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Stoichiometry

Stoichiometry. Chapters 7 and 9. Stoichiometry Objectives. Objectives: students will be able to… 1. calculate the molar mass of a compound. 2. calculate the percent composition by mass of an element in a compound.

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Stoichiometry

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  1. Stoichiometry Chapters 7 and 9

  2. Stoichiometry Objectives Objectives: students will be able to… 1. calculate the molar mass of a compound. 2. calculate the percent composition by mass of an element in a compound. 3. calculate the mass of an element present in a sample of a compound. 4. distinguish between an element's atomic mass and molar mass. 5. calculate the mass of a substance given the number of moles present. 6. calculate the number of moles present given the mass of a sample. 7. determine the mole ratio of compounds in a balanced chemical equation. 8. calculate stoichiometric quantities when given a balanced chemical equation. 9. identify the limiting reagent and calculate the quantity of products produced in such a reaction. 10. calculate theoretical yield, actual yield, and percent yield. 11. determine the empirical formula of a compound. 12. determine the molecular formula of a compound.

  3. Stoichiometry Vocabulary(12 Words) • actual yield • molar mass • percent yield • empirical formula • Mole • Stoichiometry • excess reagent • mole ratio • theoretical yield • limiting reagent • percent composition • molecular formula

  4. Stoichiometry Notes • Stoichiometry is a process used by chemists to make predictions about the quantities of substances that react and are given off as products in chemical reactions and relating those quantities to one another. • Stoichiometry relates reactants to products in amounts not masses. • Therefore the problems are worked in terms of moles not grams. • However, the information both provided and requested is often in terms of grams. • Therefore, the ability to convert between these two quantities is essential.

  5. Steps to Solve Stoichiometry Problems • Identify what you know and what you do not know and write it down. • Convert any masses given to moles. • Mass of substance (g) = Number of Moles Molar Mass of substance (g/mol) • Determine the Mole Ratio needed. • Coefficient of unknown substance in BALANCED chemical equation Coefficient of known substance in BALANCED chemical equation • mol unknown molecule # mol known molecule • # mol comes from BALANCED chemical equation NOT the problem itself. • Use the mole ratio to determine the amount of product the known reactant will produce. • Known Moles X Mole Ratio = Moles of Product • Convert moles of product to grams. • Number of Moles X Total Molar Mass = Mass of substance (g)

  6. I Basic Stoichiometry Practice 1. How many moles of sodium will react with water to produce 4.0 mol of hydrogen in the following reaction? (8.0 mol Na) 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)

  7. We Basic Stoichiometry Practice …continued 2. Aluminum will react with sulfuric acid in the following reaction. How many moles of sulfuric acid will react with 18 moles of Al? (27 mol H2SO4) 2Al(s) + 3H2SO4(l) → Al2(SO4)3(aq) + 3H2(g)

  8. We Basic Stoichiometry Practice …continued 3. Aluminum will react with sulfuric acid in the following reaction. Based on your previous answer, how many moles of each product will be produced? (27 mol H2 and 9 mol Al2(SO4)3) 2Al(s) + 3H2SO4(l) → Al2(SO4)3(aq) + 3H2(g)

  9. You Basic Stoichiometry Practice …continued 4. How many moles of lithium chloride will be formed in the reaction of chlorine with 0.046 mol of lithium bromide in the following reaction? (0.046mol LiCl) 2LiBr(aq) + Cl2(g) → 2LiCl(aq) + Br2(l)

  10. Homework • Propane burns in excess oxygen according to the following reaction: C3H8 +5O2 → 3CO2 + 4H2O • How many moles of CO2 and H2O are formed from 3.85 moles of propane? • If 0.647 moles of oxygen is used in the burning of propane, how many moles of CO2 and H2O are produced? How many moles of propane are consumed? • How many moles of ammonium sulfate can be made from the reaction of 30.0 mol of NH3 with H2SO4 according to the following equation? 2NH3 + H2SO4 → (NH4)2SO4 • Complete any unfinished worksheets.

