Stoichiometry (Chapter 12) Dr. Walker

Stoichiometry (stoy-KEY-ah-muh-tree) Definition: predicting the amounts of reactants and/or products that will be involved in a reaction

Chocolate Chip Cookie Recipe Similar to using recipes to cook When cooking, recipes require specific amounts of each ingredient to make a set amount of food. Example: Chocolate Chip Cookies 2 ½ cups flour ½ cup sugar ¾ teaspoon salt 2 eggs ½ cup butter 1 teaspoon vanilla extract 1 cup brown sugar 1 pkg chocolate chips 1 cup shortening 1 cup chopped nuts Makes 6 dozen cookies

Recipes = Chemical Formulas When cooking, recipes require specific amounts of each ingredient to make a set amount of food. Chemical reactions are the exact same thing!!! We need to know how much material can be made from existing amounts (i.e., how much food is in the pantry) OR how much we need to make a specific amount (how much of everything do I need to make 10 dozen cookies)

Chemical Reactions Chemical change involves a reorganization of the atoms in one or more substances. C2H5OH + 3O2 ® 2CO2 + 3H2O reactants products When the equation is balanced it has quantitative significance: 1 mole of ethanol reacts with 3 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water – Pay attention to the coefficients!!!!!

Chemical Reactions Chemical change involves a reorganization of the atoms in one or more substances. C2H5OH + 3O2 ® 2CO2 + 3H2O reactants products When the equation is balanced it has quantitative significance: How many moles of ethanol could react with 9 moles of oxygen? If this happened, what products would we get?

Chemical Reactions C2H5OH + 3O2 ® 2CO2 + 3H2O reactants products How many moles of oxygen could react with 9 moles of oxygen? If this happened, what products would we get? If 1 mole of ethanol reacts with 3 moles of oxygen, then it would require 3 moles of ethanol to react with 9 moles of oxygen. If 1 mole of ethanol gives 2 moles CO2 and 3 moles H2O, 3 moles of ethanol would yield 3 x 2 = 6 moles CO2 and 3 x 3 = 9 moles H2O

Mole Ratios Using the coefficients we can write mole ratios. Definition: A mole ratio gives the relative amounts of reactants and products using a fraction The coefficients you used in the recipes are basically mole ratios. You’re doing the same thing.

Using Mole Ratios Example: N2 + 3 H2  2 NH3 Give the mole ratio of N2 to NH3

Using Mole Ratios Example: 1 N2 + 3 H2  2 NH3 Give the mole ratio of N2 to NH3 Use the coefficients 1 N2 to 2 NH3

Using Mole Ratios Example: N2 + 3 H2  2 NH3 Give the mole ratio of H 2 to NH3

Using Mole Ratios Example: N2 + 3 H2  2 NH3 Give the mole ratio of H 2 to NH3 Use the coefficients 3 H2 to 2 NH3

Mole-Mole Conversions Write all Mole-Mole Conversions Example: N2 + 3 H2  2 NH3 How many moles of ammonia could be produced from 6 moles of N2 and excess H2?

Mole-Mole Conversions Example: N2 + 3 H2  2 NH3 How many moles of ammonia could be produced from 6 moles of N2 and excess H2? What are we starting with? 6 moles of N2 Whenever you have excess, you have more than enough – you won’t use it in the problem! 6 moles ????

Mole-Mole Conversions Example: N2 + 3 H2  2 NH3 How many moles of ammonia could be produced from 6 moles of N2 and excess H2? Use the coefficients in a proportion 6 moles ???? ----------- 1 mole N2 ----------- 2 moles NH3

Mole-Mole Conversions Example: N2 + 3 H2  2 NH3 How many moles of ammonia could be produced from 6 moles of N2 and excess H2? 6 moles ???? ??? = 12 moles NH3 ----------- 1 mole N2 ----------- 2 moles NH3

Another Mole-Mole Example 2 C8H18 + 25 O2 16 CO2 + 18 H2O How many moles of water can be produced from 1.2 moles of octane and excess O2?

