Lecture Notes Alan D. Earhart Southeast Community College Lincoln, NE Chapter 3 Formulas, Equations, and Moles John E. McMurry Robert C. Fay CHEMISTRY Fifth Edition Copyright © 2008 Pearson Prentice Hall, Inc.

Chapter 3/2 Balancing Chemical Equations A balanced chemical equation shows that the law of conservation of mass is adhered to. In a balanced chemical equation, the numbers and kinds of atoms on both sides of the reaction arrow are identical. 2NaCl(s)2Na(s) + Cl 2 (g) right side: 2 Na 2 Cl left side: 2 Na 2 Cl

Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/3 Balancing Chemical Equations Hg I 2 (s) + 2KNO 3 (aq)Hg(NO 3 ) 2 (aq) + 2K I (aq) right side: 1 Hg 2 I 2 K 2 N 6 O left side: 1 Hg 2 N 6 O 2 K 2 I

Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/4 Balancing Chemical Equations 2.Find suitable coefficients—the numbers placed before formulas to indicate how many formula units of each substance are required to balance the equation. 2H 2 O(l)2H 2 (g) + O 2 (g) 1.Write the unbalanced equation using the correct chemical formula for each reactant and product. H 2 O(l)H 2 (g) + O 2 (g)

Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/5 Balancing Chemical Equations 3.Reduce the coefficients to their smallest whole- number values, if necessary, by dividing them all by a common denominator. 2H 2 O(l)2H 2 (g) + O 2 (g) 4H 2 O(l)4H 2 (g) + 2O 2 (g) divide all by 2

Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/6 Balancing Chemical Equations 4.Check your answer by making sure that the numbers and kinds of atoms are the same on both sides of the equation. 2H 2 O(l)2H 2 (g) + O 2 (g) right side: 4 H 2 O left side: 4 H 2 O

Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/7 Balancing Chemical Equations Do not change subscripts when you balance a chemical equation. You are only allowed to change the coefficients. H 2 O(l)H 2 (g) + O 2 (g) unbalanced 2H 2 O(l)2H 2 (g) + O 2 (g) Chemical equation changed! H2O2(l)H2O2(l)H 2 (g) + O 2 (g) Balanced properly

Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/8 Chemical Symbols on Different Levels 2H 2 O(l)2H 2 (g) + O 2 (g) 0.56 kg of hydrogen gas react with 4.44 kg of oxygen gas to yield 5.00 kg of liquid water. macroscopic: 2 molecules of hydrogen gas react with 1 molecule of oxygen gas to yield 2 molecules of liquid water. microscopic: How can we relate the two to each other?

Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/9 2(12.0 amu) + 4(1.0 amu) = 28.0 amuC2H4:C2H4: HCl:1.0 amu amu = 36.5 amu Avogadro’s Number and the Mole Molecular Mass: Sum of atomic masses of all atoms in a molecule. Formula Mass: Sum of atomic masses of all atoms in a formula unit of any compound, molecular or ionic.

Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/10 Avogadro’s Number and the Mole Avogadro’s Number (N A ): One mole of any substance contains x formula units. One mole of any substance is equivalent to its molecular or formula mass. 1 mole = 28.0 gC2H4:C2H4: x molecules = 28.0 g 1 mole = 36.5 g x molecules = 36.5 g HCl:

Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/11 Avogadro’s Number and the Mole Avogadro’s Number (N A ): One mole of any substance contains x formula units. One mole of any substance is equivalent to its molecular or formula mass. 1 mole = 28.0 gC2H4:C2H4: 1 mole = 36.5 gHCl:

Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/12 Stoichiometry: Chemical Arithmetic Stoichiometry: The relative proportions in which elements form compounds or in which substances react. aA + bBcC + dD Moles of A Grams of A Moles of B Grams of B Mole Ratio Between A and B (Coefficients) Molar Mass of B Molar Mass of A

Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/13 Stoichiometry: Chemical Arithmetic How many grams of NaOH are needed to react with 25.0 g Cl 2 ? 2NaOH(aq) + Cl 2 (g)NaOCl(aq) + NaCl(aq) + H 2 O(l) Aqueous solutions of sodium hypochlorite (NaOCl), best known as household bleach, are prepared by reaction of sodium hydroxide with chlorine gas: Moles of Cl 2 Grams of Cl 2 Moles of NaOH Grams of NaOH Mole RatioMolar Mass

Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/14 Stoichiometry: Chemical Arithmetic How many grams of NaOH are needed to react with 25.0 g Cl 2 ? 2NaOH(aq) + Cl 2 (g)NaOCl(aq) + NaCl(aq) + H 2 O(l) Aqueous solutions of sodium hypochlorite (NaOCl), best known as household bleach, are prepared by reaction of sodium hydroxide with chlorine gas: 1 mol NaOH 40.0 g NaOH25.0 g Cl g Cl 2 1 mol Cl 2 2 mol NaOH = 28.2 g NaOH xxx

Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/15 Yields of Chemical Reactions The amount actually formed in a reaction. The amount predicted by calculations. Actual Yield: Theoretical Yield: actual yield theoretical yield x 100%Percent Yield =

Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/16 Reactions with Limiting Amounts of Reactants Limiting Reactant: The reactant that is present in limiting amount. The extent to which a chemical reaction takes place depends on the limiting reactant. Excess Reactant: Any of the other reactants still present after determination of the limiting reactant.

Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/17 Reactions with Limiting Amounts of Reactants Because water is so cheap and abundant, it is used in excess when compared to ethylene oxide. This ensures that all of the relatively expensive ethylene oxide is entirely consumed. At a high temperature, ethylene oxide reacts with water to form ethylene glycol which is an automobile antifreeze and a starting material in the preparation of polyester polymers: C 2 H 4 O(aq) + H 2 O(l)C2H6O2(l)C2H6O2(l)

Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/18 Reactions with Limiting Amounts of Reactants C 2 H 4 O(aq) + H 2 O(l)C2H6O2(l)C2H6O2(l) If 3 moles of ethylene oxide react with 5 moles of water, which reactant is limiting and which reactant is present in excess? At a high temperature, ethylene oxide reacts with water to form ethylene glycol which is an automobile antifreeze and a starting material in the preparation of polyester polymers:

Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/19 Reactions with Limiting Amounts of Reactants At a high temperature, ethylene oxide reacts with water to form ethylene glycol which is an automobile antifreeze and a starting material in the preparation of polyester polymers: C 2 H 4 O(aq) + H 2 O(l)C2H6O2(l)C2H6O2(l)

Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/20 Reactions with Limiting Amounts of Reactants Li 2 O(s) + H 2 O(g)2LiOH(s) Lithium oxide is used aboard the space shuttle to remove water from the air supply according to the equation: If 80.0 g of water are to be removed and 65.0 g of Li 2 O are available, which reactant is limiting? How many grams of excess reactant remain? How many grams of LiOH are produced?

Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/21 Reactions with Limiting Amounts of Reactants Li 2 O(s) + H 2 O(g)2LiOH(s) Which reactant is limiting? 65.0 g Li 2 O 1 mol Li 2 O 1 mol H 2 O 29.9 g Li 2 O 1 mol Li 2 O = 4.44 moles H 2 O 80.0 g H 2 O 18.0 g H 2 O 1 mol H 2 O = 2.17 moles H 2 O Amount of H 2 O given: Amount of H 2 O that will react with 65.0 g Li 2 O: Li 2 O is limiting x x x

Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/22 Reactions with Limiting Amounts of Reactants Li 2 O(s) + H 2 O(g)2LiOH(s) 80.0 g H 2 O g H 2 O = 40.9 g H 2 O 2.17 mol H 2 O 1 mol H 2 O 18.0 g H 2 O = 39.1 g H 2 O (consumed) How many grams of excess H 2 O remain? remaininginitialconsumed x

Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/23 Reactions with Limiting Amounts of Reactants Li 2 O(s) + H 2 O(g)2LiOH(s) 2.17 mol H 2 O 1 mol LiOH 23.9 g LiOH = 104 g LiOH How many grams of LiOH are produced? 1 mol H 2 O 2 mol LiOH xx

Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/24 Concentrations of Reactants in Solution: Molarity Molarity: The number of moles of a substance dissolved in each liter of solution. In practice, a solution of known molarity is prepared by weighing an appropriate amount of solute, placing it in a container called a volumetric flask, and adding enough solvent until an accurately calibrated final volume is reached. Solution: A homogeneous mixture. Solute: The dissolved substance in a solution. Solvent: The major component in a solution.

Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/25 Concentrations of Reactants in Solution: Molarity Molarity converts between mole of solute and liters of solution: molarity = moles of solute liters of solution L mol or 1.00 M 1.00 L 1.00 mol = mol of sodium chloride placed in enough water to make 1.00 L of solution would have a concentration equal to:

Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/26 Concentrations of Reactants in Solution: Molarity Molar mass C 6 H 12 O 6 = g/mol How many grams of solute would you use to prepare 1.50 L of M glucose, C 6 H 12 O 6 ? 1 mol mol180.0 g = 49.5 g 1 L 1.50 L0.250 mol = mol x x

Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/27 Diluting Concentrated Solutions dilute solutionconcentrated solution + solvent M i V i = M f V f finalinitial Since the number of moles of solute remains constant, all that changes is the volume of solution by adding more solvent.

Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/28 Diluting Concentrated Solutions Add 6.94 mL 18.0 M sulfuric acid to enough water to make mL of M solution. M i = 18.0 MM f = M V i = ? mLV f = mL = 6.94 mL 18.0 M mL V i = MiMi Mf VfMf Vf M = Sulfuric acid is normally purchased at a concentration of 18.0 M. How would you prepare mL of M aqueous H 2 SO 4 ? x

Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/29 Solution Stoichiometry aA + bBcC + dD Moles of A Volume of Solution of A Moles of B Volume of Solution of B Mole Ratio Between A and B (Coefficients) Molar Mass of B Molarity of A

Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/30 Solution Stoichiometry H 2 SO 4 (aq) + 2NaOH(aq)Na 2 SO 4 (aq) + 2H 2 O(l) What volume of M H 2 SO 4 is needed to react with 50.0 mL of M NaOH? Moles of H 2 SO 4 Volume of Solution of H 2 SO 4 Moles of NaOH Volume of Solution of NaOH Mole Ratio Between H 2 SO 4 and NaOH Molarity of NaOH Molarity of H 2 SO 4

Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/31 Solution Stoichiometry H 2 SO 4 (aq) + 2NaOH(aq)Na 2 SO 4 (aq) + 2H 2 O(l) 2 mol NaOH 1 mol H 2 SO mol H 2 SO 4 1 L solution 1 L mol 1 L 1000 mL = mol NaOH Volume of H 2 SO 4 needed: 1000 mL 1 L 10.0 mL solution (0.250 M H 2 SO 4 ) mol NaOH 50.0 mL NaOH Moles of NaOH available: x x x xx

Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/32 Titration How can you tell when the reaction is complete? HCl(aq) + NaOH(aq)NaCl(aq) + 2H 2 O(l) Titration: A procedure for determining the concentration of a solution by allowing a carefully measured volume to react with a solution of another substance (the standard solution) whose concentration is known. Once the reaction is complete you can calculate the concentration of the unknown solution.

Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/33 Titration unknown concentration solution Erlenmeyer flask buret standard solution (known concentration) An indicator is added which changes color once the reaction is complete

Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/34 Titration HCl(aq) + NaOH(aq)NaCl(aq) + 2H 2 O(l) 48.6 mL of a M NaOH solution is needed to react with 20.0 mL of an unknown HCl concentration. What is the concentration of the HCl solution? Moles of NaOH Volume of Solution of NaOH Moles of HCl Volume of Solution of HCl Mole Ratio Between NaOH and HCl Molarity of HCl Molarity of NaOH

Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/35 Titration HCl(aq) + NaOH(aq)NaCl(aq) + 2H 2 O(l) 20.0 mL solution mol HCl = M HCl Concentration of HCl solution: Moles of NaOH available: 1 L mol = mol NaOH 48.6 mL NaOH 1000 mL 1 L Moles of HCl reacted: 1 mol NaOH 1 mol HCl = mol HCl mol NaOH 1 L 1000 mL x x x x

Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/36 Percent Composition and Empirical Formulas Percent Composition: Expressed by identifying the elements present and giving the mass percent of each. Empirical Formula: It tells only the ratios of the atoms in a compound. Molecular Formula: It tells the actual numbers of atoms in a compound. It can be either the empirical formula or a multiple of it. multiple = molecular mass empirical formula mass

Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/37 Percent Composition and Empirical Formulas A colorless liquid has a composition of 84.1 % carbon and 15.9 % hydrogen by mass. Determine the empirical formula. Also, assuming the molar mass of this compound is g/mol, determine the molecular formula of this compound. MolesMass percentsMole ratios Subscripts Relative mole ratios Molar masses

Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/38 Percent Composition and Empirical Formulas Mole of carbon: Assume g of the substance: 12.0 g C 1 mol C 84.1 g C = 7.01 mol C 1.0 g H 1 mol H = 15.9 mol H Mole of hydrogen: 15.9 g H x x

Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/39 Percent Composition and Empirical Formulas Molecular formula: 7.01 C 1x4 H 2.27x4 = C 4 H = C 1 H 2.27 need whole numbers C 1 H 2.27 C 7.01 H 15.9 smallest value for the ratio C 7.01 H 15.9 Empirical formula: = multiple = C 4x2 H 9x2 = C 8 H 18

Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/40 Combustion Analysis: A compound of unknown composition (containing a combination of carbon, hydrogen, and possibly oxygen) is burned with oxygen to produce the volatile combustion products CO 2 and H 2 O, which are separated and weighed by an automated instrument called a gas chromatograph. Determining Empirical Formulas: Elemental Analysis hydrocarbon + O 2 (g)xCO 2 (g) + yH 2 O(g) carbon hydrogen

Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/41 Determining Molecular Masses: Mass Spectrometry