1. Chapter 12 Stoichiometry Mr. Mole

2. Molar Mass of Compounds Molar mass (MM) of a compound - determined by up the atomic masses of – Ex. Molar mass of CaCl 2 – MM of Calcium = 40.08 g/mol – MM of Chlorine = 35.45g g/mol – Molar Mass of CaCl 2 = 40.08 g/mol Ca + 70.9 g/mol Cl = 20 Ca 40.08a 17 Cl 35.45 Cl

3. Flowchart Atoms or Molecules Moles Mass (grams) Divide by Multiply by 6.02 X 10 23 Multiply by atomic/molar mass from periodic table Divide by atomic/molar mass from periodic table

4. Practice Calculate the Molar Mass of calcium phosphide – Formula = – Masses elements: Ca = 40.08 g/mol P = 30.97 g/mol – Ca 3 P 2

5. Calculations molar mass Avogadro’s number Grams Moles particles 22.4 L Volume Everything must go through !!!

6. Chocolate Chip Cookies!! 1 cup butter 1 cup packed brown sugar 2 eggs 2 1/2 cups all-purpose flour 2 cups semisweet chocolate chips Makes 3 dozen How many eggs are needed to make 3 dozen cookies? How many eggs to make 9 dozen cookies? How much brown sugar would I need if I had 4 eggs?

7. Cookies and Chemistry…Huh!?!? Just like chocolate chip cookies have recipes, chemists have recipes called Instead of using cups and teaspoons, we use Lastly, instead of eggs, butter, sugar, etc. we use

8. Chemistry Recipes reactions tell us how much will react to get a – like the cookie recipe – Be sure you have a balanced equation before you start! Ex: Na + Cl 2  NaCl Reaction tells us 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?

9. Practice Write the balanced reaction for hydrogen gas reacting with oxygen gas. H 2 + O 2  H 2 O – 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?

10. Stoichiometry is… Greek for “measuring elements” Pronounced “stoy kee ah muh tree” Defined as: calculations of the quantities in chemical reactions, based on a balanced equation. There are to interpret a balanced chemical equation

11. #1. In terms of Particles An Element is made of A Molecular compound (made of only nonmetals) is made up of Ionic Compounds (made of a metal and nonmetal parts) are made of

12. Example: 2H 2 + O 2 → 2H 2 O Two molecules of hydrogen and one molecule of oxygen form two molecules of water. Another example: 2Al 2 O 3  Al + 3O 2 2 Al 2 O 3 form4 Al and3 O2O2 Now read this: 2Na + 2H 2 O  2NaOH + H 2

13. #2. In terms of Moles tell us how many moles of each substance 2Al 2 O 3  Al + 3O 2 – mol Al 2 O 3, mol Al, mol O 2 2Na + 2H 2 O  2NaOH + H 2 Remember: A balanced equation is a

14. #3. In terms of Mass The Law of Conservation of Mass applies We can check mass by using moles. Be + 2F  BeF 2 1 mole Be = 2 mole F 1 mole F = 36.04 g H 2 + O 2 + reactants

15. In terms of Mass (for products) Be + 2F  BeF 2 1 moles BeF 2 47.01 g BeF 2 1 mole BeF 2 = g Be + 2F = Mass of reactants and products. must be reactant = product

16. #4. In terms of Volume At STP, 1 mol of any gas = 2H 2 + O 2   2H 2 O (2 x 22.4 L H 2 ) + (1 x 22.4 L O 2 )  (2 x 22.4 L H 2 O) ** and are ALWAYS however, molecules, formula units, moles, and volumes will not necessarily be conserved! 67.2 Liters of reactant ≠ 44.8 Liters of product!

17. Practice: Show that the following equation follows the Law of Conservation of Mass (show the atoms balance, and the mass on both sides is equal) 2Al 2 O 3  Al + 3O 2 Atoms: Al and O = Al and O Mass: Reactant: g/mol = Products: 108 g/mol + 204 g/mol

18. Practice 1). Balance the equation and interpret it in terms of atoms, moles and mass. Show that the law of conservation is observed. - N 2 + H 2 --> NH 3 Atoms: N and H = N and H Moles: N 2 and H 2 ≠ NH 3 Mass: Reactants: 28.02 g/mol + 6.06 g/mol = Products: 34.08 g/mol

19. Section 12.2 Chemical Calculations

20. Mole to Mole conversions 2Al 2 O 3  Al + 3O 2 – each time we use 2 moles of Al 2 O 3 we will also make 3 moles of O 2 2 moles Al 2 O 3 3 mole O 2 or 2 moles Al 2 O 3 3 mole O 2 This is how we can convert from mols to mols. This is why we need balanced equations.

21. Mole to Mole conversions How many moles of O 2 are produced when 3.34 moles of Al 2 O 3 decompose? 2Al 2 O 3  Al + 3O 2 3.34 mol Al 2 O 3 = If you know the amount of ANY chemical in the reaction, you can find the amount of ALL the other chemicals! Conversion factor from balanced equation

22. Practice: 2C 2 H 2 + 5O 2  4CO 2 + 2H 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 mole of H 2 O? 3.84 moles C 2 H 2 8.95 moles H 2 O

23. Steps to Calculate Stoichiometric Problems 1. Correctly the equation. 2. Convert the given amount into. 3. Set up mole. 4. Use mole ratios to calculate moles of desired chemical. 5. Convert moles back into final unit.

