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Chapter 3

Chapter 3. Stoichiometry. Mole-Mass Relationships in Chemical Systems. 3.1 The Mole. 3.2 Determining the Formula of an Unknown Compound. 3.3 Writing and Balancing Chemical Equations. 3.4 Calculating the Amounts of Reactant and Product. 3.5 Fundamentals of Solution Stoichiometry.

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Chapter 3

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  1. Chapter 3 Stoichiometry

  2. Mole-Mass Relationships in Chemical Systems 3.1 The Mole 3.2 Determining the Formula of an Unknown Compound 3.3 Writing and Balancing Chemical Equations 3.4 Calculating the Amounts of Reactant and Product 3.5 Fundamentals of Solution Stoichiometry

  3. Definition of the Mole mole - the amount of a substance that contains the same number of particles as there are atoms in exactly 12 g of carbon-12. This amount contains 6.022 x 1023 particles. The number, 6.022 x 1023, is called Avogadro’s number and is abbreviated N. One mole (1 mol) contains 6.022 x 1023 particles (to four significant figures)

  4. Counting Objects of Fixed Relative Mass Figure 3.1 12 red marbles at 7 g each = 84g 12 yellow marbles at 4 g each = 48g 55.85 g Fe = 6.022 x 1023 atoms Fe 32.07 g S = 6.022 x 1023 atoms S

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

  6. Table 3.1 Summary of Mass Terminology Term Definition Unit isotopic mass mass of an isotope of an element amu atomic mass average of the masses of the naturally occurring isotopes of an element weighted according to their abundance amu (also called atomic weight) molecular (or formula) mass sum of the atomic masses of the atoms (or ions) in a molecule (or formula unit) amu (also called molecular weight) mass of 1 mole of chemical particles (atoms, ions, molecules, formula units) g/mol molar mass (M) (also called gram-molecular weight)

  7. Information Contained in the Chemical Formula of Glucose C6H12O6 ( M = 180.16 g/mol) Table 3.2 Oxygen (O) Carbon (C) Hydrogen (H) Atoms/molecule of compound 6 atoms 12 atoms 6 atoms Moles of atoms/ mole of compound 6 moles of atoms 12 moles of atoms 6 moles of atoms Atoms/mole of compound 6 (6.022 x 1023) atoms 12 (6.022 x 1023) atoms 6 (6.022 x 1023) atoms Mass/moleculeof compound 6 (12.01 amu) = 72.06 amu 12 (1.008 amu) = 12.10 amu 6 (16.00 amu) = 96.00 amu Mass/mole of compound 72.06 g 12.10 g 96.00 g

  8. no. of grams mass (g) = no. of moles x 1 mol 1 mol no. of moles = mass (g) x no. of grams 6.022x1023 entities no. of entities = no. of moles x 1 mol 1 mol no. of moles = no. of entities x 6.022x1023 entities Interconverting Moles, Mass, and Number of Chemical Entities

  9. PROBLEM: 107.9 g Ag mol Ag mol Fe 55.85g Fe 6.022x1023atoms Fe mol Fe Calculating the Mass and the Number of Atoms in a Given Number of Moles of an Element Sample Problem 3.1 (a) Silver (Ag) is used in jewelry and tableware but no longer in US coins. How many grams of Ag are in 0.0342 mol of Ag? (b) Iron (Fe), the main component of steel, is the most important metal in industrial society. How many Fe atoms are in 95.8 g of Fe? PLAN: (a) To convert mol of Ag to g, we use the number of g Ag per mol Ag, the molar mass M. SOLUTION: 0.0342 mol Ag x = 3.69 g Ag PLAN: (b) To convert g of Fe to atoms we first find the number of moles of Fe and then convert moles to atoms. SOLUTION: 95.8 g Fe x = 1.72 mol Fe = 1.04 x1024 atoms Fe 1.72 mol Fe x

  10. mol (NH4)2CO3 6.022x1023 formula units (NH4)2CO3 96.09g (NH4)2CO3 mol (NH4)2CO3 Sample Problem 3.2 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, fire extinguishers, and smelling salts. How many formula units are in 41.6 g of ammonium carbonate? PLAN: After writing the formula for the compound, we find M by adding the masses of the elements, convert the given mass (41.6 g) to moles using M, and then convert moles to formula units using Avogadro’s number. SOLUTION: The formula is (NH4)2CO3. M = (2 x 14.01 g/mol N) + (8 x 1.008 g/mol H + (12.01 g/mol C) + (3 x 16.00 g/mol O) = 96.09 g/mol 41.6 g (NH4)2CO3 x x = 2.61 x 1023 formula units (NH4)2CO3

  11. mass % of element X = moles of X in formula x molar mass of X (g/mol) x 100 mass (g) of 1 mole of compound mass % of element X = atoms of X in formula x atomic mass of X (amu) x 100 molecular (or formula) mass of compound(amu)

  12. moles of X in one mole of compound mass (g) of X in one mole of compound Mass fraction of X Mass % of X Flow Chart of Mass Percentage Calculation Multiply by M (g/mol) of X Divide by mass (g) of one mole of compound Multiply by 100

  13. 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.

