Chapter 9
Stoichiometry

1. Introduction to Stoichiometry
   
   
   1. Composition stoichiometry - definition
      
   2. Reaction stoichiometry - definition
      
      
2. Reaction-Stoichiometry Problems
   
   
   1. use ratios from balanced equations
      
   2. mole-mole problems
      amount given is in moles, amount unknown and to find in moles
      
   3. mole-mass problems
      amount given is in moles, amount unknown and to find in grams
      
   4. mass-mole problems
      amount given is in grams, amount unknown and to find in moles
      
   5. mass-mass problems
      amount given is in grams, amount unknown and to find in grams
      
      
      1. Mole Ratio
         
         
         1. mole ratio - definition
            
         2. for the reaction 2 Al₂O₃ ---> 4 Al + 3 O₂
            
            2 moles of aluminum oxide produces 4 moles of aluminum and 3 moles of oxygen OR
            
            2 moles of aluminum oxide produces 4 moles of aluminum i.e. 2 mol Al₂O₃/ 4 mol Al OR
            
            2 moles of aluminum oxide produces 3 moles of oxygen
            i.e. 2 mol Al₂O₃/ 3 mol O₂
            
            4 moles of aluminum is produced with 3 moles of oxygen
            i.e. 4 mol Al / 3 mol O₂
            
         3. Example: How many moles of aluminum can be produced from 13.0 mol of aluminum oxide according to the balanced chemical equation: 2 Al₂O₃ ---> 4 Al + 3 O₂
            
            Solution: relationship between aluminum and aluminum oxide is 4 mol Al / 2 mol Al₂O₃
            
            Aluminum is in numerator because you are asked how many moles of aluminum can be produced i.e. it is the unknown for which you are solving
            
            13.0 mol Al₂O₃ x 4 mol Al / 2 mol Al₂O₃ = 26.0 mol Al
            
            N.B. mole ratios do not limit significant figures
            
            Homework: Web Page, 9.1: Chapter 9, Molar Ratio Problems
            
            
            Molar Mass
            
         4. molar mass - definition
            
         5. used as conversion factor in changing moles to grams and grams to moles
            
         6. use periodic table -- round to second place to right of decimal point.
            
         7. Example: convert 26 mol of aluminum to grams
            
            Solution: molar mass is 26.98 g Al / 1 mol Al
            
            26.0 mol Al x 26.98 g Al / 1 mol Al = 701 g Al
            
            Homework: Web Page, 9.2: Chapter 9, Molar Mass Problems
            
3. Ideal Stoichiometric Calculations
   
   1. must have balanced chemical equation to do stoichiometry
      
   2. problems we will be doing are for ideal conditions,
      
   3. rarely exist in the laboratory.
      
   4. assumes all reactants are completely converted to products which rarely happens.
      
   5. problems will indicate the maximum amount of a substance that can be produced from a given amount of a reactant.
      
      1. Conversions of Quantities in Moles (Mole-Mole Problems)
         
         1. revolves around balanced equation
            
         2. coefficients are used to created a mole ratio
            
         3. start calculation with given, multiply by mole ratio so that units (substances) cancel
            
         4. e.g. Sample Problem 9-1 page 281
            How many moles of lithium hydroxide are required to react with 20 mol of CO₂ according to the following balanced equation
            
            CO₂ + 2 LiOH ---> Li₂CO₃ + H₂O
            
            given
            20 mol CO₂
            x mol LiOH
            
            Solution: ratio is either 1 mol CO₂ / 2 mol LiOH or
            2 mol LiOH / 1 mol CO₂ --- from the coefficients in balanced equation
            
            start calculation with given information 20 mol CO₂
            
            20 mol CO₂ x 2 mol LiOH / 1 mol CO₂ = 40 mol LiOH
            
            Alternate: equation method
            
            Homework: Web Page, 9.3: Chapter 9, Mole-Mole Problems
            
      2. Conversions of Amounts in Moles to Mass (Mole-mass problems)
         
         1. Need two conversion factors
            
         2. moles to moles as above then moles to grams using molar mass
            
         3. Sample Problem 9-2
            
            1. Given: mass in g or glucose (C₆H₁₂O₆) = ?
               3.00 mol water
               6 H₂O + 6 CO₂ ---> C₆H₁₂O₆ + 6 O₂
               
               
            2. Solution: convert mole of water into mole of glucose then convert mole of glucose into gram of glucose
               
               _{convert moles of water to moles of glucose
               conversion factors:}
               6 mol H₂O / 1 mol C₆H₁₂O₆ or
               1 mol C₆H₁₂O₆ /6 mol H₂O
               
               3.00 mol H₂O x 1 mol C₆H₁₂O₆ /6 mol H₂O = 0.5 mol C₆H₁₂O₆
               
               convert moles of glucose to grams
               conversion factors:
               C: 6 X 12.01g = 72.06 g
               H: 12 x 1.01 g = 12.12 g
               O: 6 x 16.00 g = 96.00 g
               molar mass = 180.18g
               
               180.18 g / mole or mole / 180.18 g
               
               0.5 mole glucose X 180.18 g glucose/ mole glucose = 90.09 g
               
               
               
            3. Sample Problem 9-3 page 283
               Given: 3.00 mol water
               mass in grams of carbon dioxide = ?
               6 H₂O + 6 CO₂ ---> C₆H₁₂O₆ + 6 O<sub>2
               
               Conversion Factors:</sub>
               ratio of CO₂ to water is 6:6
               molar mass of CO₂: C: 1 x 12.01g = 12.01 g
               O: 2 x 16.00g = 32.00g
               molar mass = 44.01g
               
               3 mol water x 6 mol CO₂/6 mol water x 44.01 g CO₂/mol CO₂
               
               132 g CO₂
               
               
      3. Conversion of Mass to Amounts in Moles (Mass-Mole Problems)
         _{given the mass of one substance in grams, calculate the mass of a second substance in moles}
         
         
         1. Sample Problem 9-4 page 285
            
            Given:
            824 g of NH₃
            moles of NO = ?
            moles of water = ?
            
            4NH₃ + 5O₂ ---> 4NO + 6H₂O
            _{Conversion Factors Part A:}
            ratio of NO to NH₃ is 4:4
            ratio of water to ammonia is 6:4
            molar mass of ammonia: N: 1 x 14.01g = 14.01g
            H: 3 x 1.01 g = 3.03g
            molar mass = 17.04g
            
            824 g NH₃ x ( 1 mol NH₃/17.04 g NH₃) x (4 mol NO/4 mole ammonia) = 48.4 mol NO
            Conversion Factors Part B:
            824 g of ammonia
            moles of water = ?
            molar ratio of ammonia to water is 4:6
            
            824 g NH₃ x ( 1 mol NH₃/17.04 g NH₃) x (6 mol water/4 mol NH₃) = 72.6 mol water
            
            Homework: Web Page, 9.4: Chapter 9, Mole-Mass/Mass-Mole Problems
            
      4. Mass-Mass Calculations
         
         1. mass in grams --> amount in moles --> amount in moles --> mass in grams
            
            Sample Problem 9-5 page 286
            
            Given Information:
            Sn + 2 HF ---> SnF₂ + H₂
            30.00 g HF
            mass of SnF₂ produced = ?
            
            Conversion Factors:
            molar mass of HF: 20.01 g/mole
            molar mass of SnF_2: 156.71 g/mole
            molar ratio: SnF₂ to HF is 1:2
            
            30.00 g HF x ( 1 mol HF/20.01 g HF) x (1 mol SnF₂/2 mol HF) x (156.71 g SnF₂ / 1 mole SnF₂) = 117.5 g SnF₂
            
            Homework: Web Page, 9.5: Chapter 9, Mass-Mass Problems
   6. Limiting Reactants and Percent Yield
      Deals with forcing a reaction to produce a minimum of a product by adding an excess of one reactant. When the reactant not in excess is used up the reaction stops.
      
      Limiting reactant - definition
      Excess reactant - definition
      
      1. Sample problem 9-6 page 289
         Given: 2.0 mol HF
         4.5 mol SiO₂
         SiO₂ + 4 HF ---> SiF₄ + 2 H₂O
         limiting reactant = ?
         
         molar ratio for SiO_{2 to HF is 1:4 from the balanced equation}
         2.0 mol HF x (1 mol SiO₂/4 mol HF) = 0.5 mol SiO₂
         
         Tells us that 2 mol of HF requires 0.5 mol SiO₂ to react completely. Since we are told that we have 4.5 mol SiO₂, SiO₂ is present in excess. That means HF is the limiting reactant.
         
         Homework: Web Page, 9.6: Chapter 9, Limiting Reactant Problems
         
         
      2. Percent Yield
         theoretical yield - definition
         actual yield - definition
         percent yield - definition
         
         
      3. Sample problem 9-8 page 293
         Given: 36.8 g C₆H₆
         excess of Cl₂
         yield of C₆H₆Cl = 38.8 g
         % yield of C₆H₅Cl = ?
         
         C₆H₆ + Cl₂ ---> C₆H₅Cl + HCl
         
         molar ratio of C₆H₅Cl to benzene is 1:1
         molar mass of benzene = 78.12g/mole
         molar mass of C₆H₅Cl = 112.56 g/mole
         
         36.8 g benzene x (mole benzene/78.12g benzene) x (1 mole C₆H₅Cl / 1 mole benzene) x (112.56 g C₆H₅Cl/mole C₆H₅Cl) = 53.0 g C₆H₅Cl
         
         (actual yield / theoretical yield) x 100 = percentage yield
         (38.8 g / 53.0 g) x 100 = 73.2 % yield
         
         Homework: Web Page, 9.7: Chapter 9, Percent Yield Problems
         
         END OF NOTES
         
         Limiting reactant is the reactant that limits the amounts of the other reactants that can combine and the amount of product that can form in a chemical reaction.back
         
         Excess reactant is the substance that is not used up completely in a reaction.back
         
         Theoretical yield is the maximum amount of product that can be produced from a given amount of reactant. back
         
         Actual yield is the measured amount of a product obtained form a reaction.back
         
         Percent yield is the ratio of the actual yield to the theoretical yield multiplied by 100. Back