1. Unit 10 Stoichiometry

2. Stoichiometry Looking at quantitative relationships of the reactants and products of a chemical equation MUST use a balanced equation

3. Using Balanced Equations 2 Pedals + 3 Wheel → 1Tricycle –How many pedals and wheels would I need to make 9 Tricycles? –18 Pedals and 27 wheels You can determine the quantities of reactants and products in a chemical rxn from a balanced equation (like adjusting a recipe for more/less)

4. Relationships Derived from a Balanced Equation Iron +OxygenIron (III) oxide 4Fe (s) +3O 2 (g)2Fe 2 O 3 (s) 4 atoms Fe3 molecules O 2 2 formula units Fe 2 O 3 4 moles Fe3 moles O 2 2 moles Fe 2 O 3 223.4 g Fe96.0 g O 2 319.4 g Fe 2 O 3 319.4 g reactants319.4 g products

5. Mole Ratios Ratio of the coefficients of any two parts of a balanced chemical reaction –Example: 2 Al (s) + 3 Br 2 (l) → 2 AlBr 3 (s) You can make 6 mole ratios from the balanced equation above 2 mol Al 2 mol Al 3 mol Br 2 3 molBr 2 3 mol Br 2 2 mol AlBr 2 mol Al 2 mol AlBr 2 mol AlBr 2 mol AlBr 2 mol Al 3 mol Br 2

6. Mole to Mole Conversions If you start with a given amount of moles of one reactant, how many moles of the product will be created? Just like finding out how many cookies you can make from 5 eggs… Mole ratios can be used as conversion factors First, write a balanced equation. Multiply moles given by the correct mole ratio to find what you’re looking for

7. Mole to Mole conversions How many moles of O 2 are produced when 3.34 moles of Al 2 O 3 decompose? 3.34 moles Al 2 O 3 2 moles Al 2 O 3 3 mole O 2 = 5.01 moles O 2 2 Al 2 O 3  4Al + 3O 2

8. Practice 2C 2 H 2 + 5 O 2  4CO 2 + 2 H 2 O If 3.84 moles of C 2 H 2 are burned, how many moles of O 2 are needed? How many moles of C 2 H 2 are needed to produce 8.95 moles of H 2 O? If 2.47 moles of C 2 H 2 are burned, how many moles of CO 2 are formed?

9. You can now… Calculate the number of moles of a reactant needed to create a given product. Calculate the number of moles of a product that will be created if you start with a given amount of a reactant

10. One step further… You can also calculate the mass of reactant needed or product created if given a certain number of moles These are called MOLE-MASS CONVERSIONS

11. Mole-Mass Conversions You will be given an amount in moles (either reactant or product) Multiply it by the correct mole ratio Then convert to mass by multiplying by the molar mass of the compound or element you’re looking for

12. How many grams of NaCl will be produced when 1.25 mol of Cl 2 reacts? 2 Na + Cl 2 → 2 NaCl 1.25mol Cl 2 2mol NaCl 58.5g NaCl = 146 g 1mol Cl 2 1mol NaCl (mole ratio) (molar mass)

13. Let’s add one more step… You can do the exact same thing if your given is in GRAMS instead of MOLES This is called a MASS-MASS CONVERSION You will be given an amount in grams

14. Mass-mass conversions Convert mass to moles of given by using the molar mass Next multiply that number by the correct mole ratio Then multiply by the molar mass of the compound or element that the problem is asking for.