Another Mole-Mole Example 2 C8H18 + 25 O2 16 CO2 + 18 H2O How many moles of water can be produced from 1.2 moles of octane and excess O2? 1.2 moles C8H18 ????? ----------------- 2 moles C8H18 ---------------- 18 moles H2O

Another Mole-Mole Example 2 C8H18 + 25 O2 16 CO2 + 18 H2O How many moles of water can be produced from 1.2 moles of octane and excess O2? 1.2 moles C8H18 ????? ----------------- 2 moles C8H18 ---------------- 18 moles H2O ???? = 10.8 moles H2O

You try… 2 Al + 3 H2SO4 Al2(SO4)3 + 3 H2 How many moles of hydrogen can be produced from 0.24 moles of aluminum? 2 Al2O3 4 Al + 3 O2 How many moles of aluminum can be made from 0.35 moles of Al2O3

You try… 2 Al + 3 H2SO4 Al2(SO4)3 + 3 H2 How many moles of hydrogen can be produced from 0.24 moles of aluminum? 2 Al2O3 4 Al + 3 O2 How many moles of aluminum can be made from 0.35 moles of Al2O3 0.24 moles Al ????? ----------------- 2 moles Al ---------------- 3 moles H2 ???? = 0.36 moles H2 0.35 moles Al2O3 ????? ----------------- 2 moles Al2O3 ---------------- 4 moles Al ???? = 0.70 moles Al

The Next Step… Mass-Mass Stoichiometry When we measure materials for chemical reactions, the scales measure in grams, not moles Why can’t we just multiply the original mass times the mole ratio?

The Next Step… Mass-Mass Stoichiometry When we measure materials for chemical reactions, the scales measure in grams, not moles Why can’t we just multiply the original mass times the mole ratio? We must account for the difference in molar masses between the two materials we are dealing with

Proportion Method “Plug and chug” Plug in the known numbers Solve for the unknown

Mass-Mass Example C3H8 + 5 O2 3 CO2 + 4 H2O How many grams of H2O could be produced from the combustion of 100 g of propane?

Mass-Mass Example C3H8 + 5 O2 3 CO2 + 4 H2O How many grams of H2O could be produced from the combustion of 100 g of propane? 100 g ???? Mass (start) = 100 g Coefficient (start) = 1 Molar Mass (start) = 44.11 g/mole Mass (finish) = unknown Coefficient (finish) = 4 Molar Mass (start) = 18.02 g/mole Remember – cross multiply, then divide for solving proportions (100 g x 4 moles x 18.02 g/mole) / (1 mole x 44.11 g/mole) = ???

Mass-Mass Example C3H8 + 5 O2 3 CO2 + 4 H2O How many grams of H2O could be produced from the combustion of 100 g of propane? 100 g ???? Grams H2O 100 g C3H8 1 mole C3H8 x 44.11 g/mole C3H8 4 mole H2O x 18.02 g/mole H2O Mass (start) = 100 g Coefficient (start) = 1 Molar Mass (start) = 44.11 g/mole Mass (finish) = 163.41 g (answer!!) Coefficient (finish) = 4 Molar Mass (start) = 18.02 g/mole

Some helpful hints… Only use the materials provided in the problem The compounds not listed are irrelevant to the problem, even if they’re in the reaction The coefficients and the molar masses are separate Do not count the coefficient in your molar mass!!! If you see the word “excess”, cross out that chemical – you will not use it!!! The word excess just means you’ve got plenty of material to react with the amounts that you have.

Another Example 6 Na + Fe2O3 3 Na2O + 2 Fe How many grams of Fe can be produced by the reaction of 10 g Na and excess iron oxide?

Another Example 6 Na + Fe2O3 3 Na2O + 2 Fe How many grams of Fe can be produced by the reaction of 10 g Na and excess iron oxide? 10 g ????

Another Example 6 Na + Fe2O3 3 Na2O + 2 Fe How many grams of Fe can be produced by the reaction of 10 g Na and excess iron oxide? Grams Fe 10 g Na 2 mole Fe x 55.85 g/mole Fe 6 mole Na x 22.99 g/mole Na

Another Example 6 Na + Fe2O3 3 Na2O + 2 Fe How many grams of Fe can be produced by the reaction of 10 g Na and excess iron oxide? Grams Fe 10 g Na 2 mole Fe x 55.85 g/mole Fe 6 mole Na x 22.99 g/mole Na Grams Fe = 8.10 g (answer!!)

Volume-Volume Stoichiometry It is difficult to measure gases with a mass, so they are usually measured with a volume Instead of using the molar mass, use 22.4 L 1 mole of any gas = 22.4 L volume (at 0oC and 1 atm pressure, known as STP)

Example N2 + 3 H2 2 NH3 How many liters of ammonia can be made from 15 liters of hydrogen and excess nitrogen?

