1. Stoichiometry I Equations, The Mole, & Chemical Formulas Chapter 3

2. 3.1 Chemical Equations Objectives Write the balanced equations for chemical reactions given chemical formulas of reactants and products Identify an acid and a base Identify and balance chemical equations for neutralization reactions, combustion reactions, and oxidation-reduction reactions Assign oxidation numbers to elements in simple compounds

3. 3.1 Chemical Equations Chemical equations describe the identities and relative amounts of reactants and products in a chemical reaction Ex: 2 Mg + O 2 → 2 MgO reactants yield products

4. 3.1 Chemical Equations Note use of symbols such as (s) for solid, (l) for liquid, (g) for gas, and (aq) for aqueous solution Note the Law of Conservation of Matter The amount of magnesium entering the reaction is the same as the amount being produced The amount of oxygen entering the reaction is the same as the amount being produced

5. 3.1 Chemical Equations Balancing Chemical Equations The relative amounts of each species in a chemical equation can be adjusted by changing the number that precedes the chemical formula. This number is referred to as the coefficient. Coefficients must be assigned so that the smallest correct set of whole numbers is used Never alter the subscripts in any substances when writing balanced equations!

6. 3.1 Chemical Equations Try to balance the following: C 3 H 8 + O 2 → CO 2 + H 2 O C 4 H 10 + O 2 → CO 2 + H 2 O NaOH + Al + H 2 O → H 2 + NaAlO 2

7. 3.1 Chemical Equations Try to balance the following: C 3 H 8 + 5 O 2 → 3 CO 2 + 4 H 2 O 2 C 4 H 10 + 13 O 2 → 8 CO 2 + 10 H 2 O 2 NaOH + 2 Al + 2 H 2 O → 3 H 2 + 2 NaAlO 2

8. 3.1 Chemical Equations Try to balance the following: Na 2 CO 3 + HNO 3 → CO 2 + H 2 O + NaNO 3 Ca 3 P 2 + H 2 O → PH 3 + Ca(OH) 2

9. 3.1 Chemical Equations Try to balance the following: Na 2 CO 3 + 2 HNO 3 → CO 2 + H 2 O + 2 NaNO 3 Ca 3 P 2 + 6 H 2 O → 2 PH 3 + 3 Ca(OH) 2

10. 3.1 Chemical Equations Try to balance the following: IBr reacts with ammonia to yield nitrogen triiodide and ammonium bromide. Write the balanced equation. 3 IBr + 4 NH 3 → NI 3 + 3 NH 4 Br

11. Types of Chemical Reactions Neutralization Reactions Combustion Reactions Oxidation-Reduction Reactions Precipitation (will cover in Ch 4)

12. 3.1 Chemical Equations Neutralization reactions The reaction of an acid and a base to yield water and a salt Acid = any substance that reacts with water to yield the hydrogen cation (H + ) Ionization = the process by which a neutral substance separates into its component ions in aqueous solution

13. 3.1 Chemical Equations Example of an ionization… HCl (l) → H + (aq) + Cl - (aq) or HCl (l) + H 2 O (l) → H 3 O + (aq) + Cl - (aq)

14. 3.1 Chemical Equations Some common acids are listed in Table 3.1 on p. 97 in the textbook

15. Table 3-1, p. 97

16. 3.1 Chemical Equations Neutralization reactions cont’d …. Base = any substance that provides the hydroxide anion (OH) - Note that this is not the only definition of a base. Later in this course we will revise this simplistic definition NaOH (s) → Na + (aq) + OH - (aq)

17. 3.1 Chemical Equations Note that when you write an equation that contains charged species, the sum of the charges on each side of the equation must be the same, as well as the numbers of atoms of each element NaOH (s) → Na + (aq) + OH - (aq)

18. 3.1 Chemical Equations Neutralization Reactions Ex 1: HCl (aq) + NaOH (aq) → H 2 O (l) + NaCl (aq) Ex 2: H 2 SO 4 (aq) + Ca(OH) 2 → H 2 O (l) + CaSO 4 (s) * Note that in each example, the reaction of an acid and a base produces water and a salt H + + OH - → H 2 O Na + + Cl - → NaCl Salt = an ionic compound composed of a cation from a base and an anion from an acid

19. 3.1 Chemical Equations Sample Questions…. Write the equation for the reaction of nitric acid with magnesium hydroxide ….. Write the equation for the reaction of hydrochloric acid with calcium hydroxide…

