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Chapter 3—Stoichiometry of Formulas and Equations

Chapter 3—Stoichiometry of Formulas and Equations. AP Chemistry. The Mole. mole - the amount of a substance that contains the same number of entities as there are atoms in exactly 12g of carbon-12. 1 mole = Avogadro’s Number = N A = 6.022x10 23. Oxygen 32.00 g.

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Chapter 3—Stoichiometry of Formulas and Equations

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  1. Chapter 3—Stoichiometry of Formulas and Equations AP Chemistry

  2. The Mole mole - the amount of a substance that contains the same number of entities as there are atoms in exactly 12g of carbon-12. 1 mole = Avogadro’s Number = NA = 6.022x1023

  3. Oxygen 32.00 g One Mole of Common Substances Water 18.02 g CaCO3 100.09 g Figure 3.2 Copper 63.55 g

  4. PROBLEM: Sample Problems 3.1 & 3.2 Calculating the Mass and the Number of Atoms in a Given Number of Moles of an Element (a) Silver (Ag) is used in jewelry and tableware but no longer in U.S. coins. How many grams of Ag are in 0.0342mol of Ag? (b) Iron (Fe), the main component of steel, is the most important metal in industrial society. How many Fe atoms are in 95.8g of Fe?

  5. Sample Problem 3.3 Calculating the Moles and Number of Formula Units in a Given Mass of a Compound PROBLEM: Ammonium carbonate is white solid that decomposes with warming. Among its many uses, it is a component of baking powder, first extinguishers, and smelling salts. How many formula units (molecules) are in 41.6g of ammonium carbonate?

  6. Mass % of element X = moles of X in formula x molar mass of X (amu) x 100 molecular (or formula) mass of compound (amu) Mass % of element X = atoms of X in formula x atomic mass of X (amu) x 100 molecular (or formula) mass of compound(amu)

  7. PROBLEM: Glucose (C6H12O6) is the most important nutrient in the living cell for generating chemical potential energy. Sample Problem 3.4 Calculating the Mass Percents and Masses of Elements in a Sample of Compound (a) What is the mass percent of each element in glucose? (b) How many grams of carbon are in 16.55g of glucose?

  8. Empirical and Molecular Formulas Empirical Formula - The simplest formula for a compound that agrees with the elemental analysis and gives rise to the smallest set of whole numbers of atoms. Molecular Formula - The formula of the compound as it exists, it may be a multiple of the empirical formula.

  9. Sample Problem 3.4 Determining the Empirical Formula from Masses of Elements PROBLEM: Elemental analysis of a sample of an ionic compound gave the following results: 2.82g of Na, 4.35g of Cl, and 7.83g of O. What are the empirical formula and name of the compound?

  10. Sample Problem 3.6 Determining a Molecular Formula from Elemental Analysis and Molar Mass PROBLEM: During physical activity. lactic acid (M=90.08g/mol) forms in muscle tissue and is responsible for muscle soreness. Elemental anaylsis shows that it contains 40.0 mass% C, 6.71 mass% H, and 53.3 mass% O. (a) Determine the empirical formula of lactic acid. (b) Determine the molecular formula.

  11. How can scientists determine the empirical & molecular formulas?  One way to determine the empirical formula of a compound is to combust it in excess oxygen and analyze the amount of carbon dioxide and water is produced.

  12. Vitamin C Linus Pauling 1901-1994

  13. PROBLEM: Vitamin C (M=176.12g/mol) is a compound of C,H, and O found in many natural sources especially citrus fruits. When a 1.000-g sample of vitamin C is placed in a combustion chamber and burned, the following data are obtained: mass of CO2 absorber after combustion =85.35g mass of CO2 absorber before combustion =83.85g mass of H2O absorber after combustion =37.96g mass of H2O absorber before combustion =37.55g What is the molecular formula of vitamin C? Sample Problem 3.6 Determining a Molecular Formula from Combustion Analysis

  14. Sample Problem: a) Write an unbalanced chemical equation, identifying the reactants and products. The equation cannot yet be balanced because vitamin C can only be represented as CxHyOz, where x, y, and z are integers to be determined at a later step in the problem solution. CxHyOz + O2 CO2 + H2O

  15. b) What is the reactant source of carbon in the carbon dioxide product? CxHyOz c) What is the reactant source of the hydrogen in the water product? CxHyOz d) What is the reactant source of the oxygen in the carbon dioxide and water products? CxHyOz & O2

  16. CO2 85.35g-83.85g = 1.50g H2O 37.96g-37.55g = 0.41g e) How many grams of carbon were in the original vitamin C sample? 0.409 g C 1.50 g CO2 1 mol CO2 1 mol C 12 g C 44 g CO2 1 mol CO2 1 mol C f) How many grams of hydrogen were in the sample? 0.41 g H2O 1 mol H2O 2 mol H 1 g H 0.046 g H 18 g H2O 1 mol H2O 1 mol H