Example N2 + 3 H2 2 NH3 How many liters of ammonia can be made from 15 liters of hydrogen and excess nitrogen? Liters NH3 15 L 2 mole NH3 x 22.4 L/mole NH3 3 mole H2 x 22.4 L/mole H2 Answer = 10 liters NH3

Mixed Stoichiometry Sometimes, we need to start with one unit and finish with another. If you’re converting between mass and moles, use the molar mass If you’re converting between volume and moles, use 1 mole = 22.4 L If the conversion involves molecules, use Avogadro’s number!

Mixed Example 2 KClO3 (s) 2 KCl (s) + 3 O2 (g) How many liters of oxygen are produced from the decomposition of 244 g KClO3?

Mixed Example 2 KClO3 (s) 2 KCl (s) + 3 O2 (g) How many liters of oxygen are produced from the decomposition of 244 g KClO3? Remember, since we’re dealing with liters of oxygen, we’ll use 22.4 L instead of the molar mass.

Mixed Example 2 KClO3 (s) 2 KCl (s) + 3 O2 (g) How many liters of oxygen at STP are produced from the decomposition of 244 g KClO3? 244 g KClO3 Liters O2 3 mole O2 x 22.4 L O2 2 mole KClO3 x 122.55 g/mole KClO3

Mixed Example 2 KClO3 (s) 2 KCl (s) + 3 O2 (g) How many liters of oxygen at STP are produced from the decomposition of 244 g KClO3? 244 g KClO3 Liters O2 3 mole O2 x 22.4 L O2 2 mole KClO3 x 122.55 g/mole KClO3 Liters O2 = 66.90 L (answer!!)

Another Mixed Example 2 Ca (s) + O2 (g) 2 CaO How many grams of calcium are required to completely react with 4.48 liters of O2 at STP?

Another Mixed Example 2 Ca (s) + O2 (g) 2 CaO How many grams of calcium are required to completely react with 4.48 liters of O2 at STP? X Grams Ca 4.48 L O2 2 mole Ca x 40.08 g/mole Ca 1 mole O2 x 22.4 L/mole O2 X = 16.03 g Ca

Other Possible Mixing Sometimes you may start with moles and go to mass or liters How will THAT work in the proportion? If you are dealing with MOLES for a particular compound, simply leave out the second number on the bottom.

Mole-Mass Example How many grams of iron could be made from 6 moles of carbon?

Mole-Mass Example How many grams of iron could be made from 6 moles of carbon? Grams Fe 6 moles C 2 mole Fe x 55.85 g Fe 3 mole C Grams Fe = 223.4 g Fe Notice that since we’re only dealing with moles, you don’t need a second number on the bottom!

Volume – Mole Example How many moles of B5H9 are required to react with 100 L of O2?

Volume – Mole Example Moles B5H9 100 L O2 How many moles of B5H9 are required to react with 100 L of O2? 2 moles B5H9 12 x 22.4 L Notice that since we’re only dealing with moles, you don’t need a second number on the bottom! Moles B5H9 = 0.74 moles

Percent Yield Theoretical Yield Actual Yield The amount of material you can make from a reaction if it works perfectly….in theory Everything you have calculated so far has been a theoretical yield Yes, you’ve been doing THIS already Actual Yield The amount of stuff you actually make in a reaction

Percent Yield Percent Yield Notice (again) that we multiply by 100 to turn a decimal to a percentage The percent yield should NEVER be > 100% If this occurs, you either made a math error or your sample is impure (wet).

Percent Yield Example (Easy) Some problems will give you a theoretical and percent yield A student produces 50 g of NaCl in a reaction yield where the theoretical yield is 60 g

Percent Yield Example (Easy) Some problems will give you a theoretical and percent yield A student produces 50 g of NaCl in a reaction yield where the theoretical yield is 60 g 50 g ----------- 60 g x 100 = 83.3%

Percent Yield Example (Not As Easy…) Calculate theoretical yield first Then, plug into actual/theoretical x 100

Percent Yield Example (Not As Easy…) X Grams KCl 45.8 g K2CO3 2 mole KCl x 74.55 g/mole KCl 1 mole K2CO3 x 138.03 g/mole K2CO3 46.3 g Theoretical Yield = 49.5 g KCl ----------- 49.5 g x 100 = 93.5 %

3rd Percent Yield Example 2 Al(s) + 3 Cl2 (g) 2 AlCl3 (s) 40.5 g of aluminum is reacted with excess chlorine gas to produce 180.0 g of aluminum chloride. Give the percent yield of this process.