  14. mol Na mol O mol Cl 22.99 g Na 16.00 g O 35.45 g Cl 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: 2.82 g of Na, 4.35 g of Cl, and 7.83 g of O. What are the empirical formula and name of the compound? PLAN: Find the relative number of moles of each element; divide by the lowest mole amount to find the relative mole ratios (empirical formula). SOLUTION: 2.82 g Na x = 0.123 mol Na 4.35 g Cl x = 0.123 mol Cl 7.83 g O x = 0.489 mol O Na1 Cl1 O3.98 NaClO4 NaClO4 is sodium perchlorate.

  15. Sample Problem 3.5 Determining a Molecular Formula from Elemental Analysis and Molar Mass PROBLEM: During physical activity, lactic acid (M = 90.08 g/mol) forms in muscle tissue and is responsible for muscle soreness. Elemental analysis 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.

  16. mol C mol H mol O 12.01g C 1.008 g H 16.00 g O 3.33 3.33 3.33 90.08 g molar mass of lactate 30.03 g mass of CH2O Sample Problem 3.5 (continued) SOLUTION: Assuming 100 g of lactic acid, the constituents are: 40.0 g C x 6.71 g H x 53.3 g O x 3.33 mol C 6.66 mol H 3.33 mol O C3.33 H6.66 O3.33 CH2O = empirical formula C3H6O3 is the molecular formula 3

  17. m 2 m 2 CnHm + (n+ ) O2 = nCO(g) + H2O(g) Combustion Apparatus for the Determination of Formulas of Organic Compounds Figure 3.4

  18. PROBLEM: Vitamin C (M = 176.12 g/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 = 85.35 g mass of CO2 absorber before combustion = 83.85 g mass of H2O absorber after combustion = 37.96 g mass of H2O absorber before combustion = 37.55 g What is the molecular formula of vitamin C? Sample Problem 3.6 Determining a Molecular Formula from Combustion Analysis PLAN: difference (after-before) = mass of oxidized element find the mass of each element in its combustion product preliminary formula empirical formula molecular formula find the moles

  19. CO2: 85.35 g - 83.85 g = 1.50 g H2O: 37.96 g - 37.55 g = 0.41 g 12.01 g C 2.016 g H 44.01 g CO2 18.02 g H2O 12.01 g C 1.008 g H 16.00g O 0.409 g C 0.046 g H 0.545 g O 176.12 g/mol 88.06 g Sample Problem 3.6 (continued) SOLUTION: There are 12.01 g C per mol CO2 1.50 g CO2x = 0.409 g C There are 2.016 g H per mol H2O 0.41 g H2O x = 0.046 g H O must be the difference: 1.000 g - (0.409 + 0.046) = 0.545 = 0.0341 mol C = 0.0461 mol H = 0.0341 mol O C1H1.3O1 C3H4O3 = 2.000 C6H8O6

  20. Table 3.4 Two Compounds with Molecular Formula C2H6O Property Ethanol Dimethyl Ether 46.07 M (g/mol) 46.07 colorless color colorless -138.5 0C melting point -117 0C boiling point 78.5 0C -25 0C density (20 0C) 0.789 g/mL (liquid) 0.00195 g/mL (gas) use intoxicant in alcoholic beverages refrigeration structural formulas and space-filling models

  21. Formation of HF gas: macroscopic and molecular levels Figure 3.6

  22. A three-level view of the chemical reaction in a flashbulb Figure 3.7

  23. PLAN: SOLUTION: C8H18 + O2 CO2 + H2O C8H18 + O2 CO2 + H2O 2C8H18 + 25O216CO2 + 18H2O balance the atoms adjust the coefficients check the atom balance 2C8H18(l) + 25O2 (g) 16CO2 (g) + 18H2O (g) specify states of matter Sample Problem 3.7 Balancing Chemical Equations PROBLEM: Within the cylinders of a car’s engine, the hydrocarbon, octane (C8H18), which is one of the many components of gasoline, mixes with oxygen from air and burns to form carbon dioxide and water vapor. Write a balanced equation for this reaction. translate the statement 8 CO2 + 9 H2O C8H18 + O2 25/2 8 9