  17. g) How many grams of oxygen were in the sample? O must be the difference: 1.000g - (0.409 + 0.049) = 0.545 g O

  18. h) What is the mole ratio of carbon to hydrogen to oxygen in vitamin C? (divide by the smallest number of moles) 1 mol H 0.046 g H = 0.046 mol H = 1 g H 1 mol O 0.545 g O = = 0.0341 mol C 16 g O 1 mol C 0.0341 mol C 0.409 g C = = 12 g C 0.0341 mol C 0.0341 mol O 0.046 mol H : : 0.0341 mol 0.0341 mol 0.0341 mol 1 C : 1.3 H : 1 O

  19. i) What is the empirical formula of vitamin C? 1 C : 1.3 H : 1 O X 3 3 C : 4 H : 3 O Empirical Formula: C3H4O3

  20. j) The molar mass of vitamin C is 176 g/mol. What is the molecular formula? In other words, what are the values of x, y, and z in CxHyOz? 1st: Find Molar Mass of empirical Formula, C3H4O3 3 x 12 + 4 x 1 + 3 x 16 = 88 g/mol 2nd: Divide Molar Mass of Molecular by Molar Mass of Empirical 176 g/mol = 2 88 g/mol 3rd: Multiply subscripts of empirical by your answer in Step 2 C3H4O3x 2 = C6H8O6

  21. Table 3.4 Two Compounds with Molecular Formula C2H6O Property Ethanol Dimethyl Ether 46.07 M(g/mol) 46.07 Colorless Color Colorless -138.50C Melting Point -1170C Boiling Point 78.50C -250C Density at 200C 0.789g/mL(liquid) 0.00195g/mL(gas) Use intoxicant in alcoholic beverages in refrigeration Structural formulas and space-filling model

  22. Sample Problem 3.10 Calculating Amounts of Reactants and Products PROBLEM: In a lifetime, the average American uses 1750lb(794g) of copper in coins, plumbing, and wiring. Copper is obtained from sulfide ores, such as chalcocite, or copper(I) sulfide, by a multistage process. After an initial grinding step, the first stage is to “roast” the ore (heat it strongly with oxygen gas) to form powdered copper(I) oxide and gaseous sulfur dioxide. (a) How many moles of oxygen are required to roast 10.0mol of copper(I) sulfide? (b) How many grams of sulfur dioxide are formed when 10.0mol of copper(I) sulfide is roasted? (c) How many kilograms of oxygen are required to form 2.86Kg of copper(I) oxide?

  23. Summary of the Mass-Mole-Number Relationships in a Chemical Reaction Figure 3.8

  24. Calculations Involving a Limiting Reactant: when reactants are mixed in stoichiometric quantities: they are mixed in exactly the correct amounts so the reactants are used up at the same time

  25. when reactants are NOT mixed in stoichiometric quantities: limiting reactant (reagent): the reactant that is not used up the reactant used up and limits the amount of products formed excess reactant (reagent):

  26. An Ice Cream Sundae Analogy for Limiting Reactions Figure 3.9

  27. Sample Problem: View the video clip on iron smelting showing the reaction of carbon monoxide and iron (III) oxide. Write a balanced chemical equation for the reaction in the space below. Fe2O3 + 3 CO  2 Fe + 3 CO2

  28. 60.0 g Fe2O3 1 mol Fe2O3 2 mol Fe 55.85 g Fe = 42.0 g Fe 159.7 g Fe2O3 1 mol Fe2O3 1 mol Fe 100.0 g CO 1 mol CO 2 mol Fe 55.85 g Fe = 133.0 g Fe 28 g CO 3 mol CO 1 mol Fe If you start with 60.0 g of iron (III) oxide and 100.0 g of carbon monoxide, how many grams of iron can be produced? Fe2O3 + 3 CO  2 Fe + 3 CO2 1st: Convert both reactants to grams of Fe

  29. 60.0 g Fe2O3 1 mol Fe2O3 2 mol Fe 55.85 g Fe = 42.0 g Fe 159.7 g Fe2O3 1 mol Fe2O3 1 mol Fe 100.0 g CO 1 mol CO 2 mol Fe 55.85 g Fe = 133.0 g Fe 28 g CO 3 mol CO 1 mol Fe 2nd: The limiting reactant is the one that makes the smallest amount of Fe LIMITING REACTANT!!!

  30. Industrial processors typically use an excessive amount of CO in this reaction. Can you think of two reasons why they would want to do this? All of the Fe2O3 would be converted into Fe CO is cheap

  31. Theoretical Yield: The amount of product that should be formed when the limiting reactant is all used up  Actual Yield: The amount actually produced Actual Yield X 100 = ______ % Percent Yield = Theoretical Yield

  32. Actual Yield X 100 = ______ % Percent Yield = Theoretical Yield 38.00 g Fe X 100 = 42.0 g Fe If you do the reaction above and obtain 38.00 g of Fe product, what is the percent yield? 90.5 %

  33. How many grams of the excess reactant remains?

  34. PROPERTIES OF SOLUTIONS Concentration Solute Solvent concentration moles of solute n Molarity (M) = = Liters of solution V

  35. Sample Problem 3.16 Calculating Mass of Solute in a Given Volume of Solution PROBLEM: How many grams of solute are in 1.75L of 0.460M sodium monohydrogen phosphate?

