Stoichiometry APPLICATION: PURPOSE: We are applying concepts learned in Chapters 5, 6 & 7 Naming/Formulas Molar Conversions Balancing Chemical Equations PURPOSE: To understand how chemical equations are used to express ratios of reactants and products

Stoichiometry =The calculation of quantities in a chemical reaction. From Greek: “Stoichio” = matter or chemicals “Metry” = measure =The calculation of quantities in a chemical reaction. Kinds of quantities measured: Particles (atoms, molecules, formula units) Moles Mass Volume

Stoichiometry Why is it useful? Who uses stoichiometry? To determine the mass, volume or number of particles needed for a particular reaction. To determine the mass, volume or number of particles produced by a particular reaction. Who uses stoichiometry? Chemists, Chemical engineers, Engineers, Rocket scientists, Environmental scientists…. + criminals making illegal drugs, terrorists making bombs…

Stoichiometry Why are balanced chemical equations useful? Look at the ratios of all reactants and products One molecule of nitrogen reacts with three molecules of hydrogen to produce two molecules of ammonia 1:3:2 ratio What happens if you have 10 molecules of N2? 1 10 3 30 2 20 NH3 (g) N2(g) + H2(g)

Stoichiometry 1 3 2 Mass: NH3 (g) N2(g) + H2(g) Reactants = 34.0 g All balanced chemical reactions must follow the Law of Conservation of Mass 1 mol N2 = 28.0 g 3 mol H2 = 6.0 g 2 mol NH3 = 34.0 g 1 3 2 NH3 (g) N2(g) + H2(g) MASS IS ALWAYS CONSERVED IN CHEMICAL REACTIONS! Reactants = 34.0 g Products = 34.0 g

VOLUME IS NOT USUALLY CONSERVED! Stoichiometry Volume: At STP (standard temperature and pressure); 1 mol of any gas = 22.4 L Is volume conserved? 1 mol N2 = 22.4 L 3 mol H2 = 67.2 L (3 x 22.4) 2 mol NH3 = 44.8 L (2 x 22.4) 1 3 2 NH3 (g) N2(g) + H2(g) Reactants = 89.6 L Products = 44.8 L VOLUME IS NOT USUALLY CONSERVED!

MOLES ARE NOT USUALLY CONSERVED! Stoichiometry Moles: The coefficients in a balanced chemical equation represent the number of moles of that substance Are # of Moles conserved? 1 mol N2 3 mol H2 2 mol NH3 1 3 2 NH3 (g) N2(g) + H2(g) Reactants = 4 mol MOLES ARE NOT USUALLY CONSERVED! Products = 2 mol

Stoichiometry Moles: All coefficients are related to the number of moles required for a reaction. One mole of nitrogen reacts with three moles of hydrogen to produce two moles of ammonia 1 3 2 NH3 (g) N2(g) + H2(g) KEY CONCEPT: Coefficients show relative numbers of moles involved in a reaction = the molar ratio

Stoichiometry 1 3 2 Mole-Mole Calculations NH3 (g) N2(g) + H2(g) Balanced chemical equations are valuable because they shoe the relative ratio of reactants to products. You can now relate the # of moles of a reactant(s) to find the # of moles of a product(s) 1 3 2 NH3 (g) N2(g) + H2(g)

Stoichiometry 1 3 2 Sample Problem 1: How many moles of ammonia are produced when 1.2 moles of nitrogen reacts with hydrogen? 1 3 2 NH3 (g) N2(g) + H2(g)

Stoichiometry 1 3 2 Example: Step 1: Step 2: NH3 (g) N2(g) + H2(g) Look at balanced chemical equation 1 mol of N2 produces 2 mol NH3 Step 2: Use a conversion factor Convert mol N2 to mol NH3 1 3 2 NH3 (g) N2(g) + H2(g)

Stoichiometry 1 3 2 Example: How much NH3 can be made from 1.2 mol N2? NH3 (g) N2(g) + H2(g) 1.2 mol N2 2 mol NH3 = 2.4 mol NH3 1 mol N2 Mole ratio

Stoichiometry 1 3 2 Sample Problem 2: NH3 (g) N2(g) + H2(g) Calculate the number of moles of nitrogen required to make 9.54 mol of NH3. 1 3 2 NH3 (g) N2(g) + H2(g) 9.54 mol NH3 1 mol N2 = 4.77 mol N2 2 mol NH3

Stoichiometry 1 3 2 Sample Problem 3: NH3 (g) N2(g) + H2(g) Calculate the number of moles of hydrogen required to make 9.54 mol of NH3. 1 3 2 NH3 (g) N2(g) + H2(g) 9.54 mol NH3 3 mol H2 = 14.3 mol H2 2 mol NH3

Stoichiometry Mass-Mass Calculations: Mass (grams) A  moles A When scientist conduct a lab experiment, they measure the amount of a particular substance using grams…not moles Remember, you can convert the mass of a substance into moles (1-step conversion) Mass (grams) A  moles A

mass A  moles A  moles B  mass B Stoichiometry Mass-Mass Calculations: Additionally, it is possible to convert between masses of reactants and products. The mole ratio (obtained from the balanced chemical equation) is the key to convert from the mass of substance A to mass of substance B Remember, the balanced chemical equation provides the mole ratio! mass A  moles A  moles B  mass B

Stoichiometry 1 3 2 Mass-Mass Example #1: NH3 (g) N2(g) + H2(g) Calculate the number of grams of NH3 produced by the reaction of 6.80 g of hydrogen with an excess of nitrogen. Use the balanced equation above. 1 3 2 NH3 (g) N2(g) + H2(g)

mass H2  moles H2  moles NH3  mass NH3 Stoichiometry Mass-Mass Example #1: Step 1: What are we calculating? Grams H2  grams NH3 Step 2: What is the path we need to take? 1 3 2 NH3 (g) N2(g) + H2(g) mass H2  moles H2  moles NH3  mass NH3

Stoichiometry 1 3 2 Mass-Mass Example #1: NH3 (g) N2(g) + H2(g) Step 3: Complete conversions 1 3 2 NH3 (g) N2(g) + H2(g) 6.80 g H2 1 mol H2 2 mol NH3 17.0 g NH3 1 mol NH3 2.0 g H2 3 mol H2 = 38.5 g NH3

Stoichiometry 1 3 2 Mass-Mass Example #2: NH3 (g) N2(g) + H2(g) How many grams of nitrogen are needed to produce the 38.5 g of NH3 produced in the previous example? 1 3 2 NH3 (g) N2(g) + H2(g) 38.5 g NH3 1 mol NH3 1 mol N2 28.0 g N2 1 mol N2 17.0 g NH3 2 mol NH3 = 31.7 g N2

Stoichiometry Mass-Mass Example #3: What is the total amount of mass (in grams) of the reactants in the example balanced chemical equation? Is mass conserved in a chemical reaction??? Mass of Reactants = Mass of Products

Mass of Reactants = Mass of Products Stoichiometry Mass-Mass Example #3: Mass of Reactants = Mass of Products 31.7 g N2 + ___ g H2  38.5 g NH3 ANSWER  38.5 – 31.7 = 6.8 g H2

Stoichiometry Other Stoichiometric Calculations: You can use the same steps as before to calculate the following… Mass-Volume Volume-Volume Particle-Mass The first step is to ALWAYS convert to moles

Stoichiometry Other Stoichiometric Calculations: Choose the appropriate road map… particles A particles B mass A mass B moles A moles B volume A volume B

Limiting Reagent Limiting Reagent Chemical equations are like recipes… The reactants are combined to make the products Reactions take place based upon the mole ratios expressed in the balanced chemical equation

A reaction can only occur until the limiting reagent is used up!! Limits how much product will form = determines the amount of product that can be formed in a reaction Excess Reagent: The substance(s) that is leftover because they are in excess = there is more than enough to react with the limiting reagent A reaction can only occur until the limiting reagent is used up!!

