1. John E. McMurry Robert C. Fay Lecture Notes Alan D. Earhart Southeast Community College Lincoln, NE General Chemistry: Atoms First Chapter 6 Mass Relationships in Chemical Reactions Copyright © 2010 Pearson Prentice Hall, Inc.

2. Chapter 6/2 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

3. Chapter 6/3 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

4. Chapter 6/4 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)

5. Chapter 6/5 Balancing Chemical Equations 3.Reduce the coefficients to their smallest whole- number values, if necessary, by dividing them all by a common divisor. 2H 2 O(l)2H 2 (g) + O 2 (g) 4H 2 O(l)4H 2 (g) + 2O 2 (g) divide all by 2

6. Chapter 6/6 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

7. Chapter 6/7 Chemical Symbols on a Different Level 2H 2 O(l)2H 2 (g) + O 2 (g) 2 molecules of hydrogen gas react with 1 molecule of oxygen gas to yield 2 molecules of liquid water. microscopic:

8. Chapter 6/8 Chemical Symbols on a Different Level 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:

9. Chapter 6/9 Chemical Arithmetic: Stoichiometry

10. Chapter 6/10 2(12.0 amu) + 4(1.0 amu) = 28.0 amuC2H4:C2H4: Chemical Arithmetic: Stoichiometry HCl:1.0 amu + 35.5 amu = 36.5 amu 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.

11. Chapter 6/11 1 mole = 28.0 gC2H4:C2H4: 6.022 x 10 23 molecules = 28.0 g Chemical Arithmetic: Stoichiometry One mole of any substance is equivalent to its molecular or formula mass. 1 mole = 36.5 g 6.022 x 10 23 molecules = 36.5 g HCl:

12. Chapter 6/12 Chemical Arithmetic: Stoichiometry How many moles of chlorine gas, Cl 2, are in 25.0 g? 25.0 g Cl 2 70.9 g Cl 2 1 mol Cl 2 x= 0.353 mol Cl 2

13. Chapter 6/13 Chemical Arithmetic: Stoichiometry How many grams of sodium hypochlorite, NaOCl, are in 0.705 mol? 0.705 mol NaOCl 1 mol NaOCl 40.0 g NaOCl x= 28.2 g NaOCl

14. Chapter 6/14 Chemical Arithmetic: Stoichiometry 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

15. Chapter 6/15 Chemical Arithmetic: Stoichiometry 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

16. Chapter 6/16 Chemical Arithmetic: Stoichiometry 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 2 70.9 g Cl 2 1 mol Cl 2 2 mol NaOH = 28.2 g NaOH xxx

17. Chapter 6/17 Yields of Chemical Reactions The amount actually formed in a reaction. The amount predicted by calculations. Actual Yield: Theoretical Yield: Actual yield of product Theoretical yield of product x 100%Percent Yield =

18. Chapter 6/18 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.

19. Chapter 6/19 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)

20. Chapter 6/20 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:

21. Chapter 6/21 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)

22. Chapter 6/22 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?

23. Chapter 6/23 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

24. Chapter 6/24 Reactions with Limiting Amounts of Reactants Li 2 O(s) + H 2 O(g)2LiOH(s) 80.0 g H 2 O - 39.1 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

25. Chapter 6/25 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

26. Chapter 6/26 Concentrations of Reactants in Solution: Molarity Molarity (M): 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.

27. Chapter 6/27

28. Chapter 6/28 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 = 1.00 1.00 mol of sodium chloride placed in enough water to make 1.00 L of solution would have a concentration equal to:

29. Chapter 6/29 Concentrations of Reactants in Solution: Molarity Molar mass C 6 H 12 O 6 = 180.0 g/mol How many grams of solute would you use to prepare 1.50 L of 0.250 M glucose, C 6 H 12 O 6 ? 1 mol 0.275 mol180.0 g = 49.5 g 1 L 1.50 L0.250 mol = 0.275 mol x x

30. Chapter 6/30

31. Chapter 6/31 Diluting Concentrated Solutions dilute solutionconcentrated solution + solvent Since the number of moles of solute remains constant, all that changes is the volume of solution by adding more solvent. M i x V i = M f x V f finalinitial

32. Chapter 6/32 Diluting Concentrated Solutions Add 6.94 mL 18.0 M sulfuric acid to enough water to make 250.0 mL of 0.500 M solution. M i = 18.0 MM f = 0.500 M V i = ? mLV f = 250.0 mL = 6.94 mL 18.0 M 250.0 mL V i = VfVf 0.500 M = Sulfuric acid is normally purchased at a concentration of 18.0 M. How would you prepare 250.0 mL of 0.500 M aqueous H 2 SO 4 ? x MiMi MfMf x

33. Chapter 6/33 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

34. Chapter 6/34 Solution Stoichiometry H 2 SO 4 (aq) + 2NaOH(aq)Na 2 SO 4 (aq) + 2H 2 O(l) What volume of 0.250 M H 2 SO 4 is needed to react with 50.0 mL of 0.100 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

35. Chapter 6/35 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 4 0.250 mol H 2 SO 4 1 L solution 1 L 0.100 mol 1 L 1000 mL = 0.00500 mol NaOH Volume of H 2 SO 4 needed: 1000 mL 1 L 10.0 mL solution (0.250 M H 2 SO 4 ) 0.00500 mol NaOH 50.0 mL NaOH Moles of NaOH available: x x x xx

36. Chapter 6/36 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.

37. Chapter 6/37 Titration unknown concentration solution Erlenmeyer flask buret standard solution (known concentration) An indicator is added which changes color once the reaction is complete

39. Chapter 6/39 Titration HCl(aq) + NaOH(aq)NaCl(aq) + 2H 2 O(l) 48.6 mL of a 0.100 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

40. Chapter 6/40 Titration HCl(aq) + NaOH(aq)NaCl(aq) + 2H 2 O(l) 20.0 mL solution 0.00486 mol HCl = 0.243 M HCl Concentration of HCl solution: Moles of NaOH available: 1 L 0.100 mol = 0.00486 mol NaOH 48.6 mL NaOH 1000 mL 1 L Moles of HCl reacted: 1 mol NaOH 1 mol HCl = 0.00486 mol HCl 0.00486 mol NaOH 1 L 1000 mL x x x x

41. Chapter 6/41 Percent Composition and Empirical Formulas Percent Composition: Expressed by identifying the elements present and giving the mass percent of each. Empirical Formula: It tells the smallest whole-number ratios of 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

42. Chapter 6/42 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 114.2 g/mol, determine the molecular formula of this compound. MolesMass percentsMole ratios Subscripts Relative mole ratios Molar masses

43. Chapter 6/43 Percent Composition and Empirical Formulas Mole of carbon: Assume 100.0 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

44. Chapter 6/44 Percent Composition and Empirical Formulas Molecular formula: 7.01 C 1x4 H 2.27x4 = C 4 H 9 7.01 = 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: = 2 57.0 114.2 multiple = C 4x2 H 9x2 = C 8 H 18

45. Chapter 6/45 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