1. Chapter 9 Stoichiometry Pages 298 - 318

2. Intro to Stoichiometry All stoichiometric calculations start with a __________________________. To solve, you will need a given amount which may be given in moles or grams. Then you will need a conversion factor. You will need to be proficient at setting up a dimensional analysis problem to correctly calculate a quantity. A ______________ is a conversion factor that relates the amounts in moles of any two substances involved in a chemical reaction. EXAMPLE) page 300. 2Al 2 O 3 (l)  4Al(s) + 3O 2 (g) **remember** the coefficient in front of each substance indicates the number of moles present for each substance in the chemical reaction **remember** the coefficient in front of each substance indicates the number of moles present for each substance in the chemical reaction

3. Intro to Stoichiometry What are the SIX possible mole ratios for the above chemical equation? 1 & 2: The ratio between a single reactant (2Al 2 O 3 – Aluminum oxide) and one of two reactants (4Al or 3O 2 ). Let’s take the reactant and the product aluminum metal first. The mole ratios would be: 2 mole Al2O3 OR 4 mole Al 4 mole Al 2 mole Al 2 O 3 **remember** since a mole ratio is a conversion factor, it can be “flipped” as demonstrated above, depending on what quantity you are calculating. Now you take a few moments to determine the other FOUR mole ratios that can be determined from the chemical equation.

4. Intro to Stoichiometry PRACTICE: Determine the amount in moles of aluminum that can be produced from 13.0 mol of aluminum oxide (chemical formula?). What mole ratio will be needed to make this determination? Note that aluminum and aluminum oxide are the two substances that are being considered in this stoichiometric calculation. This means that we need the mole ratio between these two substances. So our conversion factor then will be: 4 mol AlOR2 mol Al 2 O 3 2 mol Al 2 O 3 4 mol Al But which of the two ratios will be used? To determine this, we need to examine the given quantity. In this case, our given is 13.0 mol Al 2 O 3. Now what?? DIMENSIONAL ANALYSIS TIME!!

5. Intro to Stoichiometry 13.0 mol Al 2 O 3 This is our GIVEN 1 **remember** our conversion factor must be set up so that the units in our given are canceled so units in the conversion factor are on the “bottom” or in the “denominator”. 13.0 mol Al2O3X 4 mol Al 1 2 mol Al 2 O 3 Note what units are canceled and what units are left after canceling as a result of dimensional analysis. What is the solution to this stoichiometric calculation?

6. Intro to Stoichiometry Molar mass is the mass, in grams, of one__________________. Molar mass is a CONVERSION FACTOR. What is the molar mass of one mole of aluminum oxide (Al 2 O 3 )? Of one mole of aluminum metal (Al)? Of one mole of oxygen gas (O 2 )? PRACTICE: Take a few minutes to calculate these molar masses. Ready for a pop quiz? Your results? Now take a few minutes to find the number of grams of aluminum equivalent to 26.0 mol of aluminum.

7. Intro to Stoichiometry Start with GIVEN. Use your conversion factor. Cancel units accordingly. Make your calculation. Present your result (don’t forget units!). I’M GRADING YOU ON THIS CALCULATION! PLEASE MAKE AN ATTEMPT. OR TAKE A 1…YOUR CHOICE. FILE YOUR CALCULATION IN YOUR BELLWORK. I WILL CHECK NEXT NOTEBOOK CHECK.

8. Ideal Stoichiometric Calculations The _____________ chemical equation is the key step in all stoichiometric calculations because the mole ratio is obtained directly from it. Under ideal conditions, all reactants are converted into ___________. Solving these types of calculations requires PRACTICE! PRACTICE: In a spacecraft, the carbon dioxide (CO 2 ) exhaled by astronauts can be removed by its reaction with lithium hydroxide (LiOH) according to the following chemical equation: CO 2 (g) + 2LiOH(s)  Li 2 CO 3 (s) + H 2 O(l) Is the above equation BALANCED?? What happens if carbon dioxide gas (CO 2 ) builds up in a spacecraft of limited oxygen gas (O 2 )?

9. Ideal Stoichiometric Calculations PROBLEM STATEMENT: How many “moles” of lithium hydroxide (LiOH) are required to react with 20 mol CO 2 – which is the average amount exhaled by a person in a day. GIVEN: amount of CO 2 = 20 moles UNKNOWN: amount of LiOH in moles WHAT IS NEEDED TO MAKE THIS CALCULATION? Since your unknown is asking for your calculation to be in “moles” then you only need a mole ratio! So then, what mole ratio is needed? This will be your CONVERSION FACTOR. So, set up your calculation with given first over 1 and your conversion factor which is your mole ratio. You have a few minutes. Do this. You will be GRADED on this calculation. File your calculation with your bellwork – I will check this on the next notebook check.

