1. STOICHIOMETRY USING THE REACTION EQUATION LIKE A RECIPE

2. Cool word, but it’s all Greek to me! Stoicheion = element Metron = measure STOICHIOMETRY…measuring and calculating the chemical elements YOU WILL NEED YOUR CALCULATOR EVERY DAY!

3. Nearly everything we use is manufactured from chemicals. –Soaps, shampoos, conditioners, cd’s, cosmetics, medications, and clothes. For a manufacturer to make a profit the cost of making any of these items can’t be more than the money paid for them. Chemical processes carried out in industry must be economical, this is where balanced equations help. USING EQUATIONS

4. Equations are a chemist’s recipe. –Eqs tell chemists what amounts of reactants to mix and what amounts of products to expect. When you know the quantity of one substance in a rxn, you can calculate the quantity of any other substance consumed or created in the rxn. –Quantity meaning the amount of a substance in grams, liters, molecules, formula units, or moles. USING EQUATIONS

5. The calculation of quantities in chem- ical reactions is called stoichiometry. Imagine you are in charge of manu- facturing for Rugged Rider Bicycle Company. The business plan for Rugged Rider requires the production of 128 custom-made bikes each day. You are responsible for insuring that there are enough parts at the start of each day. USING EQUATIONS

6. Assume that the major components of the bike are the frame (F), the seat (S), the wheels (W), the handlebars (H), and the pedals (P). The finished bike has a “formula” of FSW 2 HP 2. The balanced equation for the production of 1 bike is. USING EQUATIONS F +S+2W+H+2P  FSW 2 HP 2

8. Now in a 5 day workweek, Rugged Riders is scheduled to make 640 bikes. How many wheels should be in the plant on Monday morning to make these bikes? What do we know? –Number of bikes = 640 bikes –1 FSW 2 HP 2 =2W (balanced eqn) What is unknown? –# of wheels = ? wheels USING EQUATIONS

9. The connection between wheels and bikes is 2 wheels per bike. We can use this information as a conversion factor to do the calculation. 640 FSW 2 HP 2 1 FSW 2 HP 2 2 W = 1280 wheels.

10. We can make the same kinds of connections from a chemical rxn eqn. N 2 (g) + 3H 2 (g)  2NH 3 (g) The key is the “coefficient ratio”.

11. COEFFICIENT RATIOS or MOLE RATIOS N 2 (g) + 3H 2 (g)  2NH 3 (g) 1 mol N2 or 3 mol H2 3 mol H2 1 mol N2 1 mol N2 or 2 mol NH3 2 mol NH3 1 mol N2 3 mol H2 or 2 mol NH3 2 mol NH3 3 mol H2

12. –The coefficients of the balanced chemical equation indicate the numbers of moles of reactants and products in a chemical rxn. 1 mole of N 2 reacts with 3 moles of H 2 to produce 2 moles of NH 3. –N 2 and H 2 will always react to form ammonia in this 1:3:2 ratio of moles. So if you started with 10 moles of N 2 it would take 30 moles of H 2 and would produce 20 moles of NH 3

13. Using the coefficients, from the balan- ced rxn equation to make connections between reactants and products, is the most important information that a rxn equation provides. –Using this information, you can calculate the amounts of the reactants involved and the amount of product you might expect. –Any calculation done with the next process is a theoretical number, the real world isn’t always perfect.

14. Using the coefficients of balanced rxn equations and our knowledge of mole conversions we can perform powerful calculations. A.K.A. stoichiometry. A balanced rxn equation is essential for all calculations involving amounts of reactants and products. –If you know the number of moles of 1 substance, the balanced eqn allows you to calc. the number of moles of all other substances in a rxn equation.

15. MOLE – MOLE EXAMPLE The following rxn shows the synthesis of aluminum oxide. 3O 2 (g) + 4Al(s)  2Al 2 O 3 (s) If you only had 1.8 mols of Al how much product could you make? Given: 1.8 moles of Al Uknown: ____ moles of Al 2 O 3 3O 2 (g) + 4Al(s)  2Al 2 O 3 (s)

16. Solve for the unknown: 1.8 mol Al 4 mol Al 2 mol Al 2 O 3 = 0.90mol Al 2 O 3 MOLE – MOLE EXAMPLE 3O 2 (g) + 4Al(s)  2Al 2 O 3 (s) Mole Ratio

17. MOLE – MOLE EXAMPLE 2 The following rxn shows the synthesis of aluminum oxide. 3O 2 (g) + 4Al(s)  2Al 2 O 3 (s) If you wanted to produce 24 mols of product how many mols of each reactant would you need? Given: 24 moles of Al 2 O 3 Uknown: ____ moles of Al ____ moles of O 2

18. Solve for the unknowns: 24 mol Al 2 O 3 2 mol Al 2 O 3 4 mol Al = 48 mol Al MOLE – MOLE EXAMPLE 2 3O 2 (g) + 4Al(s)  2Al 2 O 3 (s) 24 mol Al 2 O 3 2 mol Al 2 O 3 3 mol O 2 = 36 mol O 2

19. Cross the Mole-Mole Bridge With thanks to Rossini, Willam Tell, and the Lone Ranger William Tell Overture Guitar Hiho, Stoichiometry...Away! Backwards

21. MASS – MASS CALCULAT’NS No lab balance measures moles directly, generally mass is the unit of choice. From the mass of 1 reactant or prod- uct, the mass of any other reactant or product in a given chemical equation can be calculated, provided you have a balanced rxn equation. As in mole-mole calcs, the unknown can be either a reactant or a product.