  11. End Day 1

  12. Homework Answers • Propane burns in excess oxygen according to the following reaction: C3H8 +5O2 → 3CO2 + 4H2O • How many moles of CO2 and H2O are formed from 3.85 moles of propane? CO2 = 11.55 mol and H20 = 15.4 mol • If 0.647 moles of oxygen is used in the burning of propane, how many moles of CO2 and H2O are produced? How many moles of propane are consumed? CO2 = 0.3882 mol and H20 = 0.5176 mol and C3H8 = 0.1294 mol • How many moles of ammonium sulfate can be made from the reaction of 30.0 mol of NH3 with H2SO4 according to the following equation? 2NH3 + H2SO4 → (NH4)2SO4 (NH4)2SO4 = 15.0 mol

  13. I Stoichiometry with Conversions Practice 1. Potassium chlorate is sometimes decomposed in the laboratory to generate oxygen. The reaction is 2KClO3(s) → 2KCl(s) + 3O2(g). What mass of KClO3 do you need to produce 0.50 mol O2? 40.85 g KClO3

  14. We Stoichiometry with Conversions Practice …continued 2. How many moles of aluminum will be produced from 30.0 kg Al2O3 in the following reaction: 2Al2O3→4Al + 3O2? 1176.9321 mol Al

  15. We Stoichiometry with Conversions Practice …continued 3. Methane burns in air by the following reaction: CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) What mass of water is produced by burning 500.00 g of methane? 1122.7414 g H2O

  16. You Stoichiometry with Conversions Practice …continued 4. Using the reaction 2Ag(s) + Cl2(g) → 2AgCl(s) Calculate the mass of silver needed to react with chlorine to produce 84.0 grams of silver chloride. 63.2227g Ag

  17. Homework • Modern Chemistry Book page 320 questions 6-13 • Whoosh Bottle Questions • Complete any unfinished worksheets.

  18. End Day 2

  19. Homework Answers(Modern chemistry Book pg. 320 questions 6-13) • O2= 15.75 mol, CO2 =9 mol, and H2O= 13.5 mol • Na= 574.75g, Cl2= 886.25g • CO= 75.1409 mol, F3= 50.094 mol, and CO2= 75.1410 mol • CO= 87.3953 kg and H2= 12.6053 kg • NO2= 1,104.24g and NO= 720.24g • NaOH= 42.0359 mol, Na2CO3= 21.0180 mol, and CO2= 21.0180 mol • NaBr=1,389.015g and AgBr= 2,534.895g • NaHCO3= 3.0584 mol, CO2= 3.0584 mol, Na2SO4= 1.5292 mol, and H2O= 3.0584 mol

  20. Limiting Reactants • Substances can only react in a chemical reaction until the reactants are used up. • Reactions only occur when reactants are in a specific ratio to one another and in multiples thereof. • If 1 pencil reacts with 4 pieces of paper, then 2 pencils will react with 8 pieces of paper. 3 pencils will react with 12 pieces of paper etc. at a continuous ratio of 1:4. • If I have 32 sheets of paper, how many pencils do I need to use them all up? • If I have 1236 pencils and 5000 sheets of paper, which one will be used up first? • The one used up first is the limiting reactant.

  21. Steps to Solve Limiting Reactants Problems • Identify what you know and what you do not know and write it down. • Convert any masses given to moles. • Mass of substance (g) = Number of Moles Molar Mass of substance (g/mol) • Determine the Mole Ratio needed. • Coefficient of unknown substance in BALANCED chemical equation Coefficient of known substance in BALANCED chemical equation • mol unknown molecule # mol known molecule • # mol comes from BALANCED chemical equation NOT the problem itself. • Use the mole ratio to determine the amount of product the known reactant will produce. • Known Moles X Mole Ratio = Moles of Product • Convert moles of product to grams. • Number of Moles X Total Molar Mass = Mass of substance (g)

  22. Steps to Solve Limiting Reactant Problems …continued • Compare the amount of “Unknown” calculated and compare it to the amount of “Unknown” given in the problem. • If there is more “Unknown” calculated than given, then it is the limiting reactant. • If there is more “Unknown” given than calculated, then it is the excess reactant and the “Known” is the limiting reactant. Extension: If the problem continues to ask specifically how much one of the products in the equation will be formed by the reaction…Steps 1-6 Same as Above and then: • Determine your new Mole Ratio using the Limiting Reactant identified above. • Coefficient of unknown substance in BALANCED chemical equation Coefficient of Limiting Reactant in BALANCED chemical equation • mol unknown molecule # mol Limiting Reactant • # mol comes from BALANCED chemical equation NOT the problem itself. • Use the mole ratio to determine the amount of substance the Limiting Reactant will produce. • Known Moles of Limiting Reactant X Mole Ratio = Moles of Unknown • Convert moles of product to grams. • Number of Moles X Total Molar Mass = Mass of substance (g)

  23. I Limiting Reactants Practice 1. Calcium hydroxide, used to neutralize acid spills, reacts with hydrochloric acid according to the following equation: Ca(OH)2 + 2HCl → CaCl2 + 2H2O. If you have spilled 6.3 mol of HCl and put 2.8 mol of Ca(OH)2 on it, which substance is the limiting reactant?

