2. Stoichiometry

3. The Mole: Review A counting unit A counting unit Similar to a dozen, except instead of 12, it’s 602,000,000,000,000,000,000,000 Similar to a dozen, except instead of 12, it’s 602,000,000,000,000,000,000,000 6.02 X 10 23 (in scientific notation) 6.02 X 10 23 (in scientific notation) This number is named in honor of Amedeo Avogadro (1776 – 1856) This number is named in honor of Amedeo Avogadro (1776 – 1856)

4. Review of Molar Mass of Compounds The molar mass (MM) of a compound is determined the same way, except now you add up all the atomic masses for the molecule (or compound) The molar mass (MM) of a compound is determined the same way, except now you add up all the atomic masses for the molecule (or compound) Ex. Molar mass of CaCl 2 Ex. Molar mass of CaCl 2 Avg. Atomic mass of Calcium = 40.08g Avg. Atomic mass of Calcium = 40.08g Avg. Atomic mass of Chlorine = 35.45g Avg. Atomic mass of Chlorine = 35.45g Molar Mass of calcium chloride = 40.08 g/mol Ca + (2 X 35.45) g/mol Cl  110.98 g/mol CaCl 2 Molar Mass of calcium chloride = 40.08 g/mol Ca + (2 X 35.45) g/mol Cl  110.98 g/mol CaCl 2 20 Ca 40.08a 17 Cl 35.45 Cl

5. The Mole Map Mole Mass Gas Volume @ STP # Particles x x x    22.4L 6.02 x 10 23 Molar Mass

6. Chocolate Chip Cookies!! 1 cup butter 1/2 cup white sugar 1 cup packed brown sugar 1 teaspoon vanilla extract 2 eggs 2 1/2 cups all-purpose flour 1 teaspoon baking soda 1 teaspoon salt 2 cups semisweet chocolate chips Makes 3 dozen How many eggs are needed to make 3 dozen cookies? How much butter is needed for the amount of chocolate chips used? How many eggs would we need to make 9 dozen cookies? How much brown sugar would I need if I had 1 ½ cups white sugar?

7. Cookies and Chemistry…Huh!?!? Just like chocolate chip cookies have recipes, chemists have recipes as well Just like chocolate chip cookies have recipes, chemists have recipes as well Instead of calling them recipes, we call them reaction equations Instead of calling them recipes, we call them reaction equations Furthermore, instead of using cups and teaspoons, we use moles Furthermore, instead of using cups and teaspoons, we use moles Lastly, instead of eggs, butter, sugar, etc. we use chemical compounds as ingredients Lastly, instead of eggs, butter, sugar, etc. we use chemical compounds as ingredients

8. Chemistry Recipes Looking at a reaction tells us how much of something you need to react with something else to get a product (like the cookie recipe) Looking at a reaction tells us how much of something you need to react with something else to get a product (like the cookie recipe) Be sure you have a balanced reaction before you start! Be sure you have a balanced reaction before you start! Example: 2 Na + Cl 2  2 NaCl Example: 2 Na + Cl 2  2 NaCl This reaction tells us that by mixing 2 moles of sodium with 1 mole of chlorine we will get 2 moles of sodium chloride This reaction tells us that 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? 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. Write the balanced reaction for hydrogen gas reacting with oxygen gas. 2 H 2 + O 2  2 H 2 O 2 H 2 + O 2  2 H 2 O How many moles of reactants are needed? How many moles of reactants are needed? What if we wanted 4 moles of water? 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 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? What if we had 50 moles of hydrogen, how much oxygen would we need and how much water produced? 2 mol H 2 1 mol O 2 4 mol H 2 2 mol O 2 6 mol H 2, 6 mol H 2 O 25 mol O 2, 50 mol H 2 O

10. Mole Ratios These mole ratios can be used to calculate the moles of one chemical from the given amount of a different chemical These mole ratios can be used to calculate the moles of one chemical from the given amount of a different chemical Example: How many moles of chlorine is needed to react with 5 moles of sodium (without any sodium left over)? Example: How many moles of chlorine is needed to react with 5 moles of sodium (without any sodium left over)? 2 Na + Cl 2  2 NaCl 5 moles Na 1 mol Cl 2 2 mol Na = 2.5 moles Cl 2

11. Mole-Mole Conversions How many moles of sodium chloride will be produced if you react 2.6 moles of chlorine gas with an excess (more than you need) of sodium metal? How many moles of sodium chloride will be produced if you react 2.6 moles of chlorine gas with an excess (more than you need) of sodium metal? 2 Na + Cl 2  2 NaCl

