1. Copyright  2011 Pearson Education, Inc. Chapter 3 Molecules, Compounds, and Chemical Equations Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA Chemistry: A Molecular Approach, 2nd Ed. Nivaldo Tro

2. