1. Chapter 12 Stoichiometry

2. What is Stoichiometry? Stoichiometry is at the heart of the production of many things you use in your daily life. Soap, tires, fertilizer, gasoline, deodorant, and chocolate bars are just a few commodities you use that are chemically engineered, or produced through chemical reactions. Chemically engineered commodities all rely on stoichiometry for their production.  But what is stoichiometry? Stoichiometry is the calculation of quantities in chemical equations. Given a chemical reaction, stoichiometry tells us what quantity of each reactant we need in order to get enough of our desired product. Because of its real- life applications in chemical engineering as well as research, stoichiometry is one of the most important and fundamental topics in chemistry.

3. Sect. 12.1 What is Stoichiometry?  Mole-Mass Relationships in Chemical Reactions  Chemical reactions STOP when one of the reactants is used up.

4.  Stoichiometry – study of quantitative relationships between amounts of reactants and products in a chemical reaction Based on law of conservation of mass- that matter is neither created nor destroyed in a chemical reaction. Based on law of conservation of mass- that matter is neither created nor destroyed in a chemical reaction.

5.  Chemical bonds in reactants break and new chemical bonds form to produce products.  Mass of the reactants equals the mass of the products.

6.  Stoichiometry and the balanced chemical equation  Coefficients of a balanced equation tell how many representative particles and also how many moles of each substance are involved in the chemical reaction

7. 4 Fe + 3O 2  2 Fe 2 O 3  Describe this reaction in terms of representative particles, moles, and grams

8. Interpreting Chemical Equations in terms of representative particles, moles, and mass Iron + Oxygen  Iron(III) oxide Balanced 4 Fe + 3O 2  2 Fe 2 O 3 Representativeparticles 4 atoms Fe + 3 molecules O 2  2 formula units Fe 2 O 3 Moles 4 moles Fe + 3 moles O 2  2 moles Fe 2 O 3 Mass 223.4 g Fe +96.0 g O 2  319.4 g Fe 2 O 3 Show Masses are equal 319.4 g reactants  319.4 g products law of conservation of mass

9.  The mass of the reactants equals the mass of products as predicted by the law of conservation of mass.

10.  Mole ratio – ratio between the number of moles of any 2 substances in a balanced chemical equation  Mole ratios are the key to calculations based on a chemical equation.

11. 2 Al + 3 Br 2  2 AlBr 3  # of possible ratios = 3 parts of the equation x 2 other parts = 6 mole ratios 2 mol Al2 mol Al3 mol Br 2 2 mol Al2 mol Al3 mol Br 2 3 mol Br 2 2 mol AlBr 3 2 mol Al 3 mol Br 2 2 mol AlBr 3 2 mol Al 3 mol Br 2 2 mol AlBr 3 2 mol AlBr 3 2 mol AlBr 3 2 mol Al3 mol Br 2

12. 12.2 Stoichiometric Calculations A. Mole to Mole conversion:  Step 1: Begin with a balanced chemical equation  Step 2: Identify the substance that you know, and the substance that you need to determine.  Step 3: Write mole ratios  Step 4: Do the mole to mole conversions

13.  Example:  Potassium reacts in water to produce Potassium hydroxide and hydrogen gas.  Determine the moles of hydrogen produced when 0.0400 moles of potassium is used.  Step 1: Write the balanced equation  2K + 2H 2 O  2KOH + H 2

14.  Step 2: Identify the substance that you know, and the substance you need to determine.  2K + 2H 2 O  2KOH + H 2 0.0400 mol K? need to Known determine Unknown  Step 3: Write the mole ratios needed 1 mol H 2 1 mol H 2 2 mol K 2 mol K

15.  Step 4: Mole to mole conversion. We use conversion factors to cancel units.  Moles of known x Moles of unknown = moles of Moles of known unknown  0.0400 mol K x 1 mol H 2 = 0.0200 mol H 2 2 mol K 2 mol K

16. Mole to Mass conversion  Example: What mass of NaCl produced from 1.25 moles of Chlorine reacting with Sodium.  Step 1: Write the balanced chemical equation and identify the known and unknown substances.  2Na + Cl 2  2NaCl  1.25 mols Cl 2 ? g Unknown

17.  Step 2: Write the mole ratio that relates the moles of the 2 substances. 2 mol NaCl 1 mol Cl 2 1 mol Cl 2  Step 3: Multiply the number of moles by the mole ratio. 1.25 mol Cl 2 x 2 mol NaCl = 2.50 mol NaCl 1 mol Cl 2 1 mol Cl 2  Step 4: Multiply mol by the molar mass of the substance 2.50 mol NaCl x 58.44 g NaCl = 146 g NaCl 1 mol NaCl 1 mol NaCl

18. Mass to Mass Conversion  Example:  Determine the mass of water produced from the decomposition of 25.0 g of ammonium nitrate into water and N 2 O gas.  Step 1: Write the balanced chemical equation and identify the known and unknown substances.  NH 4 NO 3  N 2 O + H 2 O  25.0 g Known? Unknown

19.  Step 2: Convert grams to moles. 25.0 g NH 4 NO 3 x 1 mol NH 4 NO 3 = 0.312 mol NH 4 NO 3 80.04 g NH 4 NO 3 Step 3: Determine the mole ratio. 2 mol H 2 O__ 2 mol H 2 O__ 1 mol NH 4 NO 3

20.  Step 4: Multiply mol of substance determined by the mole ratio. 0.312 mol NH 4 NO 3 X 2 mol H 2 O___ = 0.624mol H 2 O 1 mol NH 4 NO 3 1 mol NH 4 NO 3 Step 5: Calculate the mass using molar mass. 0.624 mol H 2 O X 18.02 g H 2 O = 11.2 g H 2 O 1 mol H 2 O 1 mol H 2 O

21.  pg 363 flow chart of steps  Practice Problems: pg 359 9, 10  Pg 360 11,12  Pg 362 13,14

22. Sect. 12.4: % Yield  Calculations are done to determine the success of chemical reactions.  Most reactions never succeed in producing the predicted amount of product.

23. 3. Not every reaction goes cleanly or completely: a. Stop before all reactants are used up. b. Liquids may adhere to the containers or evaporate b. Liquids may adhere to the containers or evaporate c.Solid product may be left on the filter paper or lost in the purification process. d. Products other than the intended ones may form.

24. How Much Product?  Theoretical yield – maximum amount of product that can be produced from a given amount of reactant calculated value…typically mass- mass calculated value…typically mass- mass

25.  Actual yield – amount of product produced when a reaction is carried out must be given in the problem must be given in the problem (found in the experiment)

26.  percent yield – ratio of actual yield to theoretical yield expressed as a percent Shows efficiency of a reaction Shows efficiency of a reaction % yield = actual x 100 % yield = actual x 100 theoretical theoretical

27. 30 g actual yield from the experiment X 100 = 60 g theoretical yield calculated 50% percent yield

28. Limiting Reagent (reactant)  The reactant in a chemical reaction that limits the quantity of a product obtained.

29. Calculating limiting reagent 1. Convert atoms to moles or grams to moles. 2. Calculate the amount of product. Convert moles of reactant to moles of product. 3. Choose the reactant that produces the least amount of product.

30. Excess Reagent  The reactant that is left over after the reaction is complete.

31. Calculating Excess reagent  1. Calculate the # of moles of the excess reactant.  2. Subtract the amount determined in step 1 from the original molar amount present.  3. Change moles to grams if needed.