12. Mole-Mass Conversions Most of the time in chemistry, the amounts are given in grams instead of moles Most of the time in chemistry, the amounts are given in grams instead of moles We still go through moles and use the mole ratio, but now we also use molar mass to get to grams We still go through moles and use the mole ratio, but now we also use molar mass to get to grams Example: How many grams of chlorine are required to react completely with 5.00 moles of sodium to produce sodium chloride? Example: How many grams of chlorine are required to react completely 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 = 177g Cl 2

13. Practice Calculate the mass in grams of Iodine required to react completely with 0.50 moles of aluminum. Calculate the mass in grams of Iodine required to react completely with 0.50 moles of aluminum. 2 Al + 3 I 2  2 AlI 3

14. Mass-Mole We can also start with mass and convert to moles of product or another reactant We can also start with mass and convert to moles of product or another reactant We use molar mass and the mole ratio to get to moles of the compound of interest 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 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 2 C 2 H 6 + 7 O 2  4 CO 2 + 6 H 2 0 10.0 g H 2 O 1 mol H 2 O 2 mol C 2 H 6 18.0 g H 2 O 6 mol H 2 0 = 0.185 mol C 2 H 6

15. Practice Calculate how many moles of oxygen are required to make 10.0 g of aluminum oxide Calculate how many moles of oxygen are required to make 10.0 g of aluminum oxide 4 Al + 3 O 2  2 Al 2 O 3 4 Al + 3 O 2  2 Al 2 O 3

16. Mass-Mass Conversions Most often we are given a starting mass and want to find out the mass of a product we will get (called theoretical yield) or how much of another reactant we need to completely react with it (no leftover ingredients!) Most often we are given a starting mass and want to find out the mass of a product we will get (called theoretical yield) or how much of another reactant we need to completely react with it (no leftover ingredients!) Now we must go from grams to moles, mole ratio, and back to grams of compound we are interested in Now we must go from grams to moles, mole ratio, and back to grams of compound we are interested in

17. Mass-Mass Conversion Ex. Calculate how many grams of ammonia are produced when you react 2.00g of nitrogen with excess hydrogen. Ex. Calculate how many grams of ammonia are produced when you react 2.00g of nitrogen with excess hydrogen. N 2 + 3 H 2  2 NH 3 N 2 + 3 H 2  2 NH 3

18. Practice How many grams of calcium nitride are produced when 2.00 g of calcium reacts with an excess of nitrogen? How many grams of calcium nitride are produced when 2.00 g of calcium reacts with an excess of nitrogen? 3 Ca + N 2  Ca 3 N 2 3 Ca + N 2  Ca 3 N 2

19. The Cheese Sandwich Analogy

20. Limiting Reactant: Cookies 1 cup butter 1/2 cup white sugar 1 cup packed brown sugar 1 teaspoon vanilla extract 2 eggs 2 1/2 cups all-purpose flour 1 teaspoon baking soda 1 teaspoon salt 2 cups semisweet chocolate chips Makes 3 dozen If we had the specified amount of ingredients listed, could we make 4 dozen cookies? What if we had 6 eggs and twice as much of everything else, could we make 9 dozen cookies? What if we only had one egg, could we make 3 dozen cookies?

21. Limiting Reactant Most of the time in chemistry we have more of one reactant than we need to completely use up other reactant. Most of the time in chemistry we have more of one reactant than we need to completely use up other reactant. That reactant is said to be in excess (there is too much). That reactant is said to be in excess (there is too much). The other reactant limits how much product we get. Once it runs out, the reaction s. This is called the limiting reactant. The other reactant limits how much product we get. Once it runs out, the reaction s. This is called the limiting reactant.

22. Limiting Reactant To find the correct answer, we have to try all of the reactants. We have to calculate how much of a product we can get from each of the reactants to determine which reactant is the limiting one. To find the correct answer, we have to try all of the reactants. We have to calculate how much of a product we can get from each of the reactants to determine which reactant is the limiting one. The lower amount of a product is the correct answer. The lower amount of a product is the correct answer. The reactant that makes the least amount of product is the limiting reactant. Once you determine the limiting reactant, you should ALWAYS start with it! The reactant that makes the least amount of product is the limiting reactant. Once you determine the limiting reactant, you should ALWAYS start with it! Be sure to pick a product! You can’t compare to see which is greater and which is lower unless the product is the same! Be sure to pick a product! You can’t compare to see which is greater and which is lower unless the product is the same!

