions available: 8 F3 S12 E1 M 2 F1 S 2 E ½ M 4 batches3 batches6 batches2 batches Another Recipe Scenario

34.  We will run out of MILK first, so this is our limiting reagent! We can tell because this is the one with the smallest fraction. Use the limiting reagent to determine the amount of product. Another Recipe Scenario

35.  1 M x C 2 F + 1 S + 2 E + ½ M  30 C x = 60 cookies Another Recipe Scenario

36.  Identify the limiting reagent in the reaction below if you have 18.1 g of CaCl 2 and 0.41 mol of NaOH. CaCl 2 + 2 NaOH  Ca(OH) 2 + 2 NaCl Limiting Reagents

37.  Identify the limiting reagent in the reaction below if you have 18.1 g of CaCl 2 and 0.41 mol of NaOH. 18.1 g 0.41 mol CaCl 2 + 2 NaOH  Ca(OH) 2 + 2 NaCl 110.986 g 2 mol Which is a smaller fraction? 0.163 0.205 Limiting Reagents

38.  Identify the limiting reagent in the reaction below if you have 18.1 g of CaCl 2 and 0.41 mol of NaOH. 18.1 g CaCl 2 + 2 NaOH  Ca(OH) 2 + 2 NaCl 110.986 g You can then use the CaCl 2 numbers to calculate how much product was made. Limiting Reagents

39.   When I buy Betty Crocker cookie mix, it says that I will get 3 dozen 2-inch cookies if I follow their recipe.  However, I usually only get 28 cookies! Why? What are my sources of error? One last Recipe Scenario

40.   Cookies aren’t exactly 2-inch diameter when I make them.  Some of the mix stayed in the bowl.  The baking process changes the cookies chemically.  Betty Crocker lied to me! Common Student Answers

41.   In cooking and in chemistry, we never get the full amount we expected to make.  In other words, we never get 100% yield! Actual Yield Theoretical Yield What this demonstrates % Yield = x 100

42.  If I actually produce 28 cookies when the recipe says I should get 36 cookies, what is my % yield? Actual yield = 28 cookies Theoretical yield = 36 cookies (28/36) x 100 = 78 % Calculating % yield

43.  http://dfwminicast.weebly.com Feb. 20, 2016 (Saturday) Wakeland High School in Frisco, TX Presenters get free registration! Interested in presenting at Mini-CAST?

44.  Thanks!