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3 Chapter Formulas, Equations, and Moles Chemistry, 4th Edition McMurry/Fay Dr. Paul Charlesworth Michigan Technological University Balancing Chemical Equations • 01 A balanced chemical equation represents the conversion of the reactants to products such that the number of atoms of each element is conserved. reactants → products limestone → quicklime + gas Calcium carbonate → calcium oxide + carbon dioxide CaCO3(s) → CaO(s) + CO2(g) Chapter 03 Slide 2 Prentice Hall ©2004 Balancing Chemical Equations • Balancing Equations: Write the unbalanced equation: A2B A2 + B2 • Use coefficients to indicate how many formula 02 units are required to balance the equation: 2 A2 + B2 2 A2B Chapter 03 Slide 3 Prentice Hall ©2004 Chapter 03: Formulas , Equations, and Moles Balancing Chemical Equations 03 • Balancing Equations - Method 01: • Balance those atoms which occur in only one compound on each side • Balance the remaining atoms • Reduce coefficients to smallest whole integers • Check your answer Chapter 03 Slide 4 Prentice Hall ©2004 Balancing Chemical Equations 04 • Balancing Equations - Method 02: • Identify most complex compound • Balance this compound by placing 1 before it • Balance remaining compounds using fractions • Multiply fractions to obtain integers Chapter 03 Slide 5 Prentice Hall ©2004 Balancing Chemical Equations • 05 Balance the following equations: C6H12O6 C2H6O + CO2 Fe + O2 Fe2O3 NH3 + Cl2 N2H4 + NH4Cl KClO3 + C12H22O11 KCl + CO2 + H2O Chapter 03 Slide 6 Prentice Hall ©2004 Chapter 03: Formulas , Equations, and Moles Balancing Chemical Equations • 06 Write a balanced equation for the reaction of element A (red spheres) with element B (green spheres) as represented below: Chapter 03 Slide 7 Prentice Hall ©2004 Atomic and Molecular Mass 1 amu = Mass: 01 mass of carbon −12 atom 12 proton = 1.00728 amu neutron = 1.0086 amu electron = 0.0005486 12C atom = 12.00000 amu 13C atom = 13.00335 amu Chapter 03 Slide 8 Prentice Hall ©2004 Atomic and Molecular Mass 02 • The atomic masses as tabulated in the periodic table are the averages of the naturally occurring isotopes. • Mass of C = average of 12C and 13C = 0.9889 x 12 amu + 0.0111 x 13.0034 amu = 12.011 amu Chapter 03 Slide 9 Prentice Hall ©2004 Chapter 03: Formulas , Equations, and Moles Atomic and Molecular Mass • The mass of a molecule is just the sum of the masses of the atoms making up the molecule. • m(C2H4O2) = 2·mC + 4·mH + 2·mO 03 = 2·(12.01) + 4·(1.01) + 2·(16.00) = 60.06 amu Chapter 03 Slide 10 Prentice Hall ©2004 Avogadro and the Mole 01 • One mole of a substance is the gram mass value equal to the amu mass of the substance. • One mole of any substance contains 6.02 x 1023 units of that substance. • Avogadro’s Number (NA, 6.022 x 1023) is the numerical value assigned to the unit, 1 mole. Chapter 03 Slide 11 Prentice Hall ©2004 Avogadro and the Mole • 02 The Mole: Allows us to make comparisons between substances that have different masses. Chapter 03 Slide 12 Prentice Hall ©2004 Chapter 03: Formulas , Equations, and Moles Avogadro and the Mole • 03 Calculate the molar mass of the following: Fe2O3 (Rust) C6H8O7 (Citric acid) C16H18N2O4 (Penicillin G) • Balance the following, and determine how many moles of CO will react with 0.500 moles of Fe2O3. Fe2O3(s) + CO(g) Fe(s) + CO2(g) Chapter 03 Slide 13 Prentice Hall ©2004 Avogadro and the Mole • 04 Methionine, an amino acid used by organisms to make proteins, is represented below. Write