23. Limiting Reactant: Example 10.0g of aluminum reacts with 35.0 grams of chlorine gas to produce aluminum chloride. Which reactant is limiting, which is in excess, and how much product is produced? 10.0g of aluminum reacts with 35.0 grams of chlorine gas to produce aluminum chloride. Which reactant is limiting, which is in excess, and how much product is produced? 2 Al + 3 Cl 2  2 AlCl 3 Start with Al: Start with Al: Now Cl 2 : Now Cl 2 : 10.0 g Al 1 mol Al 2 mol AlCl 3 133.5 g AlCl 3 27.0 g Al 2 mol Al 1 mol AlCl 3 = 49.4g AlCl 3 35.0g Cl 2 1 mol Cl 2 2 mol AlCl 3 133.5 g AlCl 3 71.0 g Cl 2 3 mol Cl 2 1 mol AlCl 3 = 43.9g AlCl 3 Limiting Reactant

24. LR Example Continued We get 49.4g of aluminum chloride from the given amount of aluminum, but only 43.9g of aluminum chloride from the given amount of chlorine. Therefore, chlorine is the limiting reactant. Once the 35.0g of chlorine is used up, the reaction comes to a complete. We get 49.4g of aluminum chloride from the given amount of aluminum, but only 43.9g of aluminum chloride from the given amount of chlorine. Therefore, chlorine is the limiting reactant. Once the 35.0g of chlorine is used up, the reaction comes to a complete.

25. Limiting Reactant Practice 15.0 g of potassium reacts with 15.0 g of iodine. Calculate which reactant is limiting and how much product is made. 15.0 g of potassium reacts with 15.0 g of iodine. Calculate which reactant is limiting and how much product is made.

26. Finding the Amount of Excess By calculating the amount of the excess reactant needed to completely react with the limiting reactant, we can subtract that amount from the given amount to find the amount of excess. By calculating the amount of the excess 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 previous problem? Can we find the amount of excess potassium in the previous problem?

27. Finding Excess Practice 15.0 g of potassium reacts with 15.0 g of iodine. 2 K + I 2  2 KI 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, and 19.6 g of potassium iodide are produced. We found that Iodine is the limiting reactant, and 19.6 g of potassium iodide are produced. 15.0 g I 2 1 mol I 2 2 mol K 39.1 g K 254 g I 2 1 mol I 2 1 mol K = 4.62 g K USED! 15.0 g K – 4.62 g K = 10.38 g K EXCESS 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!

28. The Myth! MYTH – A tooth left in a bottle of coca-cola overnight will dissolve. MYTH – A tooth left in a bottle of coca-cola overnight will dissolve. The message behind the myth: The message behind the myth: Colas are bad for you. Colas are bad for you. Coke rots your insides. Coke rots your insides.

29. The Myth! This myth is quite FALSE. Coke cannot come close to dissolving a tooth over night. This myth is quite FALSE. Coke cannot come close to dissolving a tooth over night. Coke (and other sodas) are very acidic, but so are “healthy” drinks, like orange juice. Coke (and other sodas) are very acidic, but so are “healthy” drinks, like orange juice.

30. The Myth! If you leave a tooth, nail, penny, etc. in a (pressurized) bottle of Coke long enough (*many* days or months), the acids dissolve it. If you leave a tooth, nail, penny, etc. in a (pressurized) bottle of Coke long enough (*many* days or months), the acids dissolve it. You could do the same with OJ or vinegar, too. You could do the same with OJ or vinegar, too. Most of us don’t walk around for days with mouthfuls of soda, OJ, or vinegar, though. Most of us don’t walk around for days with mouthfuls of soda, OJ, or vinegar, though.

31. The Myth! Here’s the facts: Here’s the facts: Sodas are less acidic than your stomach acids. Sodas are less acidic than your stomach acids. That’s why drinking them doesn’t hurt you. That’s why drinking them doesn’t hurt you. Drinking large amounts of sugared soda can be a huge source of empty calories, which isn’t healthy. But the soda won’t “eat you alive.” Drinking large amounts of sugared soda can be a huge source of empty calories, which isn’t healthy. But the soda won’t “eat you alive.”

32. From Last Time We are trying to answer: We are trying to answer: If you have 100 g of hydrogen sulfate, how many grams of magnesium hydroxide do you need for a complete chemical reaction? If you have 100 g of hydrogen sulfate, how many grams of magnesium hydroxide do you need for a complete chemical reaction?