15. For example... If 10.1 g of Fe are added to a solution of Copper (II) Sulfate, how much solid copper would form? Fe + CuSO 4  Fe 2 (SO 4 ) 3 + Cu 2Fe + 3CuSO 4  Fe 2 (SO 4 ) 3 + 3Cu 10.1 g Fe 55.85 g Fe 1 mol Fe 2 mol Fe 3 mol Cu 1 mol Cu 63.55 g Cu =17.2 g Cu

16. How many grams of water will be produced when 25.0 g of NH 4 NO 3 decomposes? NH 4 NO 3 → N 2 O + 2H 2 O 25.0g NH 4 NO 3 x 1molNH 4 NO 3 x 2mol H 2 O x 18.0gH 2 O = 11.2g 80.0gNH 4 NO 3 1mol NH 4 NO 3 1mol H 2 O (given)(molar mass) (mole ratio) (molar mass)

17. Mass-Mass Conversions These problems are also known as “gram-mole-mole-gram” problems for obvious reasons…

18. In a perfect world… We would have the exact amount of each reactant needed for a chemical equation 2H 2 + O 2  2H 2 O This would be a COMPLETE reaction & uses ALL of the given reactants

19. But in reality… Our measurements may not be so precise or we may not have that perfect amount on hand So the reaction will continue until one of the reactants runs out There will be “leftovers” of the other reactant(s)

20. The reactant that runs out first is called the LIMITING REAGENT or LIMITING REACTANT This is what determines how much product you can make

21. An example… A recipe calls for 2 eggs, 2 cups of sugar, and 1 cup of flour You have full bags of sugar and flour, but only 1 egg The egg is the limiting reactant & ultimately determines how much you can make

22. Which is the limiting reactant? 1.Make sure you have a balanced equation. 2.For each reactant, find out how many moles you have and how much product would be produced. 3.The reactant that produces the smaller amount is the limiting reactant.

23. Determine the limiting reactant if you have 25.0 g of P 4 and 50.0 g of O 2 in the following equation. P 4 + O 2 → P 4 O 10 25.0 g P 4 mole P 4 g P 4 1 124 mole P 4 O 10 mole P 4 1111 = 0.202 mole P 4 O 10 produced

24. Determine the limiting reactant if you have 25.0 g of P 4 and 50.0 g of O 2 in the following equation. P 4 + O 2 → P 4 O 10 50.0 g O 2 mole O 2 g O 2 1 32 mole P 4 O 10 mole O 2 1515 = 0.313 mole P 4 O 10 produced

25. So… 25.0 g of P 4 could produce 0.202 moles of P 4 O 10 50.0 g of O 2 could produce 0.313 moles of P 4 O 10 0.202 < 0.313 P 4 is the limiting reagent

26. Determining Amount of Product Since we know the limiting reagent, we know how many moles of the product will be produced. Convert that to grams and we know the actual amount of product produced.

27. P 4 + 5O 2 → P 4 O 10 The limiting reactant was P 4. We calculated that we could produce 0.202 mole with 25 g of P 4. 0.202 mole 284 g 1 mole = 57.37 g P 4 O 10 produced

28. Percent Yield Percent Yield = Actual Yield x 100% Theoretical Yield Actual Yield = what you experimentally determine in lab Theoretical Yield = how much you should produce in a perfect world

29. How to determine percent yield 1.Determine theoretical yield (using given in a gram-mole- mole-gram problem) 2.Calculate percent yield using formula

30. 2AgNO 3 + K 2 CrO 4 → Ag 2 CrO 4 + 2KNO 3 In a lab experiment, you complete the reaction above and produce 0.455 g of Ag 2 CrO 4. You started with 0.500 g of AgNO 3. What is the percent yield of Ag 2 CrO 4 ?

31. 2AgNO 3 + K 2 CrO 4 → Ag 2 CrO 4 + 2KNO 3 STEP 1 - Determine Theoretical Yield 0.50 g AgNO 3 mole AgNO 3 g AgNO 3 1 169 1 mole Ag 2 CrO 4 2 moles AgNO 3 331.78 g Ag 2 CrO 4 1 mole Ag 2 CrO 4 = 0.488 g Ag 2 CrO 4

32. 2AgNO 3 + K 2 CrO 4 → Ag 2 CrO 4 + 2KNO 3 STEP 2 – Calculate Percent Yield 0.455 g 0.488 g x 100%= 93.2%