Limiting Reagent 2 1 1 1 Limiting Reagent  S’mores Gc2MCh Gc + M + Ch Graham Cracker = Gc Marshmallow = M Chocolate = Ch S’more = Gc2MCh To make one s’more…you need to have 2 graham crackers, 1 marshmallow, and 1 piece of chocolate. 2 1 1 1 Gc2MCh Gc + M + Ch

2 1 1 1 Limiting Reagent  S’mores Gc2MCh Gc + M + Ch How many s’mores can you make if you have 14 (mol) graham crackers, 10 pieces of chocolate, and 8 marshmallows? 2 1 1 1 Gc2MCh Gc + M + Ch 14 mol Gc 1 mol Gc2MCh = 7.0 mol Gc2MCh 2 mol Gc 10 mol Ch 1 mol Gc2MCh = 10.0 mol Gc2MCh 1 mol Ch 8 mol M 1 mol Gc2MCh = 8 mol Gc2MCh 1 mol M

2 1 1 1 Limiting Reagent  S’mores Gc2MCh Gc + M + Ch How many s’mores can you make if you have 14 (mol) graham crackers, 10 pieces of chocolate, and 8 marshmallows? Graham crackers  limiting reagent THE LIMITING AGENT WILL DETERMINE HOW MUCH PRODUCT WILL BE CREATED! 2 1 1 1 Gc2MCh Gc + M + Ch 14 mol Gc 1 mol Gc2MCh = 7.0 mol Gc2MCh 2 mol Gc

Limiting Reagent 1 3 2 Limiting Reagent NH3 (g) N2(g) + H2(g) What would happen if 2 mol of N2 reacted with 3 mol of H2? NEED: Nitrogen-Hydrogen ratio  1:3 HAVE: Nitrogen-Hydrogen ratio  2:3 What reactant is limiting? What reactant is in excess? 1 3 2 NH3 (g) N2(g) + H2(g)

Limiting Reagent 1 3 2 Limiting Reagent NH3 (g) N2(g) + H2(g) 6 mol H2 2 mol N2 6 mol H2 needed 1 mol N2 But don’t have! 3 mol H2 1 mol N2 1 mol N2 needed have! 3 mol H2 There is not enough hydrogen for 2 mol N2, so: hydrogen is the limiting reagent!!

2 1 2 Limiting Reagent—Example #1 NaCl (s) Na(s) + Cl2(g) What will occur when 7.20 mol of Na reacts with 3.5 mol of Cl2? What is the limiting reagent? How many moles of NaCl are produced? How much of the excess reagent remains unreacted? 2 1 2 NaCl (s) Na(s) + Cl2(g)

2 1 2 Limiting Reagent—Example #1 NaCl (s) Na(s) + Cl2(g) What will occur when 7.20 mol of Na reacts with 3.5 mol of Cl2? Choose one of the reactants and convert to the amount of moles of the other reactant. 2 1 2 NaCl (s) Na(s) + Cl2(g) 7.20 mol Na 1 mol Cl2 = 3.6 mol Cl2 2 mol Na Does the value you calculated exceed the amount presented in the problem???

2 1 2 Limiting Reagent—Example #1 NaCl (s) Na(s) + Cl2(g) What will occur when 7.20 mol of Na reacts with 3.5 mol of Cl2? Use the given amount (from the problem) of the limiting reagent…calculate moles of NaCl. 2 1 2 NaCl (s) Na(s) + Cl2(g) moles Cl2  moles NaCl 3.5 mol Cl2 2 mol NaCl = 7.0 mol NaCl 1 mol Cl2

2 1 2 Limiting Reagent—Example #1 NaCl (s) Na(s) + Cl2(g) What will occur when 7.20 mol of Na reacts with 3.5 mol of Cl2? The amount of excess reagent is the difference between the given amount of Na and the amount of Na needed in the reaction. 2 1 2 NaCl (s) Na(s) + Cl2(g) moles Cl2 moles Na 3.5 mol Cl2 2 mol Na = 7.0 mol Na 1 mol Cl2