the formula for methionine and calculate its molar mass. (red = O; gray = C; blue = N; yellow = S; ivory = H) Chapter 03 Slide 14 Prentice Hall ©2004 Stoichiometry • 01 Stoichiometry: Relates the moles of products and reactants to each other and to measurable quantities. Chapter 03 Slide 15 Prentice Hall ©2004 Chapter 03: Formulas , Equations, and Moles Stoichiometry • 02 Aqueous solutions of NaOCl (household bleach) are prepared by the reaction of NaOH with Cl2: 2 NaOH(aq) + Cl2(g) → NaOCl(aq) + NaCl(aq) + H2O(l) • How many grams of NaOH are needed to react with 25.0 g of Cl2? Chapter 03 Slide 16 Prentice Hall ©2004 Stoichiometry • 03 Aspirin is prepared by reaction of salicylic acid (C7H6O3) with acetic anhydride (C4H6O3) to form aspirin (C9H8O4) and acetic acid (CH3CO2H). Use this information to determine the mass of acetic anhydride required to react with 4.50 g of salicylic acid. How many grams of aspirin will result? How many grams of acetic acid will be produced as a by-product? Chapter 03 Slide 17 Prentice Hall ©2004 Stoichiometry • 04 Yields of Chemical Reactions: If the actual amount of product formed in a reaction is less than the theoretical amount, we can calculate a percentage yield. % yield = Actual product yield × 100% Theoretica l product yield Chapter 03 Slide 18 Prentice Hall ©2004 Chapter 03: Formulas , Equations, and Moles Stoichiometry • 05 Dichloromethane (CH2Cl2) is prepared by reaction of methane (CH4) with chlorine (Cl2) giving hydrogen chloride as a by-product. How many grams of dichloromethane result from the reaction of 1.85 kg of methane if the yield is 43.1%? Chapter 03 Slide 19 Prentice Hall ©2004 Stoichiometry • 06 Limiting Reagents: The extent to which a reaction takes place depends on the reactant that is present in limiting amounts—the limiting reagent. Chapter 03 Slide 20 Prentice Hall ©2004 Stoichiometry • 07 In this figure, ethylene oxide is the limiting reagent. Chapter 03 Slide 21 Prentice Hall ©2004 Chapter 03: Formulas , Equations, and Moles Stoichiometry • 08 Limiting Reagent Calculation: Lithium oxide is a drying agent used on the space shuttle. If 80.0 kg of water is to be removed and 65 kg of lithium oxide is available, which reactant is limiting? Li2O(s) + H2O(l) → • MM(Li2O) = 29.88 g/mol • MM(H2O) = 18.02 g/mol 2 LiOH(s) Chapter 03 Slide 22 Prentice Hall ©2004 Stoichiometry • 10 Limiting Reagent Calculation: Cisplatin is an anticancer agent prepared as follows: K2PtCl4 + 2 NH3 → Pt(NH3)2Cl2 + 2 KCl • If 10.0 g of K2PtCl4 and 10.0 g of NH3 are allowed to react: (a) which is the limiting reagent? (b) How many grams of the excess reagent are consumed? (c) How many grams of cisplatin are formed? • MM(K2PtCl4) = 415.08 g/mol MM(NH3) = 18.04 g/mol Chapter 03 Slide 23 Prentice Hall ©2004 Solution Concentrations • 01 Concentration allow us to measure out a specific number of moles of a compound by measuring the mass or volume of a solution. Chapter 03 Slide 24 Prentice Hall ©2004 Chapter 03: Formulas , Equations, and Moles Solution Concentrations • Molarity: The most useful way of expressing the amount of a substance dissolved in a solution is with molarity. Molarity (M) = • 02 Moles of solute Liters of solvent It is important to note that the final volume of solution must be used, not volume of solvent. Chapter 03 Slide 25 Prentice Hall ©2004 Solution Concentrations • 03 How many moles of solute are present in 125 mL of 0.20 M NaHCO3? • How many grams of solute would you use to prepare 500.0 