  24. We Limiting Reactants Practice …continued 2. Aluminum oxidizes according to the following equation: 4Al + 3O2 → 2Al2O3. 0.048 mol of powdered Al is placed into a container containing 0.030 mol O2. What is the limiting reactant?

  25. We Limiting Reactants Practice …continued 3. Chlorine can replace bromine in bromide compounds forming a chloride compound and elemental bromine. 2KBr + Cl2 → 2KCl + Br2 is an example of this reaction. When 0.855 g of Cl2 and 3.205 g of KBr are mixed in solution, which is the limiting reactant? How many grams of Br2 are formed?

  26. You Limiting Reactants Practice …continued 4. A process by which zirconium metal can be produced from the mineral zirconium(IV) orthosilicate, ZrSiO4 , starts by reacting it with chlorine gas to form zirconium(IV) chloride. ZrSiO4 + 2Cl2 → ZrCl4 + SiO2 + O2 What mass of ZrCl4 can be produced if 862 g of ZrSiO4 and 950.0 g of Cl2 are available? (You must first determine the limiting reactant.)

  27. End Day 3

  28. Homework Answers(????) • = • = • = • = • = • = • = • =

  29. Percent Yield • In theory, everything is perfect. People do what is right, even if it hurts them, simply because that is what is right. If a person works hard toward a goal, one day he will achieve it. If I buy expensive, energy efficient light bulbs that are supposed to last for 5 years, I will only have to buy light bulbs once every 5 years. • Unfortunately, we do not live in a theoretical world. In our world, things go wrong. And as we’ve seen with many other things, as goes the world, goes chemistry. • Therefore, our perfectly balanced chemical equations that say if we put in so many grams/moles of X we will get back so many grams/moles of Y are victims of our imperfect world also. In the real world, we will only get a percentage of the Y out that the chemical equation predicted for us.

  30. Percent Yield…continued • Chemical Equation produces →Theoretical (perfect) Yield • Real World produces → Actual (imperfect) Yield • The actual yield can never be greater than the theoretical yield. • The actual Yield can never be more than 100% of the theoretical yield. • The discrepancy is not due to an incorrect equation. It is due to imperfections in the process such as: • Reactants or products leak out, especially when they are gases. • The reactants are not 100% pure. • Some product is lost when it is purified. • There are also many chemical reasons, including: • The products decompose back into reactants (as with the ammonia process). • The products react to form different substances. • Some of the reactants react in ways other than the one shown in the equation. • These are called side reactions. • The reaction occurs very slowly. This is especially true of reactions involving organic substances.

  31. Steps to Solve Percent Yield Problems • Identify the Limiting Reactant in the reaction. • Calculate the Theoretical Yield of the substance in question. • Identify what you know and what you do not know and write it down. • Convert any masses given to moles. • Mass of substance (g) = Number of Moles Molar Mass of substance (g/mol) • Determine the Mole Ratio needed. • Coefficient of unknown substance in BALANCED chemical equation Coefficient of Limiting Reactant in BALANCED chemical equation • # mol unknown molecule # mol Limiting Reactant • # mol comes from BALANCED chemical equation NOT the problem itself. • Use the mole ratio to determine the amount of product the known reactant will produce. • Known Moles of Limiting Reactant X Mole Ratio = Moles of Unknown • Convert moles of product to grams. • # of Moles Calculated X Total Molar Mass (g/mol) = Theoretical Mass of substance (g) • Determine % Yield • Mass of Actual Yield (g) X 100 = % Yield Mass of Theoretical Yield (g)

  32. We Percent Yield Practice 1. Dichlorine monoxide, Cl2O, is sometimes used as a powerful chlorinating agent in research. It can be produced by passing chlorine gas over heated mercury (II) oxide according to the following equation: HgO + Cl2 → HgCl2 + Cl2O What is the % yield, if the quantity of reactants is sufficient to produce 0.86g of Cl2O but only 0.71g is obtained?