  24. Sample Problem 3.8 Calculating Amounts of Reactants and Products PROBLEM: In a lifetime, the average American uses 1750 lb (794 kg) 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 10.0 mol of copper(I) sulfide? (b) How many grams of sulfur dioxide are formed when 10.0 mol of copper(I) sulfide are roasted? (c) How many kilograms of oxygen are required to form 2.86 kg of copper(I) oxide? PLAN: write and balance equation find mols O2 find mols SO2 find mols Cu2O find g SO2 find mols O2 find kg O2

  25. 2Cu2S(s) + 3O2(g) 2Cu2O(s) + 2SO2(g) 3 mol O2 2 mol Cu2S 2 mol SO2 64.07 g SO2 2 mol Cu2S mol SO2 103g Cu2O mol Cu2O kg Cu2O 143.10 g Cu2O 3 mol O2 32.00 g O2 kg O2 2 mol Cu2O mol O2 103g O2 Sample Problem 3.8 (continued) SOLUTION: (a) 10.0 mol Cu2S x = 15.0 mol O2 (b) 10.0 mol Cu2S x x = 641 g SO2 (c) 2.86 kg Cu2O x = 20.0 mol Cu2O x 20.0 mol Cu2O x x x = 0.960 kg O2

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

  27. PROBLEM: Roasting is the first step in extracting copper from chalcocite, the ore used in the previous problem. In the next step, copper(I) oxide reacts with powdered carbon to yield copper metal and carbon monoxide gas. Write a balanced overall equation for the two-step process. 2Cu2S(s) + 3O2(g) 2Cu2O(s) + 2SO2(g) Cu2O(s) + C(s) 2Cu(s) + CO(g) 2Cu2O(s) + 2C(s) 4Cu(s) + 2CO(g) 2Cu2S(s)+3O2(g)+2C(s) 4Cu(s)+2SO2(g)+2CO(g) Sample Problem 3.9 Calculating Amounts of Reactants and Products in a Reaction Sequence PLAN: SOLUTION: write balanced equations for each step cancel reactants and products common to both sides of the equations or sum the equations

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

  29. mass of N2H4 mass of N2O4 limiting mol N2 multiply by M mol of N2H4 mol of N2O4 g N2 Sample Problem 3.10 Calculating Amounts of Reactant and Product in Reactions Involving a Limiting Reactant PROBLEM: A fuel mixture used in the early days of rocketry was composed of two liquids, hydrazine (N2H4) and dinitrogen tetraoxide (N2O4), which ignite on contact to form nitrogen gas and water vapor. How many grams of nitrogen gas (N2) form when 1.00 x 102 g of N2H4 and 2.00 x 102 g of N2O4 are mixed? PLAN: Start with a balanced chemical equation and find the number of moles of reactants; identify limiting reagent; determine grams of N2 formed. divide by M molar ratio mol of N2 mol of N2

  30. N2H4(l) + N2O4(l) N2(g) + H2O(l) mol N2H4 32.05 g N2H4 3 mol N2 28.02g N2 2 mol N2H4 mol N2 mol N2O4 92.02 g N2O4 3 mol N2 mol N2O4 Sample Problem 3.10 (continued) SOLUTION: 2 3 4 1.00 x 102g N2H4 x = 3.12 mol N2H4 N2H4 is the limiting reactant because it produces less product, N2, than does N2O4. 3.12 mol N2H4 x = 4.68 mol N2 2.00 x 102g N2O4 x = 2.17 mol N2O4 4.68 mol N2x 2.17 mol N2O4 x = 6.51 mol N2 = 131g N2

  31. Formation of side-products during chemical reactions

  32. SiO2(s) + 3C(s) SiC(s) + 2CO(g) 103g SiO2 mol SiO2 kg SiO2 60.09g SiO2 40.10 g SiC kg mol SiC 103g 51.4 kg 66.73 kg Sample Problem 3.11 Calculating Percent Yield PROBLEM: Silicon carbide (SiC) is an important ceramic material that is made by allowing sand (silicon dioxide, SiO2) to react with powdered carbon at high temperature. Carbon monoxide is also formed. When 100.0 kg of sand are processed, 51.4 kg of SiC are recovered. What is the percent yield of SiC in this process? PLAN: SOLUTION: write balanced equation x = 1664 mol SiO2 100.0 kg SiO2x find mol reactant & product mol SiO2 = mol SiC = 1664 mol find g product predicted = 66.73 kg 1664 mol SiC x x percent yield x100 = 77.0%