2 1 2 Limiting Reagent—Example #1 NaCl (s) Na(s) + Cl2(g) What will occur when 7.20 mol of Na reacts with 3.5 mol of Cl2? The amount of excess reagent is the difference between the given amount of Na and the amount of Na needed in the reaction. 2 1 2 NaCl (s) Na(s) + Cl2(g) 7.20 mol Na – 7.00 mol Na = 0.20 mol Na in excess

2 1 1 Limiting Reagent—Example #2 Cu2S (s) Cu(s) + S (g) What is the maximum number of grams of Cu2S that can be formed when 80.0 g of Cu reacts with 25.0 g of S? Find the number of moles of each reactant. 2 1 1 Cu2S (s) Cu(s) + S (g) 80.0 g Cu 1 mol Cu = 1.26 mol Cu 63.5 g Cu 25.0 g S 1 mol Cu = 0.779 mol S 32.1 g S

2 1 2 Limiting Reagent—Example #2 Cu2S (s) Cu(s) + S (g) What is the maximum number of grams of Cu2S that can be formed when 80.0 g of Cu reacts with 25.0 g of S? Use the balanced chemical equation to determine the limiting reagent. 2 1 2 Cu2S (s) Cu(s) + S (g) 1.26 mol Cu 1 mol S = 0.630 mol S 2 mol Cu

Sulfur is in excess…copper is limiting Limiting Reagent—Example #2 Use the balanced chemical equation to determine the limiting reagent. 2 1 2 Cu2S (s) Cu(s) + S (g) Compare the amount of sulfur calculated to the amount of sulfur presented in the problem. = 0.630 mol S 0.630 mol S  calculated 0.779 mol S  problem Sulfur is in excess…copper is limiting

2 1 2 Limiting Reagent—Example #2 Cu2S (s) Cu(s) + S (g) What is the maximum number of grams of Cu2S that can be formed when 80.0 g of Cu reacts with 25.0 g of S? Use the amount of limiting reagent to calculate the maximum amount of the product. 2 1 2 Cu2S (s) Cu(s) + S (g) 1.26 mol Cu 1 mol Cu2S 159.1 g Cu2S = 100.2 g Cu2S 1 mol Cu2S 2 mol Cu

Percent Yield Percent Yield: This is a comparison of how much of a given product is created in a reaction to the ideal yield of the product in an ideal reaction. Theoretical Yield: Using an equation to determine the amount of product formed during a reaction Actual Yield: The amount of a product created in an actual laboratory experiment

Percent Yield = Percent Yield: Actual Yield Percent Yield X 100% Percent yield will NEVER exceed 100% Percent yield can be less than 100% Comparison of MASS; not moles Actual Yield Percent Yield = X 100% Theoretical Yield

mass CaCO3  moles CaCO3  moles CaO  mass CaO Percent Yield  Example #1 What is the percent yield of this reaction if 34.8 g of CaCO3 is heated to give 16.4 g of CaO? Determine the actual yield of CaO from the problem 16.4 g CaO Calculate the theoretical yield of CaO. 1 CaCo3 (s) 1 1 CaO(s) + CO2 (g) mass CaCO3  moles CaCO3  moles CaO  mass CaO 34.8 g CaCO3 1 mol CaO 1 mol CaO 56.1 g CaO 1 mol CaCO3 1 mol CaO 100.1 g CaCO3 = 19.5 g CaO

1 1 1 Percent Yield  Example #1 CaCo3 (s) CaO(s) + CO2 (g) What is the percent yield of this reaction if 34.8 g of CaCO3 is heated to give 16.4 g of CaO? 1 CaCo3 (s) 1 1 CaO(s) + CO2 (g) Actual Yield X 100 Percent Yield = Theoretical Yield 16.4 g CaO X 100 Percent Yield = 19.5 g CaO = 84.1 %