1. Stoichiometry Moles, masses, reactions, oh my!

2. Stoikheion + metria From the Greek for “element” and “the process of measuring” Quantitative analysis of reactants and products Moles Molar masses Balanced equations

3. Grinch Sandwiches Ingredients per sandwich ◦ Three pieces of bread ◦ Sauerkraut (10 g) ◦ Toadstool (15 g) ◦ Arsenic sauce (5 g)

4. Balanced Equations 2H 2 + O 2  2H 2 0 ◦ Coefficients = moles ◦ 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water ◦ What if you only had 1 mole of hydrogen? ◦ What if you had 10 moles of both hydrogen and oxygen? ◦ What if you needed to make 5 moles of water?

5. Law of Conservation of Mass Convert each of the moles in the previous equation to grams Add the reactants’ masses together. Compare to the product’s mass. Be in awe!

6. Mole Ratios Equation must be balanced Relate reactants to each other, products to each other, or reactants to products

7. Molar Masses If mass is mentioned, find that substance’s molar mass

8. Making Stoichiometry Work Dimensional Analysis ◦ Number, unit, substance…always!

9. Problem #1 When 2.50 g of hydrogen gas are mixed with an excess of oxygen gas, how many grams of water should be produced?

10. Problem #1—step 1 Write a balanced equation. ◦ 2H 2 + O 2  2H 2 0

11. Problem #1—step 2 Begin with your given information and make a plan of conversions. ◦ Grams of hydrogen  moles of hydrogen  moles of water  grams of water Write it as dimensional analysis. g H 2 x mol H 2 x mol H 2 O x g H 2 O = g H 2 mol H 2 mol H 2 0

12. Problem #1—step 3 Fill in your numbers. 2.5 g H 2 x 1 mol H 2 x 2 mol H 2 O x 2.02 g H 2 2 mol H 2 18.02 g H 2 O = 1 mol H 2 0 Cross out what divides out. Calculate your answer.

13. Problem #1—step 4 2.5 g H 2 x 1 mol H 2 x 2 mol H 2 O x 2.02 g H 2 2 mol H 2 18.02g H 2 O = 22.3 g H 2 O 1 mol H 2 0 How much oxygen gas would be required?

14. Problem #2 The thermite reaction: aluminum foil reacts with ferric oxide

15. Problem #2 Wow! Now let’s assume that 0.010 g of aluminum reacted with the excess ferric oxide on the ball bearing. What mass of aluminum oxide should be produced?

16. Problem #2—Step 1 Write a balanced equation. 2Al + Fe 2 O 3  Al 2 O 3 + 2Fe

17. Problem #2—Step 2 Write your plan. g Al x mol Al x mol Al 2 O 3 x g Al 2 O 3 = g Al mol Al mol Al 2 O 3

18. Problem #2—Step 3 Place the proper numbers and calculate. 0.010 g Al x 1 mol Al x 1 mol Al 2 O 3 x 26.98 g Al 2 mol Al 101.96 g Al 2 O 3 = 1 mol Al 2 O 3

19. Problem #2—Answer 0.0189 g of Al 2 O 3 should be formed

20. Problem #3 What if only 0.0150 g of Al 2 O 3 were actually produced in the previous problem? What is the percent yield? So, you can calculate the % yield ◦ % yield = actual amt x 100 theoretical amt

21. Problem #3 Thus, the % yield calculation 0.0150 g x 100 = 79.4 % yield 0.0189 g

22. Limiting Reagent Ideally, the perfect amount of each reactant Often one reactant limits how much product can be formed—limiting reagent (LR) Often the other reactant(s) is(are) in excess (ER) You will be given at least 2 quantities of reactants

23. Grinch Sandwiches Ingredients per sandwich ◦ Three pieces of bread ◦ Sauerkraut (10 g) ◦ Toadstool (15 g) ◦ Arsenic sauce (5 g)

24. Problem #4 0.500 g of silver nitrate in solution reacts with 0.750 g of tin (IV) chloride in solution. If 0.400 g of solid silver chloride are retrieved, what is the percent yield?

25. Problem #4—Step 1 Write a balanced equation. 4AgNO 3 + SnCl 4  4AgCl + Sn(NO 3 ) 4 Write shorthand, to save time. You’ll write the real stuff in the answer. ABCD

26. Problem #4—Step 2 Ask yourself, when I have 0.500 g of AgNO 3, how many grams of SnCl 4 do I really need? This will determine the LR. Set up the plan g A x mol A x mol B x g B = g Amol A mol B

27. Problem #4—Step 3 Plug in the numbers. 0.5 g A x 1mol A x 1 mol B 169.88g A 4 mol A x 260.50 g B = 0.192 g SnCl 4 1 mol A

28. Problem #4—Step 4 Analyze and determine the LR ◦ 0.750 g of SnCl 4 are available, and only 0.192 g of it need to be used to completely react with the AgNO 3 Which substance will be in excess? Which substance will you run out of first?

29. Problem #4—Step 5 Now, use the given amount of the LR to find how much product you should be able to make. Set up your plan. g A x mol A x mol C x g C = g A mol A mol C

30. Problem #4—Step 6 Plug in your numbers and calculate. 0.5 g A x 1 mol A x 4 mol C x 169.88 g A 4 mol A 143.32 g C = 0.422 g AgCl should be 1 mol C formed

31. Problem #4—Step 7 Look at the actual amount of product that was formed, and determine percent yield using the theoretical you just calculated. 0.400 g AgCl x 100 = 94.8% yield 0.422 g AgCl

32. Practice Sit down with some bread, sauerkraut, toadstools, and arsenic sauce, and work some stoichiometry problems just for fun.