  33. PLAN: Molarity (M) is the number of moles of solute per liter of solution. 1.80 mol HBr 1000 mL 455 mL soln 1 L Sample Problem 3.12 Calculating the Molarity of a Solution PROBLEM: Hydrobromic acid (HBr) is a solution of hydrogen bromide gas in water. Calculate the molarity of a hydrobromic acid solution if 455 mL contains 1.80 moles of hydrogen bromide. SOLUTION: mol of HBr divide by volume x = 3.96 M concentration (mol/mL) HBr 103 mL = 1L molarity (mol/L) HBr

  34. 0.460 moles 1 L 141.96 g Na2HPO4 mol Na2HPO4 Sample Problem 3.13 Calculating Mass of Solute in a Given Volume of Solution PROBLEM: How many grams of solute are found in 1.75 L of a 0.460 M solution of sodium monohydrogen phosphate? Molarity is the number of moles of solute per liter of solution. Knowing the molarity and volume allows us to find the number of moles and then the number of grams of solute. The molecular formula of the solute is Na2HPO4. PLAN: volume of soln SOLUTION: multiply by M 1.75 L x = 0.805 mol Na2HPO4 moles of solute 0.805 mol Na2HPO4 multiply by M x grams of solute = 114 g Na2HPO4

  35. 0.15 mol NaCl L soln L solnconc 6 mol NaCl Sample Problem 3.14 Preparing a Dilute Solution from a Concentrated Solution PROBLEM: Isotonic saline is a 0.15 M 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 (erythrocytes). How would you prepare 0.80 L of isotonic saline from a 6.0 M stock solution? PLAN: The number of moles of solute does not change during the dilution but volume does. The new volume will be the sum of the two volumes, that is, the total final volume. Mdil x Vdil = # mol solute = Mconc x Vconc SOLUTION: 0.80 L soln x = 0.12 mol NaCl = 0.020 L soln 0.12 mol NaCl x

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

  37. Figure 3.11

  38. mass Mg(OH)2 L HCl divide by M divide by M mol Mg(OH)2 mol HCl mol ratio Sample Problem 3.15 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 HCl to form water and magnesium chloride solution. As a government chemist testing commercial antacids, you use 0.10 M HCl to simulate the acid concentration in the stomach. How many liters of “stomach acid” react with a tablet containing 0.10 g of magnesium hydroxide? PLAN: Write a balanced equation for the reaction; find the moles of Mg(OH)2; determine the mole ratio of reactants and products; use moles to convert to molarity.

  39. Mg(OH)2(s) + 2HCl(aq) MgCl2(aq) + 2H2O(l) mol Mg(OH)2 58.33 g Mg(OH)2 2 mol HCl 1 mol Mg(OH)2 1L 0.10 mol HCl Sample Problem 3.15 (continued) SOLUTION: 0.10 g Mg(OH)2 x = 1.7 x 10-3 mol Mg(OH)2 1.7 x 10-3 mol Mg(OH)2x = 3.4 x 10-3 mol HCl 3.4 x 10-3 mol HCl x = 3.4 x 10-2 L HCl

  40. Sample Problem 3.16 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, 0.050 L of 0.010 M mercury(II) nitrate reacts with 0.020 L of 0.10 M sodium sulfide. How many grams of mercury(II) sulfide form? PLAN: Write a balanced chemical equation for the reaction. Since this is a problem concerning a limiting reactant, we find the amount of product that would be made from each reactant. We then choose the reactant that gives the lesser amount of product.

  41. Hg(NO3)2(aq) + Na2S(aq) HgS(s) + 2 NaNO3(aq) 1 mol HgS 1 mol Na2S 1 mol HgS 1 mol Hg(NO3)2 232.7 g HgS 1 mol HgS Sample Problem 3.16 (continued) SOLUTION: 0.050 L Hg(NO3)2 x 0.010 mol/L x = 5.0 x 10-4 mol HgS 0.020 L Na2S x 0. 10 mol/L x = 2.0 x 10-3 mol HgS Hg(NO3)2 is the limiting reagent. 5.0 x 10-4 mol HgS x = 0.12 g HgS

  42. Laboratory Preparation of Molar Solutions Figure 3.12

  43. Key Mass/Mole Relationships Figure 3.14

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