20. 3.1 Chemical Equations Answers…. 2 HNO 3 + Mg(OH) 2 → 2 H 2 O + Mg(NO 3 ) 2 2 HCl + Ca(OH) 2 → 2 H 2 O + CaCl 2

21. 3.1 Chemical Equations Combustion reactions The process of burning, usually in the presence of oxygen Are actually a special class of oxidation- reduction reactions Combustion of organic compounds typically results in the release of a great amount of heat along with CO 2 and H 2 O Ex: C 3 H 8 + O 2 → CO 2 + H 2 O

22. C 3 H 8 + O 2 → CO 2 + H 2 O

23. 3.1 Chemical Equations Oxidation Reduction or Redox Reactions (covered in detail in Ch 18) Reactions in which electrons are transferred from one species to another One species loses electrons and is oxidized, one species gains electrons and is reduced To help you remember LEO the lion says GER OIL RIG

24. 3.1 Chemical Equations Oxidation-Reduction Reactions cont’d…. Ex: 4 Li (s) + O 2 (g) → 2Li 2 O Li goes from an oxidation state of 0 to an oxidation state of +1 / loses an electron / oxidation O goes from an oxidation state of zero to an oxidation state of -2 / gains electrons / reduction

25. 3.1 Chemical Equations Rules for Assigning Oxidation Numbers An atom in its elemental state has an oxidation number of zero. Monoatomic ions in ionic compounds have an oxidation number equal to the charge of the ion. In compounds, fluorine has the oxidation number -1; oxygen is generally -2; hydrogen combined with non- metals is generally +1, and -1 when combined with metals. Other halogens are generally -1

26. 3.1 Chemical Equations Rules for Assigning Oxidation Numbers, cont’d All other atoms are assigned oxidation numbers so that the sum of the oxidation numbers for all of the atoms in a species is equal to the charge of the species Let’s practice … Assign oxidation numbers to each atom in CO 2 NO 3 - H 2 SO 4 Ag NaClO 3

27. 3.1 Chemical Equations Identify the type of the reaction, then balance it. (a) Na(s) + Cl 2 (g)  NaCl(s) (b) (C 2 H 5 ) 2 O + O 2  CO 2 + H 2 O (c) H 2 SO 4 + Ca(OH) 2  CaSO 4 + H 2 O

28. 3.1 Chemical Equations Answer: (a) Oxidation Reduction 2Na(s) + Cl 2 (g)  2NaCl(s) (b) Combustion (C 2 H 5 ) 2 O + 6O 2  4CO 2 + 5H 2 O (c) Neutralization H 2 SO 4 + Ca(OH) 2  CaSO 4 + 2H 2 O

29. 3.2 The Mole and Molar Mass Objectives Express the amounts of substances using moles Determine the molar mass of any element or compound from its formula Use the molar mass and Avagadro’s number to interconvert between mass, moles, and numbers of atoms, ions, or molecules

30. 3.2 The Mole and Molar Mass One mole of a substance is the amount of a substance that contains as many entities as the number of atoms in exactly 12 g of the isotope 12 C The mole is a unit of quantity One mole of a substance contains 6.022 x 10 23 particles One mol of O = 6.022 x 10 23 atoms of O One mol of H 2 O = 6.022 x 10 23 molecules of H 2 O

31. One mole of some common substances Iron Water Copper Aluminum Sodium

32. 3.2 The Mole and Molar Mass The focus of this section is on completing conversions. To be able to do this you need to: Understand & memorize Avagadro’s number be able to calculate molar masses

33. p. 106

34. Relationship between atomic mass, molecular mass, or formula mass and molar mass

35. 3.2 The Mole and Molar Mass Let’s Practice….. Moles to Atoms/Molecules Conversion of Moles to Mass Conversion of Mass to Moles Conversion of Mass to Atoms/Molecules Conversion of Atoms/Molecules to Mass

36. 3.2 The Mole and Molar Mass The general format used to solve all conversion problems listed in the previous slide …. molar mass Avagadro’s of A # mass A ↔ mol A ↔ atoms or molecules A

37. Problem Solving for 3.2 The Mole and Molar Mass Be sure to complete as many odd number problems from the textbook as you feel you need to in order to master the concepts presented. Remember, the answers to all odd-numbered problems are found in an appendix in the back of the book.