mL of 1.25 M NaOH? Chapter 03 Slide 26 Prentice Hall ©2004 Solution Concentrations • Dilution: • Is the process of 04 reducing a solution’s concentration by adding more solvent. Chapter 03 Slide 27 Prentice Hall ©2004 Chapter 03: Formulas , Equations, and Moles Solution Concentrations 04 Concentrated solution + Solvent ⇒ Dilute solution Moles of solute (mol) = molarity (M) x volume (V) Mconcentrated x Vconcentrated = Mdilute x Vdilute Chapter 03 Slide 28 Prentice Hall ©2004 Solution Concentrations • 05 What volume of 18.0 M H2SO4 is required to prepare 250.0 mL of 0.500 M aqueous H2SO4? • What is the final concentration if 750 mL of 3.50 M glucose is diluted to a volume of 400.0 mL? Chapter 03 Slide 29 Prentice Hall ©2004 Solution Stoichiometry • 01 Solution Stoichiometry uses molarity as a conversion factor between volume and moles of a substance in a solution. Chapter 03 Slide 30 Prentice Hall ©2004 Chapter 03: Formulas , Equations, and Moles Solution Stoichiometry • 02 Stomach acid, a dilute solution of HCl in water, can be neutralized by reaction with sodium hydrogen carbonate, NaHCO3. How many milliliters of 0.125 M NaHCO3 solution are needed to neutralize 18.0 mL of 0.100 M HCl? Chapter 03 Slide 31 Prentice Hall ©2004 Solution Stoichiometry • Titration: • A technique for determining the concentration of a solution. 03 Chapter 03 Slide 32 Prentice Hall ©2004 Solution Stoichiometry • 04 What is the molarity of a sulfuric acid solution if a 25.0 mL sample is titrated to equivalence with 50.0 mL of 0.150 M potassium hydroxide solution? H2SO4(aq) + NaOH(aq) → KNO3(aq) + H2O(l) Chapter 03 Slide 33 Prentice Hall ©2004 Chapter 03: Formulas , Equations, and Moles Percentage Composition • 01 Percent Composition: Identifies the elements present in a compound as a mass percent of the total compound mass. • The mass percent is obtained by dividing the mass of each element by the total mass of a compound and converting to percentage. Chapter 03 Slide 34 Prentice Hall ©2004 Percentage Composition • 02 Glucose has the molecular formula C6H12O6. What is its empirical formula, and what is the percentage composition of glucose? • Saccharin has the molecular formula C7H5NO3S. What is its empirical formula, and what is the percentage composition of saccharin? Chapter 03 Slide 35 Prentice Hall ©2004 Empirical Formula • 01 The empirical formula gives the ratio of the number of atoms of each element in a compound. Compound Formula Hydrogen peroxide H2O2 Benzene C6H6 Ethylene C2H4 Propane C3H8 Chapter 03 Empirical Formula OH CH CH2 C3H8 Slide 36 Prentice Hall ©2004 Chapter 03: Formulas , Equations, and Moles Empirical Formula • A compound’s empirical formula can be determined from its percent composition. • A compound’s molecular formula is determined from the molar mass and empirical formula. 02 Chapter 03 Slide 37 Prentice Hall ©2004 Empirical Formula 03 • Combustion analysis is one of the most common methods for determining empirical formulas. • A weighed compound is burned in oxygen and its products analyzed by a gas chromatogram. • It is particularly useful for analysis of hydrocarbons. Chapter 03 Slide 38 Prentice Hall ©2004 Empirical Formula • 04 A compound was analyzed to be 82.67% carbon and 17.33% hydrogen by mass. An osmotic pressure experiment determined that its molar mass is 58.11 g/mol. • What is the empirical formula and molecular formula for the compound? Chapter 03 Slide 39 Prentice Hall ©2004 Chapter 03: Formulas , Equations, and Moles