10. Ideal Stoichiometric Calculations 20 mol CO 2 This is our GIVEN 1 What is our conversion factor? The mole ratio 2 mol LiOH - is our conversion factor! 1 mol CO 2 CALCULATION: 20 mol CO 2 X 2 mol LiOH 1 1 mol CO 2 AND? What are the results of your calculations? This is graded and to be kept with your bellwork for the next notebook check.

11. Ideal Stoichiometric Calculations CHEMISTRY APPLICATION: In photosynthesis, autotrophs use energy from the sun to produce glucose (C 6 H 12 O 6 ) and oxygen gas (O 2 ) from the reaction of carbon dioxide (CO 2 ) and water (H 2 O). PROBLEM STATEMENT: What mass, in grams, of “glucose” is produced when 3.00 mol of “water” reacts with carbon dioxide? GIVEN: amount of H 2 O = 3.00 mol UNKNOWN: mass of C 6 H 12 O 6 produced in grams So what we need is a balanced equation: CO 2 (g) + H 2 O(l)  C 6 H 12 O 6 (s) + O 2 (g) Okay, but is this balanced – does this formula equation satisfy the law of conservation of mass?

12. Ideal Stoichiometric Calculations In a word – NOPE! 6CO 2 (g) + 6H 2 O(l)  C 6 H 12 O 6 (s) + 6O 2 (g) Now are we balanced??? We will start will our given. Then we will use two conversion factors. Any time we need to go from moles to grams, we need TWO CONVERSION FACTORS – mole ratio and molar mass – in that order. 3.00 mol H 2 O X 1 mol C6H12O6 X 180.18 g C 6 H 12 O 6 = ? 1 6 mol H 2 O 1 mol C 6 H 12 O 6 Make your calculations and keep your work with your bellwork. I will check for completion and accuracy on the next notebook check!

13. Ideal Stoichiometric Calculations

14. 3.00 mol H2Othis is your GIVEN. 1 What are your conversion factors? Perhaps a mole ratio? Perhaps a molar mass? Perhaps both? 3.00 mol H2O X 6 mol CO 2 X 44.01 g CO 2 = ? 1 6 mol H2O 1 mol CO 2 What is your calculated value? Keep with your bellwork – I will check during the next notebook check. What conversion factors were used in this calculation?

15. Ideal Stoichiometric Calculations Now lets practice converting mass amounts to mole amounts. We are going to need two conversion factors, again, and they are the molar mass and the mole ratio. CHEMISTRY APPLICATION: The first step in the industrial manufacturing of nitric acid is the catalytic oxidation of ammonia. The formula equation is as follows: NH 3 (g) + O 2 (g)  NO(g) + H 2 O(g) Is this equation balanced?? Take a few minutes to attempt to balance this equation IF NEEDED………

16. Ideal Stoichiometric Calculations 4NH 3 (g) + 5O 2 (g)  4NO(g) + 6H 2 O(g) Now are we balanced? This reaction is run using 824g of NH 3 with excess oxygen. Now that we have a balanced equation and a GIVEN quantity, what is the amount of NO produced in moles? And the amount of H 2 O produced in moles? 824 g NH 3 X 1 mol NH 3 X 4 mol NO= 48.4 mol NO 1 17.04 g NH 3 4 mol NH 3 1 17.04 g NH 3 4 mol NH 3 Notice! What happens to identical units when they are found in both the numerator and the denominator in a dimensional analysis calculation? What about the amount of moles of H 2 O produced from the reaction if given 824g of NH 3 ? SEE NEXT SLIDE……………

17. Ideal Stoichiometric Calculations 4NH 3 (g) + 5O 2 (g)  4NO(g) + 6H 2 O(g) 824 g NH3X 1 mol NH 3 X 6 mol H 2 O = ? 1 17 g NH 3 4 mol NH 3 First note the units that are canceled in the dimensional analysis procedure. Second, make your calculations. Where is division? Where is multiplication? Finally, what is your result? CALCULATION: Take 824 X 6 = A Then take 17 X 4 = B Finally divide A by B. A/B = your answer! It should equal 72.5 mol H 2 O

18. Limiting Reactants and Percent Yield The ______________ is the reactant that limits the amount of the other reactant that can combine and the amount of product that can form in a chemical reaction. The substance that is not used up completely in a reaction is called the _____________. Chemical reactions are not always ideal. Most times, one or another reactant is used up during the reaction so it cannot proceed to producing the products. When calculating limiting reactant, you calculate the amount of substance (can be in moles or grams) that is produced from each reactant. Which ever of the two REACTANT QUANTITIES is least in value is your limiting reactant and the lesser amount of product is your theoretical yield.