33. Summary of Steps (From Last Time) Step 1: Calculate the molecular weight of hydrogen sulfate (98 g/mol) Step 1: Calculate the molecular weight of hydrogen sulfate (98 g/mol) Step 2: Calculate the molecular weight of magnesium hydroxide (58 g/mol) Step 2: Calculate the molecular weight of magnesium hydroxide (58 g/mol) Step 3: Use the balanced equation to figure out how many moles of magnesium hydroxide you need to react with each mole of hydrogen sulfate (1:1) Step 3: Use the balanced equation to figure out how many moles of magnesium hydroxide you need to react with each mole of hydrogen sulfate (1:1) Step 4: Calculate how many moles of hydrogen sulfate we have (100 g = 1.02 moles) Step 4: Calculate how many moles of hydrogen sulfate we have (100 g = 1.02 moles)

34. The Last Steps: Step 5 – figure out how many moles of Mg(OH) 2 we need to totally react with the hydrogen sulfate. Step 5 – figure out how many moles of Mg(OH) 2 we need to totally react with the hydrogen sulfate. Every 1 mole of hydrogen sulfate needs to react with 1 mole of magnesium hydroxide. Every 1 mole of hydrogen sulfate needs to react with 1 mole of magnesium hydroxide. So if we have 1.02 moles of one, we’ll need 1.02 moles of the other. So if we have 1.02 moles of one, we’ll need 1.02 moles of the other.

35. The Last Steps Step 6 – Multiply the number of moles of magnesium hydroxide we need to use with the molecular weight of magnesium hydroxide (58 g/mol). Step 6 – Multiply the number of moles of magnesium hydroxide we need to use with the molecular weight of magnesium hydroxide (58 g/mol). 1.02 mol × 58 g/mol = 59.16 g of Mg(OH) 2. Solution: So if we have 100 g of hydrogen sulfate, and we want it to all be used up in the reaction, we will need to use 56.19 g of magnesium hydroxide. Solution: So if we have 100 g of hydrogen sulfate, and we want it to all be used up in the reaction, we will need to use 56.19 g of magnesium hydroxide.

36. Try It Yourself… One of the substances the reaction produces is water. How many grams of water will be produced if we mix 100 g of hydrogen sulfate and 59.16 g of magnesium hydroxide? One of the substances the reaction produces is water. How many grams of water will be produced if we mix 100 g of hydrogen sulfate and 59.16 g of magnesium hydroxide?

37. Try It Yourself… Did you get it right? The solution is 36.72 g of water. Did you get it right? The solution is 36.72 g of water.

38. Probably Not… If you didn’t figure it out, here’s how to do it: If you didn’t figure it out, here’s how to do it: In the problem, there are 1.02 moles of hydrogen sulfate. In the problem, there are 1.02 moles of hydrogen sulfate. According to the chemical equation, I’ll get two moles of water for every one mole of hydrogen sulfate. According to the chemical equation, I’ll get two moles of water for every one mole of hydrogen sulfate. So if I start with 1.02 moles of hydrogen sulfate, I’ll get 2.04 moles of water. So if I start with 1.02 moles of hydrogen sulfate, I’ll get 2.04 moles of water.

39. Probably Not… Water’s molecular weight is 18 g/mol: Water’s molecular weight is 18 g/mol: One oxygen atom is 16 g/mol One oxygen atom is 16 g/mol Each hydrogen atom is 1 g/mol, and there are two of them. Each hydrogen atom is 1 g/mol, and there are two of them. So 16 + (1 x 2) = 18 g/mol. So 16 + (1 x 2) = 18 g/mol. So: 18 g/mol × 2.04 mol = 36.72 g of water. So: 18 g/mol × 2.04 mol = 36.72 g of water.

40. More Practice Consider this chemical reaction: Consider this chemical reaction: CaCO 3 + HCl CaCl 2 + H 2 O + CO 2 (This is what happens when you take an antacid) Balance this equation. Balance this equation. How much calcium carbonate would you need to “use up” 3.6 g of hydrogen chloride? How much calcium carbonate would you need to “use up” 3.6 g of hydrogen chloride? You have a few minutes to work this out… You have a few minutes to work this out…

41. 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 correct answer. 4. The reactant that gave you the lowest answer is the LIMITING REACTANT. 5. The other reactant(s) are in EXCESS. 6. To find the amount of excess, subtract the amount used from the given amount. 7. 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!