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Stoichiometry

Stoichiometry. What is included in this lecture?. Review of the mole hill Review of balancing equations Introduction to molarity Finding the amounts of reactants and products based on knowns (stoichiometry) Finding and calculating limiting reagents Calculating % yield.

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Stoichiometry

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  1. Stoichiometry

  2. What is included in this lecture? • Review of the mole hill • Review of balancing equations • Introduction to molarity • Finding the amounts of reactants and products based on knowns (stoichiometry) • Finding and calculating limiting reagents • Calculating % yield

  3. Remember the Mole Hill! Divide up, multiply down. Notice we have added a new piece to the mole hill. More about that later.

  4. Remember how to write and balance reactions! Write and balance the following reaction: Copper (II) sulfate + iron  iron (III) sulfate + copper Cu+2 SO4-2= CuSO4 Fe+3 SO4-2 = Fe2(SO4)3 CuSO4 + Fe  Fe2(SO4)3 + Cu 3 2 3 Cu= 1 SO4 = 1 Fe = 1 3 Cu= 1 SO4 = 3 Fe = 2 x 3 • Try these on your own: • hydrochloric acid and sodium hydroxide combine to make salt and water. • aluminum oxide is heated to form aluminum and oxygen. • the steel on your bike is left outside in the rain and forms iron (II) oxide and hydrogen. x 3 x 2

  5. Molarity • Molarity = M • M = mol/L (#mol ÷ #L) Molarity (M) = moles of solute 1 liter solution

  6. Units of Molarity 2.0 M HCl = 2.0 moles HCl 1 L HCl solution 6.0 M HCl= 6.0 moles HCl 1 L HCl solution

  7. Molarity Calculations NaOH is used to open stopped sinks, to treat cellulose in the making of nylon, and to remove potato peels commercially. If 4.0 g NaOH are used to make 500. mL of NaOH solution, what is the molarity (M) of the solution?

  8. M = mol ÷ L Known: 4.0 g of NaOH (mass!!) MW of NaOH = 22.99 g/mol Na + 16.00 g/mol O + 1.01 g/mol H = 40.00 g/mol 500 ml solution 1L = 1000 ml 1L/1000ml x 500 ml = 0.5L (this is unit analysis!!) Plug what you know into the mole hill! Divide up! 4.0g ÷ 40.00 g/mol = 0.10 molNaOH Now you can plug it into the molarity formula: M = 0.10 mol ÷ 0.5 L Molarity = 0.2 M 40.00 g/mol 4.0 g

  9. Learning Check A KOH solution with a volume of 400 mL contains 2 mole KOH. What is the molarity of the solution? 1) 8 M 2) 5 M 3) 2 M

  10. Learning Check A glucose solution with a volume of 2.0 L contains 72 g glucose (C6H12O6). If glucose has a molar mass of 180. g/mole, what is the molarity of the glucose solution? 1) 0.20 M 2) 5.0 M 3) 36 M

  11. Learning Check How many grams of NaOH are required to prepare 400. mL of 3.0 MNaOH solution? 1) 12 g 2) 48 g 3) 300 g

  12. Stoichiometry – The Math Behind Chemical Reactions We will be using this form to complete stoichiometry problems. You may use this form at any time. The following slides will walk you through how to fill it out.

  13. What is stoichiometry? • The study of the relationships between amounts of products and reactants. • Imagine the following reaction: 2 slices of bread + 1 slice of ham + 1 slice of cheese  1 ham sandwich No matter how you put the 4 things together, you only get one sandwich!

  14. Chemistry is the same… • 2H2 + O2 2 H2O • If we want to make 2 moles of water, we will have to start with 2 moles of hydrogen and 1 mole of oxygen. Remember: Coefficients are the number of MOLES you have of each chemical in the reaction! You get the number of moles from a balanced reaction. • What if we want to make 4 moles of water? You only have 2 moles of water in the reaction. • Fortunately, this doesn’t require you learn anything new…except maybe filling out that chart from a few slides ago.

  15. How do you make 4 moles of water from hydrogen and oxygen? Notice that I only filled out the information I needed…unless the problem tells you to fill in the entire chart, only use what you need! Hydrogen Oxygen Water Multiply down… 2 x 2 = 4 2H2 O2 2H2O Multiply down… 1 x 2 = 2 Divide up… 4 ÷ 2 = 2 Multiply down… 2 mol x 32.00 g/mol = 64g Multiply down… 4 mol x 2.02 g/mol = 8.08 g 4 mol 2 mol 4 mol 8.08 g 64g Answer: react 8.08 g of hydrogen gas with 32 g of oxygen gas to get 4 mols of water. 2.02 g/mol 32 g/mol