38. 3.3 Chemical Formulas Objectives Calculate the mass percentage of each element in a compound from the chemical formula of that compound Calculate the mass of each element present in a sample from elemental analysis data such as that produced by combustion analysis Determine the empirical formula of a compound from mass or mass percentage data Use molar mass to determine a molecular formula from an empirical formula

39. 3.3 Chemical Formulas When a new compound is discovered, one of the first tests done may be determining its percent composition….

40. 3.3 Chemical Formulas Chemical Formulas/Percent Composition Use the formula of the compound to calculate the mass of each element in the compound and use those numbers to calculate percentage composition.

41. Formula to mass percent Mass percent to formula Subscripts in formula Masses of elements and compound Percent composition Atomic masses Mass of element Mass of compound x 100% Composition (mass or mass %) Molar masses of elements Moles of each element Empirical formula Divide by smallest number Helpful steps for solving a % composition problem

42. 3.3 Chemical Formulas What is the mass percentage composition of H 2 C 2 O 4 ? Answer H = (2/90) x 100% = 2.22% C = (24/90) x 100% = 26.67% O = (64/90) x 100% = 70.11%

43. 3.3 Chemical Formulas One way to obtain % composition data and empirical formulas for organic compounds is to complete a combustion analysis This process involves burning a sample in excess oxygen and determining the amount of CO 2 and H 2 O generated The percentage of carbon and hydrogen in the sample can be calculated from the measured masses of CO 2 and H 2 O

44. 3.3 Chemical Formulas A Combustion Train The sample burns in excess O 2 H 2 O is trapped by the CaCl 2 and CO 2 is trapped by the NaOH.

45. 3.3 Chemical Formulas Combustion of a 0.2000 gram sample of a compound made up of only carbon, hydrogen, and oxygen yields 0.200 g of H 2 O and 0.4880 g of CO 2. Calculate the mass and mass percentage of each element present in the 0.2000 g sample. A: 0.1332 g C, 0.0224 g H, and 0.0444 g O 66.6 % C, 11.2% H, and 22.2% O

46. 3.3 Chemical Formulas Once percent composition data has been obtained from combustion analysis, empirical and molecular formulas can be calculated Empirical Formulas give the smallest whole number ratio of elements in a compound Ex: HO Molecular Formulas give the true ratio Ex: H 2 O 2

47. Molar masses of elements Formula to mass percent Mass percent to formula Subscripts in formula Masses of elements and compound Percent composition Atomic masses Mass of element Mass of compound x 100% Composition (mass or mass %) Moles of each element Empirical formula Divide by smallest number Steps to help you determine an empirical formula

48. 3.3 Chemical Formulas Empirical Formula = CH 3 What is the empirical formula of a compound that contains 0.799 g C and 0.201 g H in a 1.000 g sample?

49. 3.3 Chemical Formulas What is the empirical formula of a compound that is 59.9% Ti and 40.1% O? Answer: TiO 2 Ti = 59/47.87 = 1.23 ; 1.23/1.23 = 1 O = 40.1/16.00 = 2.51; 2.51/1.23 ~ 2

50. 3.3 Chemical Formulas Combustion of a 0.2000 gram sample of a compound made up of only carbon, hydrogen, and oxygen yields 0.200 g of H 2 O and 0.4880 g of CO 2. Calculate the empirical formula of the 0.2000 g sample. Earlier data: 0.1332 g C, 0.0224 g H, and 0.0444 g O 66.6 % C, 11.2% H, and 22.2% O So empirical formula = C 4 H 8 O

51. 3.3 Chemical Formulas As mentioned earlier, empirical formulas simply tell you the smallest whole number ratio of elements in a compound Empirical formulas may be the same as the molecular formula of a compound or they may be different. To calculate a molecular formula, the molar mass of a compound must be known

52. 3.3 Chemical Formulas/Molecular Formulas The molecular formula must be a whole number multiple of the empirical formula. If the empirical formula is CH 2, the molecular formula is (CH 2 ) n where The molecular mass must be measured experimentally.