19. Limiting Reactants and Percent Yield SiO 2 (s) + 4HF(g)  SiF 4 (g) + 2H 2 O(l) PROBLEM STATEMENT: If 6.0 mol HF is added to 4.5 mol SiO 2, which is the limiting reactant? 1 st step: determine which of the products contains part of BOTH reactants. In this case, SiF 4 has both SiO 2 & HF so it is the product that you want to “run” with your stoichiometric operation. 2 nd step: pick one of the products and begin calculating. Your conversion factor is mole ratio because you are given moles. Let’s pick SiO 2 first. What mole ratio will we need? What is our given? 3 rd step: dimensional analysis – don’t forget units.

20. Limiting Reactants and Percent Yield SiO 2 (s) + 4HF(g)  SiF 4 (g) + 2H 2 O(l) Take a few minutes to attempt this calculation. Remember, its only a mole-to-mole calculation! We are starting with given moles of SiO 2 and ending with unknown moles of SiF 4. (HINT) GIVEN = 4.5 mol SiO 2 UNKNOWN = mole ratio needed? UNKNOWN = limiting reactant THIS WILL BE CHECKED AND GRADED DURING NOTEBOOK CHECK.

21. Limiting Reactants and Percent Yield SiO 2 (s) + 4HF(g)  SiF 4 (g) + 2H 2 O(l) CALCULATION: 4.5 mol SiO 2 X 1 mol SiF 4 = 4.5 mol SiF 4 1 1 mol SiO 2 1 1 mol SiO 2 Now lets run the other reactant: 6.0 mol HF X 1 mol SiF 4 = 1.5 mol SiF 4 1 4 mol HF 1 4 mol HF So from running both reactants to the desired product, we can see that the limiting reactant in this reaction is HF because it produces fewer moles of desired product (SiF 4 ) than SiO 2

22. Limiting Reactants and Percent Yield To calculate a percent yield, you must first calculate a theoretical yield from a balanced equation, a given quantity and an unknown (or desired) unit of substance. CHEMISTRY APPLICATION: Chlorobenzene, C 6 H 5 Cl, is used in the production of many important chemicals, such as aspirin, dyes, and disinfectants. One industrial method of preparing chlorobenzene is to react benzene, C 6 H 6 with chlorine gas. C 6 H 6 (l) + Cl 2 (g)  C 6 H 5 Cl(l) + HCl(g) First of all, are we balanced??

23. Limiting Reactants and Percent Yield C 6 H 6 (l) + Cl 2 (g)  C 6 H 5 Cl(l) + HCl(g) PROBLEM STATEMENT: When 36.8 g C 6 H 6 react with an excess of Cl 2, the actual yield of C 6 H 5 Cl is 38.8 g. What is the percentage yield of C 6 H 5 Cl? We are doing a mass-to-mass conversion because theoretical yield must be converted from moles to mass. This will require 3 conversions: first a molar mass factor, second a mole ratio factor and finally another molar mass factor. GIVEN: 36.8 g C 6 H 6

24. Limiting Reactants and Percent Yield C 6 H 6 (l) + Cl 2 (g)  C 6 H 5 Cl(l) + HCl(g) CONVERSION FACTORS: molar mass of C 6 H 6 = _________________ molar mass of C 6 H 6 = _________________ mole ratio between C 6 H 5 Cl and C 6 H 6 = ______________________ molar mass of C 6 H 5 Cl = _______________________ UNKNOWN: mass of C 6 H 5 Cl produced from given mass of C 6 H 6 TAKE A FEW MINUTES TO ATTEMPT THIS CALCULATION. THIS WILL BE CHECKED…

25. Limiting Reactants and Percent Yield CALCULATION: 36.8 g C 6 H 6 X 1 mol C 6 H 6 X 1 mol C6H5Cl X 112.6 g C 6 H 5 Cl = ? 1 78.12 g C 6 H 6 1 mol C 6 H 6 1 mol C 6 H 5 Cl 1 78.12 g C 6 H 6 1 mol C 6 H 6 1 mol C 6 H 5 Cl 4143.68 = 78.12 ` 53.0 g C6H5Cl = theoretical yield. % YIELD = Actual yield X 100% = 38.8 g X 100% = ? Theoretical yield 53.0 g Theoretical yield 53.0 g