  33. You Percent Yield Practice …continued 2. Acetylene, C2H2, can be used as an industrial starting material for the production of many organic compounds. Sometimes, it is first brominated to form 1,1,2,2-tetrabromoethane, CHBr2CHBr2, which can then be reacted in many different ways to make other substances. The equation for the bromination of acetylene follows: C2H2 + 2Br2 → CHBr2CHBr2 If 72.0g of C2H2 reacts with excess bromine and 729g of the product is recovered, what is the % yield of the reaction?

  34. Percent Error Notes Percent error is a calculation of how efficient your lab technique is. Just like % Yield, it compares your Theoretical yield to your Actual yield. However, the focus is on what was not produced instead of what was produced. Steps to Solve % error Problems: • Identify your Actual and Theoretical Yields • Mass of Theoretical Yield(g) - Mass of Actual Yield (g)X 100 = Mass of Theoretical Yield (g)

  35. We Percent Error Practice 1. Dichlorine monoxide, Cl2O, is sometimes used as a powerful chlorinating agent in research. It can be produced by passing chlorine gas over heated mercury (II) oxide according to the following equation: HgO + Cl2 → HgCl2 + Cl2O What is the % yield, if the quantity of reactants is sufficient to produce 0.86g of Cl2O but only 0.71g is obtained?

  36. You Percent Error Practice …continued 2. Acetylene, C2H2, can be used as an industrial starting material for the production of many organic compounds. Sometimes, it is first brominated to form 1,1,2,2-tetrabromoethane, CHBr2CHBr2, which can then be reacted in many different ways to make other substances. The equation for the bromination of acetylene follows: C2H2 + 2Br2 → CHBr2CHBr2 If 72.0g of C2H2 reacts with excess bromine and 729g of the product is recovered, what is the % yield of the reaction?

  37. Homework • Modern Chemistry Book page 321 questions 22-30 • Modern Chemistry Book page 318 Practice Questions 1 and 2 • Whoosh Bottle Questions • Al Leftovers Lab Part II

  38. End Day 4

  39. Homework Answers(????) • = • = • = • = • = • = • = • =

  40. Empirical and Molecular Formula Notes An Empirical Formula is a chemical formula for a compound that reflects the atoms contained in the compound in their smallest whole-number ratio. • Examples: CH, BH, Ba(NO3)2 A Molecular Formula is the actual formula for a molecular compound. • Examples: C2H2, B12H12, Ba3(NO3)6 • An empirical formula may not be a correct molecular formula.

  41. Steps to Solve Empirical and Molecular Formula Problems Determining the correct Empirical Formula • Look at the formula subscripts. Are they in their smallest whole-number ratio? • If yes, then you are done. • If no, then divide each subscript by their largest common denominator. • Example: C2H6F4 has en empirical formula of CH3F2since all of the original subscripts were divisible by 2.

  42. Steps to Solve Empirical and Molecular Formula Problems…continued Determining the correct Molecular Formula • Determine the molar mass of the empirical formula. • Divide the molar mass of the sample by the molar mass of the empirical formula. • Multiply the subscripts of the empirical formula by the answer. • Example: the empirical formula of the sample is CH and the sample molar mass is 104.16 g/mol. • CH has a molar mass of 13.03 g/mol • (104.16 g/mol ) / (13.02 g/mol)= 8 • The Molecular Formula is C8H8

  43. I Empirical and Molecular Formula Practice Determine the empirical formula. 1. C6H6 C3H3 Determine the molecular formula. 2. A compound with an empirical formula of C2OH4 and a molar mass of 88 grams per mole. What is the molecular formula of this compound? C4O2H8

  44. We Empirical and Molecular Formula Practice …continued Determine the empirical formula. 3. C6H18C4H9 Determine the molecular formula. 4. A compound with an empirical formula of C4H4O and a molar mass of 408 grams per mole. What is the molecular formula of this compound? C8H8O2

  45. You Empirical and Molecular Formula Practice …continued Determine the empirical formula. 5. C14H42O14CH3O C4H9 Determine the molecular formula. 6. A compound with an empirical formula of CFBrO and a molar mass of 636.75 grams per mole. What is the molecular formula of this compound? C2F2Br2O2

  46. End Day 5

  47. Homework Answers(????) • = • = • = • = • = • = • = • =

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