  16. Let’s try another one… Fill in the entire chart to make NaCl from 12 grams of sodium. sodium chlorine sodium chloride We don’t need these lines since there are only 3 things in the reaction. 2Na Cl2 2NaCl 0.52 mol 1. Divide up the mole hill… 12g ÷ 22.99 g/mol 2. Multiply down the mole hill… 0.52 mol x 6.0221023mia/mol 12 g solid 3.131023 atoms 22.99 g/mol 70.90 g/mol 58.44 g/mol

  17. Let’s try another one… Fill in the entire chart to make NaCl from 12 grams of sodium. sodium chlorine sodium chloride 3. Divide up… 0.52 ÷ 2 = 0.26 2Na Cl2 2NaCl 4. Multiply down… 0.26 x 1 = 0.26 mol 5. Multiply down the mole hill… 0.26 mol x 70.90 g/mol 0.52 mol 0.26 mol 6. Multiply down the mole hill… 0.26 mol x 22.414 L/mol 7. Multiply down the mole hill… 0.26 mol x 6.022  1023mia/mol 12 g 18.43 g solid 4.79 L 3.131023 atoms 1.571023 molecules 22.99 g/mol 70.90 g/mol 58.44 g/mol

  18. Let’s try another one… Fill in the entire chart to make NaCl from 12 grams of sodium. sodium chlorine sodium chloride Divide up… 0.26 ÷ 1 = 0.26 Multiply down… 0.26 x 2 = 0.52 2Na Cl2 2NaCl 0.52 mol Multiply down the mole hill… 0.52 mol x 58.44 g/mol 0.26 mol 0.52mol 30.39 g Multiply down the mole hill… 0.52 mol x 6.0221023mia/mol 12 g 18.43 g solid 4.79 L solid 3.131023 atoms 1.571023 molecules 3.131023 molecules 22.99 g/mol 70.90 g/mol 58.44 g/mol

  19. A completed stoichiometry chart… Fill in the entire chart to make NaCl from 12 grams of sodium. sodium chlorine sodium chloride 2Na Cl2 2NaCl 0.52 mol 0.26 mol 0.52mol 30.39 g 12 g 18.43 g solid 4.79 L solid 3.131023 atoms 1.571023 molecules 3.131023 molecules 22.99 g/mol 70.90 g/mol 58.44 g/mol

  20. Try one yourself… (This will be solved on the voiced version of the lecture.) Fill in the chart for the reaction of 5 g of carbon tetrahydride with excess oxygen to make carbon dioxide and water.

  21. Finding the Limiting Reagent Your reaction is limited by how much you have. If you do not have enough of one of the chemicals, you can only run the reaction until you run out of the chemical that has the smallest amount.

  22. Limiting Reagent Problems • Do the stoichiometry the same way you would for any other problem. • Balance chemical equation • Determine limiting reagent • Do two separate calculations on the stoichiometry charts for the amount of product each reactant would produce if they were the limiting reagent • The reactant that gives the lower number of moles is the limiting reagent. • The amount of product produced is the number calculated by limiting reagent

  23. Limiting Reagent Problems (cont) 5. Determination of the amount of excess reagent left over • Calculate the amount of excess reagent (ER) used in chemical reaction • Subtract the ER used from original amount of ER.

  24. Determine the limiting reagent when… you react 3g of hydrogen with 3g of oxygen to make water. Without doing anything else, I know that oxygen is the limiting reagent…there are only 0.19 mol of water made from the oxygen, as opposed to the 1.49 mol of water made from the hydrogen. You will run out of oxygen before you run out of hydrogen. hydrogen oxygen water 2 H2 O2 2H2O 1.49 mol 0.19 mol 1.49 mol 0.09 mol 3g 3g  3.39g can be made from the given reactants. 32.00 g/mol 18.02 g/mol 2.02 g/mol

  25. Calculating % Yield • Theoretical Yield: the amount of product you should get based on the stoichiometry chart. • Actual Yield: the amount of product you actually get if you were running the reaction in real life. • % Yield = actual ÷ theoretical x 100.

  26. Why is % yield important? • It comes from the fact that we don’t live in a perfect world… • Human error plays a huge part of our reactions. • We lose reactants to the air, to the sides of the container, and to anything we stir with. • We may measure amounts incorrectly. • There may be impurities in the chemicals you are using.

  27. If you start with 5 mol of aluminum chloride and react it with sodium iodide, you get 285.4g of sodium chloride and some aluminum iodide. What is the % yield of the sodium chloride? aluminum chloride sodium iodide aluminum iodide sodium chloride AlCl3 3 NaI AlI3 3NaCl 5 mol 15 mol 292.2 g Notice that I ignored everything else in the chart…it isn’t needed in this particular problem. Filling out the chart gives you the theoretical yield = 292.2g. The actual yield is given in the problem, actual yield = 285.4g. % Yield = 285.4g ÷ 292.2g x 100 = 97.67% Note: you may need to figure out limiting reagent BEFORE solving for percent yield if you are given amounts for all of the reactants. 58.44 g/mol

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