3. Stoichiometry The Meaning of the Word The word stoichiometry derives from two Greek words: stoicheion (meaning "element") and metron (meaning "measure").

4. Stoichiometry deals with calculations about the masses (sometimes volumes) of reactants and products involved in a chemical reaction. It is a very mathematical part of chemistry, so be prepared for lots of calculator use.

5. Stoichiometry is the branch of chemistry and chemical engineering that deals with the quantities of substances that enter into, and are produced by, chemical reactions. For example, when methane, CH 4, unites with oxygen, O 2, in complete combustion, 16g of methane require 64g of oxygen. At the same time 44g of carbon dioxide, CO 2, and 36g of water, H 2 O, are formed as reaction products.

6. Things to remember:  Moles & particles – everything can be compared in moles on a macroscopic scale and particles on a microscopic scale. Stoichiometry uses moles to make conversions in a balanced equation.

7. Things to remember: Why again do we use moles? Using moles is a method to count a large number of microscopic particles and also to be able to measure them out in a practical manner (using a balance to measure grams) in the lab.

8. Things to remember: Particles – 6.02 x 10 23 particles (atoms, formula units, ions, molecules, etc.) Mass/Grams – these can be converted to moles by dividing by the molar mass (found on the periodic table) Volume/liters – one mole of a gas contains 22.4 liters.

9. Things to remember:  Equations must be balanced with the correct formulas for reactants and products.

10. Liters of gas A Liters of gas B

11. Moles to Moles In a balanced equation, the mole ratios can be converted to any other equal ratio value. N 2 + 3H 2  2NH 3 The ratio of nitrogen to hydrogen is: 1:3 or 2:6 or 1.5:4.5 or 3.3:9.9 Any value can be used and the ratio will be a constant for an equation.

12. Mole to Mole problem The reaction of 2.5 moles aluminum with oxygen produces aluminum oxide. Write a balanced equation. What is the mole ratio from the balanced equation for aluminum to oxygen and aluminum to aluminum oxide? What is the mole ratio of aluminum to oxygen and aluminum to aluminum oxide when you start with 2.5 moles of aluminum.

13. Mole to Mole problem 4Al(s) + 3 O 2 (g)  2Al 2 O 3 (s) Aluminum to oxygen is 4:3 Aluminum to aluminum oxide is 4:2

14. Work Problems 5, 6, & 8 p. 241, 33-35, 37 p. 262

15. Mass to Mass Conversions 1. Convert the grams of the given substance to moles. 2. Compare the mole ratios. 3. Convert moles of unknown to grams.

16. Mass to Mass Problem Calculate the number of grams of ammonia produced by the reaction of 6.85 grams of nitrogen with excess hydrogen. Balanced equation: N 2 + 3H 2  2NH 3

17. Mass to Mass Problem Knowns Mass of nitrogen = 6.85 g of N 2 Mole ratio of N 2 to NH 3 = 1 : 2 Unknowns Moles of N 2 = 6.85g / 28.02g/mol = 0.244 mol Mole ratio of N 2 to NH 3 based on unknown = 0.244 mol N 2 x 2 = 0.489 mol NH 3 Mass of ammonia = 0.489 mol NH 3 x 17.03 g/mol = 8.33grams NH 3

19. Summary of steps 1. Balance the equation if needed. 2. Convert the grams, particles or volume of a substance into moles by dividing by your conversion factor. 3. Compare the mole ratios for the conversion substances using the balanced equation. 4. Convert the moles into what is required; grams, particles or volume by multiplying by your conversion factor.

20. Let’s do one more!!! Nitrogen monoxide and oxygen gas combine to form the brown gas nitrogen dioxide that is found in smog. How many grams of nitrogen dioxide are produced when 8.42 liters of nitrogen monoxide reacts with the oxygen in the atmosphere at STP.

21. Smog Problem 1. Write a balanced equation, including states. 2. Convert 8.42 liters of NO at STP to moles of NO. 3. Compare the mole ratios. 4. Convert moles of NO 2 to grams of NO 2.

22. Smog Problem 2NO(g) + O 2 (g)  2NO 2 (g) Since the mole ratio is 1:1, you will form 0.376 moles of nitrogen dioxide.

23. Stoichiometry Hints 1. Balance the equation first and if you can’t balance the equation, check to make sure that your formulas are correct. 2. Anything that will allow you to calculate moles can be used as the basis for solving your problem. 3. Don’t forget to use the proportions given in the balanced equation!!! 4. Take into account if you have a limiting reagent or not.

24. Problems 38, 39 & 41 should be completed at this time. Also start working the rest of the questions (42-48, 52, 54, 56, 63, 66, 68, 69). You may not be able to complete the whole question, but you can do part of it and then finish it later.