  36. Laboratory Preparation of Molar Solutions Figure 3.12

  37. Solution Preparation

  38. Dilution: no = nT O = original T = total n Recall: M = V What stays the same? MoVo = MTVT

  39. Sample Problem 3.14 Preparing a Dilute Solution from a Concentrated Solution PROBLEM: “Isotonic saline” is a 0.15M aqueous solution of NaCl that simulates the total concentration of ions found in many cellular fluids. Its uses range from a cleaning rinse for contact lenses to a washing medium for red blood cells. How would you prepare 0.80L of isotomic saline from a 6.0M stock solution?

  40. Converting a Concentrated Solution to a Dilute Solution Figure 3.13

  41. Fig. 3.11

  42. Sample Problem 3.19 Calculating Amounts of Reactants and Products for a Reaction in Solution PROBLEM: Specialized cells in the stomach release HCl to aid digestion. If they release too much, the excess can be neutralized with antacids. A common antacid contains magnesium hydroxide, which reacts with the acid to form water and magnesium chloride solution. As a government chemist testing commercial antacids, you use 0.10M HCl to simulate the acid concentration in the stomach. How many liters of “stomach acid” react with a tablet containing 0.10g of magnesium hydroxide?

  43. Mg(OH)2(s) + 2HCl(aq) MgCl2(aq) + 2H2O(l) mol Mg(OH)2 58.33g Mg(OH)2 2 mol HCl 1 mol Mg(OH)2 1L 0.10mol HCl Sample Problem 3.19 Calculating Amounts of Reactants and Products for a Reaction in Solution continued SOLUTION: 0.10g Mg(OH)2 = 1.7x10-3 mol Mg(OH)2 1.7x10-3 mol Mg(OH)2 = 3.4x10-3 mol HCl 3.4x10-3 mol HCl = 3.4x10-2 L HCl

  44. Sample Problem 3.20 Solving Limiting-Reactant Problems for Reactions in Solution PROBLEM: Mercury and its compounds have many uses, from filling teeth (as an alloy with silver, copper, and tin) to the industrial production of chlorine. Because of their toxicity, however, soluble mercury compounds, such mercury(II) nitrate, must be removed from industrial wastewater. One removal method reacts the wastewater with sodium sulfide solution to produce solid mercury(II) sulfide and sodium nitrate solution. In a laboratory simulation, 0.050L of 0.010M mercury(II) nitrate reacts with 0.020L of 0.10M sodium sulfide. How many grams of mercury(II) sulfide form?

  45. Hg(NO3)2(aq) + Na2S(aq) HgS(s) + 2NaNO3(aq) L of Hg(NO3)2 L of Na2S multiply by M multiply by M mol Hg(NO3)2 x 1mol HgS mol Na2S 1mol Na2S mol ratio x 1mol HgS mol ratio 1mol Hg(NO3)2 mol HgS mol HgS 232.7g HgS 232.7mol HgS Sample Problem 3.16 Solving Limiting-Reactant Problems for Reactions in Solution continued SOLUTION: 0.050L Hg(NO3)2 0.020L Hg(NO3)2 x 0.010 mol/L x 0. 10 mol/L = 5.0x10-4 mol HgS = 2.0x10-3 mol HgS Hg(NO3)2 is the limiting reagent. 5.0x10-4 mol HgS = 0.12g HgS

  46. Figure 3.14

  47. Demo Observations and Inferences:   Balanced Chemical Equation:    Calculations: 0.75 g KClO3 32 g O2 1 mol KClO3 3 mol O2 = 0.29 g O2 122.55 g KClO3 2 mol KClO3 1 mol O2 Obs: Purple flame, Some solid left, Exothermic Inf: Potassium present, Oxygen produced 2 KClO3 2 KCl + 3 O2 How many grams of O2 are produced when 0.75 g of KClO3 are decomposed?

  48. Demo / Sample Problem: Observations: Balanced Equation: 2 AgNO3 (aq) + Cu (s)  Ag (s) + Cu(NO3)2 (aq) 2 20 ml of 1.0 M 1.5 g

  49. 2 AgNO3 (aq) + Cu (s)  Ag (s) + Cu(NO3)2 (aq) 2 1.5 g Cu 2 mol Ag 1 mol Cu = 0.047 mol Ag 63.55 g Cu 1 mol Cu mol solute M = L solution mol solute 1 M = 0.020 L solution 0.020 mol AgNO3 2 mol Ag = 0.020 mol Ag 2 mol AgNO3 Limiting Reagent? Cu limiting n = 0.020 mol AgNO3

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