22. Acetylene gas (C 2 H 2 ) is produced by adding water to calcium carbide (CaC 2 ). How many grams of C 2 H 2 are produced by adding water to 5.00 g CaC 2 ? CaC 2 + 2H 2 O  C 2 H 2 + Ca(OH) 2 MASS – MASS CALCULAT’NS 1

23. What do we know? –Given mass = 5.0 g CaC 2 –Mole ratio: 1 mol CaC 2 = 1 mol C 2 H 2 –MM of CaC 2 = 64 g CaC 2 –MM of C 2 H 2 = 26 g C 2 H 2 What are we asked for? –grams of C 2 H 2 produced MASS – MASS CALCULAT’NS 1

24. mass A  moles A  moles B  mass B 5.0 g CaC 2 64 g CaC 2 1 mol CaC 2 MASS – MASS CALCULAT’NS 1 1mol CaC 2 1 mol C 2 H 2 26 g C 2 H 2 = 2.0 g C 2 H 2

25. You’ve recently learned that Copper will replace silver ions out of solution. You’re eyes light up with this money making opportunity. However, you decide it might be best if you did some preliminary calculations to determine to the feasibility of this get rich scheme. Copper is not very hard to find, however the largest size of Silver nitrate found in the Flinn Catalog is the 500 g size and it costs $305.91. Currently Silver sells for $9.00/ounce on the stock market. How much money could you sell your manufactured Silver for? MASS – MASS CALCULAT’NS 2

26. What do we know? – Given mass = 500. g of AgNO 3 – Mole ratio: 2 mol AgNO 3 = 2 mol Ag – MM of AgNO 3 : 169.84g = 1mol – MM of Ag: 107.87 g = 1mol – Price of Silver: $9.00 = 1 ounce – Conversion g to oz: 28.23g = 1 oz MASS – MASS CALCULAT’NS 2 Cu + 2AgNO 3  2Ag + Cu(NO 3 ) 2

27. 500. g AgNO 3 169.84gAgNO 3 1mol AgNO 3 MASS – MASS CALCULAT’NS 2 2 mol AgNO 3 2 mol Ag 1mol Ag 107.87g Ag = $101.26 =$101 28.23 g 1 oz $9.00

28. A balanced reaction equation indicates the relative numbers of moles of reactants and products. We can expand our stoichiometric calculations to include any unit of measure that is related to the mole. The given quantity can be expressed in numbers of particles, units of mass, or volumes of gases at STP. The problems can include mass- volume, volume-volume, and particle- mass calculations.

29. In any of these problems, the given quantity is first converted to moles. Then the mole ratio from the balanced eqn is used to convert from the moles of given to the number of moles of the unknown Then the moles of the unknown are converted to the units that the problem requests. The next slide summarizes these steps for all typical stoichiometric problems

31. MORE MOLE EXAMPLES How many molecules of O 2 are produced when a sample of 29.2 g of H 2 O is decomposed by electrolysis according to this balanced equation: 2H 2 O  2H 2 + O 2

32. What do we know? –Mass of H 2 O = 29.2 g H 2 O –2 mol H 2 O = 1 mol O 2 (from balanced equation) –MM of H 2 O = 18.0 g H 2 O –1 mol O 2 = 6.02x10 23 molecules of O 2 What are we asked for? – molecules of O 2 MORE MOLE EXAMPLES

33. mass A  mols A  mols B  molecules B 29.2 g H 2 O 18.0 g H 2 O 1 mol H 2 O 2 mol H 2 O 1 mol O 2 6.02x10 23 molecules O 2 = 4.88 x 10 23 molecules O 2

34. The last step in the production of nitric acid is the reaction of NO 2 with H 2 O. 3NO 2 +H 2 O  2HNO 3 +NO How many liters of NO 2 must react with water to produce 5.0x10 22 molecules of NO? MORE MOLE EXAMPLES

35. What do we know? –Molecules NO = 5.0x10 22 molecules NO –1 mol NO = 3 mol NO 2 (from balanced equation) –1 mol NO = 6.02x10 23 molecules NO –1 mol NO 2 = 22.4 L NO 2 What are we asked for? – Liters of NO 2 MORE MOLE EXAMPLES

36. molecules A  mols  mols B  volume B 5.0x10 22 mol- ecules NO 6.02x10 23 mol- ecules NO 1 mol NO 3 mol NO 2 = 5.6 L NO 2 1 mol NO 2 22.4 L NO 2