3rd Percent Yield Example 2 Al(s) + 3 Cl2 (g) 2 AlCl3 (s) 40.5 g ????? actual yield = 180.0 g 40.5 g of aluminum is reacted with excess chlorine gas to produce 180.0 g of aluminum chloride. Give the percent yield of this process.

3rd Percent Yield Example 2 Al(s) + 3 Cl2 (g) 2 AlCl3 (s) 40.5 g ????? actual yield = 180.0 g 40.5 g of aluminum is reacted with excess chlorine gas to produce 180.0 g of aluminum chloride. Give the percent yield of this process. X Grams AlCl3 40.5 g Al 2 moles AlCl3 x 133.33 g/mole AlCl3 2 mole Al x 26.98 g/mole Al 180.0 g Theoretical Yield = 200.1 g AlCl3 ----------- 200.1 g x 100 = 90.0 %

Limiting Reagents We previously compared chemical reactions to recipes If we’re lacking one ingredient, we can’t make the entire recipe. This is called the limiting reagent (material that controls how much material we can make) Example: we have 4 hamburgers, 4 slices of cheese, and 3 buns. We’re limited by the number of buns – this is our limiting reagent We have an excess number of burgers and cheese slices

Limiting Reagents Non-chemical examples https://www.chem.tamu.edu/class/majors/tutorialnotefiles/limiting.htm https://ka-perseus-images.s3.amazonaws.com/f574961ecc30ebb45ee67a64aeb3ddacdb06d1fd.svg

Limiting Reagents https://phet.colorado.edu/sims/html/reactants-products-and-leftovers/latest/reactants-products-and-leftovers_en.html http://chemistrygroupvivzara.weebly.com/uploads/1/2/7/0/12703729/927436.png?551

Limiting Reagents C3H8 + 5 O2 3 CO2 + 4 H2O How many moles of carbon dioxide can be made from 2.5 moles of propane and 10 moles of oxygen? Notice that you are given amounts for BOTH reactants!

Limiting Reagents C3H8 + 5 O2 3 CO2 + 4 H2O 2.5 moles 10 moles ?????? How many moles of carbon dioxide can be made from 2.5 moles of propane and 10 moles of oxygen? Find the moles of CO2 from both reactants!

Limiting Reagents C3H8 + 5 O2 3 CO2 + 4 H2O 2.5 moles 10 moles ?????? How many moles of carbon dioxide can be made from 2.5 moles of propane and 10 moles of oxygen? 2.5 moles C3H8 ???? From C3H8 ---------------- 1 mole C3H8 = ----------- 3 moles CO2 ??? = 7.5 moles CO2 10 moles O2 ???? From O2 ---------------- 5 moles O2 = ----------- 3 moles CO2 ??? = 6 moles CO2

Limiting Reagents C3H8 + 5 O2 3 CO2 + 4 H2O 2.5 moles 10 moles ?????? How many moles of carbon dioxide can be made from 2.5 moles of propane and 10 moles of oxygen? 2.5 moles C3H8 ???? From C3H8 ---------------- 1 mole C3H8 = ----------- 3 moles CO2 ??? = 7.5 moles CO2 10 moles O2 ???? From O2 ---------------- 5 moles O2 = ----------- 3 moles CO2 ??? = 6 moles CO2 Limiting reagent Gave less product!

Limiting Reagents Mass-Mass Problem CaO (s) + CO2 (g) CaCO3(s) How many grams of calcium carbonate are produced from 2.8 g calcium oxide and 4.48 L carbon dioxide at STP? 2.8 g CaO 4.48 L CO2 ?????? X Grams CaCO3 2.8 g CaO 1 mole CaO x 56.08 g/mole CaO 1 moles CaCO3 x 100.09 g/mole CaCO3 X = 5.00 g CaCO3 X Grams CaCO3 4.48 L CO2 1 mole CO2 x 22.4 L/mole CO2 1 moles CaCO3 x 100.09 g/mole CaCO3 X = 20.02 g CaCO3

Terms To Know, Skills To Master Mole Ratio Percent Yield Limiting Reagent Skills to master Calculate mole ratios Perform mole-mole, mass-mass, mixed stoichiometry calculations Determine percent yield from a reaction Determine the limiting reagent of a reaction