25. Mole-Mass Conversions Most of the time in chemistry, the amounts are given in grams instead of moles We still use the mole ratio, but now we also use to get to Example: How many grams of chlorine are required to react with 5.00 moles of sodium to produce sodium chloride? 2 Na + Cl 2  2 NaCl 5.00 moles Na 1 mol Cl 2 70.90g Cl 2 2 mol Na 1 mol Cl 2

26. Practice Calculate the mass in grams of Iodine required to react completely with 0.50 moles of aluminum. Al + 3I  AlI 3 0.50 mol Al

27. Mass-Mole We can also start with mass and convert to moles We use molar mass and the mole ratio to get to moles of the compound of interest – Calculate the number of moles of ethane (C 2 H 6 ) needed to produce 10.0 g of water – 2 C 2 H 6 + 7 O 2  4 CO 2 + 6 H 2 0 10.0 g H 2 O

28. Practice Calculate how many moles of oxygen are required to make 10.0 g of aluminum oxide 2Al + 3O  Al 2 O 3 10.0 g Al 2 O 3

29. Mass-Mass Problem: 6.50 grams of aluminum reacts with oxygen. How many grams of aluminum oxide are formed? 4Al + 3O 2  2Al 2 O 3 = 6.50 g Al ? g Al 2 O 3 1 mol Al 26.98 g Al (6.50 x 1 x 2 x 101.96) ÷ (26.98 x 4 x 1) =

30. Another example: If 10.1 g of Fe are added to a solution of Copper (II) Sulfate, how many grams of solid copper would form? 2Fe + 3CuSO 4  Fe 2 (SO 4 ) 3 + 3Cu 10.1 g Fe1 mol Fe 55.85 g Fe 1 mol Cu 63.55 g Cu

31. Volume-Volume Calculations: How many liters of CH 4 at STP are required to completely react with 17.5 L of O 2 ? CH 4 + 2O 2  CO 2 + 2H 2 O 17.5 L O 2 22.4 L O 2 1 mol O 2 2 mol O 2 1 mol CH 4 22.4 L CH 4 22.4 L O 2 1 mol O 2 1 mol CH 4 22.4 L CH 4

32. Section 12.3 Limiting Reagent & Percent Yield OBJECTIVES: – Identify the limiting reagent in a reaction.

33. “Limiting” Reagent If you are given one dozen loaves of bread, a gallon of mustard, and three pieces of salami, how many salami sandwiches can you make? The is the reactant you run out of first. The is the one you have left over. The limiting reagent how much product you can make

34. How do you find out which is limited? The chemical that makes the amount of is the “limiting reagent”. Limiting reagent problems will give you 2 amounts of chemicals You must do stoichiometry problems; one for each reagent given. If 2 products are given, pick one and use it for both calculations

35. If 10.6 g of copper reacts with 3.83 g sulfur, how many grams of the product (copper (I) sulfide) will be formed? 2Cu + S  Cu 2 S 10.6 g Cu 63.55g Cu 1 mol Cu 2 mol Cu 1 mol Cu 2 S 159.16 g Cu 2 S = 13.3 g Cu 2 S 3.83 g S 32.06g S 1 mol S 1 mol Cu 2 S 159.16 g Cu 2 S = 19.0 g Cu 2 S

36. Finding the Amount of Excess By calculating the amount of the 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 next problem?

37. Finding Excess Practice 15.0 g of potassium reacts with 15.0 g of iodine. 2 K + I 2  2 KI We found that Iodine is the limiting reactant. 15.0 g I 2 1 mol I 2 39.1 g K 254 g I 2 1 mol K 15.0 g K – 4.62 g K = 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!

38. Another example: If 10.3 g of aluminum are reacted with 51.7 g of CuSO 4, how much copper (grams) will be produced? 2Al + 3CuSO 4 → 3Cu + Al 2 (SO 4 ) 3 10.3 g Al 26.98 g Al 1 mol Al 1 mol Cu 63.55 g Cu 51.7 g CuSO 4 159.62 g CuSO 4 1 mol CuSO 4 1 mol Cu 63.55 g Cu

39. How much excess reactant do we have from the last problem? CuSO 4 limiting reactant Excess Al = 51.7 g CuSO 4 159.62 g CuSO 4 1 mol CuSO 4 1 mol Al 26.98 g Al

40. Limiting Reactant: Recap 1.You can recognize a limiting reactant problem because there is MORE THAN ONE GIVEN AMOUNT. 2.Convert ALL of the reactants to the SAME product (pick any product you choose.) 3.The lowest answer is the limiting reactant 4.The other reactant(s) are in EXCESS. 5.To find the amount of excess, subtract the amount used from the given amount. 6.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!

41. The Concept of: A little different type of yield than you had in Driver’s Education class.

42. What is Yield? Yield is the amount of product made in a chemical reaction. There are three types: 1. Theoretical yield- what the balanced equation tells be made actually 2. yield- what you actually get in the lab when the chemicals are mixed 3. Percent yield 3. Percent yield =

43. Example: produced 6.78 g of copper is produced when 3.92 g of Al are reacted with excess copper (II) sulfate. 2Al + 3 CuSO 4  Al 2 (SO 4 ) 3 + 3Cu actual What is the actual yield? What is the theoretical yield? What is the percent yield? 3.92 g Al 26.98 g Al 1 mol Al 1 mol Cu 63.55 mol Cu X 100

44. Details on Yield Percent yield tells us how “efficient” a reaction is. Percent yield can not be bigger than 100 %. Theoretical yield will always be larger than actual yield! – Why? Due to impure reactants; competing side reactions; loss of product in filtering or transferring between containers; measuring