53. 3.3 Chemical Formulas Empirical & Molecular Formulas, cont’d An empirical formula calculation based on a combustion analysis experiment shows a new compound has the empirical formula C 4 H 8 O. A mass spectrometry experiment determines that the molar mass of the compound is 216 g/mol. What is the molecular formula? A: C 12 H 24 O 3

54. 3.4 Mass Relationships in Chemical Equations Objectives Calculate the mass of a product formed or a reactant consumed in a chemical reaction Determine theoretical yields from reaction data

55. 3.4 Mass Relationships in Equations 3.2 The Mole and Molar Mass

56. 3.4 Mass Relationships in Chemical Equations Based on stoichiometric principles, given the amount of one substance in a chemical equation we can calculate the amount of a second substance that is needed to react with the first, or we can calculate the amount of product produced assuming a second reactant is in excess, or given an amount of product produced, we can calculate how much of each reactant was needed

57. 3.4 Mass Relationships in Chemical Equations The schematic on the following slide should help you get from point A to point B no matter what type of stoichiometry problem you are presented with ….

58. 3.4 Mass Relationships in Chemical Equations molar mass ratio of molar mass of A coefficients of B in balanced equation mass A ↔ mol A ↔ mol B ↔ mass B Note that A does not have to be a reactant and B a product. Rather A represents the given quantity and B the unknown quantity that needs to be calculated

59. 3.4 Mass Relationships in Chemical Equations Important definition … Theoretical yield = the maximum amount of product that can be produced from a given amount of reactants in a chemical reaction

60. 3.4 Mass Relationships in Chemical Equations Calculate the mass of oxygen needed to react with 4.1 g sulfur dioxide to produce sulfur trioxide. Calculate the theoretical yield of sulfur trioxide that will form. A: 2SO 2 + O 2 → 2SO 3 O 2 = 1.0 g ; SO 3 = 5.1 g

61. 3.5 Limiting Reactants Objectives Identify the limiting reactant in a chemical reaction and use it to determine theoretical yields of products that form in a chemical reaction Determine the percent yields in chemical reactions

62. 3.5 Limiting Reactants What happens when reactants are not present in stoichiometric amounts? Oftentimes, one reactant may be in excess but the other reactant is in limited supply. Ex: 3 hotdogs/2buns or 3 mol Na/1mol Cl 2

63. 3.5 Limiting Reactants The limiting reactant determines the maximum amount of product that can be formed in the reaction. The limiting reactant is determined by calculating the number of moles of product formed from the given quantity of each reactant. The reactant that yields the least amount of product is the limiting reactant.

64. 3.5 Limiting Reactants N 2 + H 2 → NH 3

65. 3.5 Limiting Reactants Antimony + Cl 2 → SbCl 3

66. 3.5 Limiting Reactants A reaction is conducted with 20.0 g H 2 and 99.8 g O 2 to produce water. State the limiting reactant. Calculate the mass, in grams, of H 2 O that can be produced from this reaction.

67. An example of a flow chart for solving a limited reactant problem

68. A reaction is conducted with 20.0 g H 2 and 99.8 g O 2 to produce water. State the limiting reactant. Calculate the mass, in grams, of H 2 O that can be produced from this reaction. Answer: O 2 / 112 g H 2 O

69. 3.5 Limiting Reactants An added level of difficulty based on the previous problem….. A reaction is conducted with 20.0 g H 2 and 99.8 g O 2 to produce water. Now figure out how much excess reagent remains at the end of the reaction? Answer: H 2 is in excess. The amount of H 2 consumed is 12.5 g. The amount of H 2 in excess is 7.5 g.

70. 3.5 Limiting Reactants Given the following equation, calculate the mass in grams of AlCl 3 that can be produced from 4.40 g Al and 12.0 g Cl 2. Al + Cl 2 → AlCl 3

71. 3.5 Limiting Reactants Given the following equation, calculate the mass in grams of AlCl 3 that can be produced from 4.40 g Al and 12.0 g Cl 2. Al + Cl 2 → AlCl 3 Did you balance the equation first? Answer: 15.0 g AlCl 3

72. 3.5 Limiting Reactants Very few chemical reactions actually go to completion. One of the practical goals in commercial chemical operations is to determine how to increase the yield of those reactions that do not go to completion.

73. 3.5 Limiting Reactants Actual and Percent Yield Theoretical yield = the maximum amount of product that can be produced from a given amount of reactants in a chemical reaction The actual yield = the amount of product isolated at the completion of a reaction percent yield = actual yield x 100% theoretical yield

74. 3.5 Limiting Reactants What is the percent yield if 2.4 g NH 3 is obtained from the reaction of 0.64 g H 2 with excess N 2 ? A: 67%

75. Chapter 3 Review Question … In an experiment, 44 g NH 3 is mixed with 120 g O 2 and 73 g NO is isolated. Given the following equation: NH 3 + O 2 → NO + H 2 O what is the percent yield? A: 94%