25. Limiting Reagents Limiting reagents deter- mine or limit the amount of product that can be formed in a reaction. The reaction occurs until the limiting reagent is used up. The other reactant(s) will be in excess. Whatever is not used up will be the excess reagent.

26. Limiting Reagents The important component is to again convert everything into moles and compare mole ratios to determine which substance is used up and/or what is in excess.

27. Let’s try a limiting reagent problem When sodium reacts with chlorine gas, sodium chloride is produced. Suppose that you react 4.5 moles of sodium with 3.1 moles of chlorine. 1. What is the limiting reagent? 2. How many moles of salt are produced?

28. Sodium Chloride Problem 1. Balanced equation: 2Na + Cl 2  2NaCl 2. Mole ratio is 2 Na to 1 Cl 2. To use all 4.5 moles of sodium it would take ________ moles of chlorine. We have 3.1 moles of Cl 2. To use all 3.1 moles of chlorine it would take ________ moles of sodium. We have 4.5 moles of Na. What is the limiting reagent? 6.2 2.25

29. Sodium Chloride Problem Since the sodium limits the reaction, the number of moles of sodium chloride produced will also be 4.5 moles. How much of the chlorine will be in excess? Since you will use 2.25 moles of chlorine the excess will be 0.85 moles, the difference between what you use and what you have.

30. Just a reminder of possible set up of problems 1. Convert from the known g, L, or atoms to moles of known by dividing by the conversion factor needed. 2. Compare the mole ratios of known and unknown to find the moles of unknown. 3. Convert from the moles of unknown by multiplying by the conversion factor to find g, L or atoms of unknown.

31. Percent Yield  On paper we can predict amounts of reactants and products, this is the theoretical yield of a reaction.  In the lab, reality strikes! Labs do not give 100% yield, the actual yield.  It’s like your correct score on an assignment.

32. Percent Yield What would cause a lab to not have 100% yield? Keep in mind some of the lab processes that we have done involving measurements.

33. Percent Yield It a lot like a batting average or you grade in a class.

34. Percent Yield – Let’s try a problem MMethane combusts with oxygen to produce water and carbon dioxide. WWhat is the theoretical yield of water in this reaction if 23.4 grams of methane reacts in excess oxygen? WWhat is the percent yield if you produce 50.0 grams of water?

35. Percent Yield – Let’s try a problem CH 4 (g) + 2O 2 (g)  2H 2 O(g) + CO 2 (g) 23.4 g  1.46 moles methane 1 methane to 2 waters means 2.92 moles of water 2.92 moles of water  52.6 grams

36. What is the percent error then? You are right, it’s the left over amount. 4.9%

37. Percent Error It’s like the number that you missed on an assignment. It’s the absolute value. Finish the rest of the problems.

38. Work problems p. 262 #42-48, 52, 54, 56, 63, 66, 68, 69

39. Credits http://www.nitrogenorder.org/lessons/s toich.shtml http://www.nitrogenorder.org/lessons/s toich.shtml http://pages.prodigy.com/chemistry/hin ts.htm http://pages.prodigy.com/chemistry/hin ts.htm http://www.chemtutor.com/mols.htm http://staff.chsd218.org/hlr/brodemusc hem/equation.htm http://staff.chsd218.org/hlr/brodemusc hem/equation.htm http://dbhs.wvusd.k12.ca.us/Stoichiom etry/What-is-Stoichiometry.html http://dbhs.wvusd.k12.ca.us/Stoichiom etry/What-is-Stoichiometry.html http://science.widener.edu/svb/pset/sto ichio.html http://science.widener.edu/svb/pset/sto ichio.html

41. The Hamburger

42. First the hamburger analogy My recipe for a bacon double cheeseburger is: 1 hamburger bun 2 hamburger patties 2 slices of cheese 4 strips of bacon Ketchup & mustard

43. Based on this recipe: 1. If I have five bacon double cheeseburgers: a. How many hamburger buns do I have? b. How many hamburger patties do I have? c. How many slices of cheese do I have? d. How many strips of bacon do I have?

44. 1. How many bacon double cheeseburgers can you make if you start with: a. 1 bun, 2 patties, 2 slices of cheese, 4 strips of bacon b. 2 buns, 4 patties, 4 slices of cheese, 8 strips of bacon c. 1 dozen buns, 2 dozen patties, 2 dozen slices of cheese, 4 dozen strips of bacon d. 1 mole bun, 2 mole patties, 2 mole slices of cheese, 4 mole strips of bacon e. 10 buns, 20 patties, 2 slices of cheese, 40 strips of bacon

45. 1. If you had fixings for 100 bacon double cheeseburgers, but when you were cooking you ruined 10 of them. What percentage of the bacon double cheeseburgers do you actually make?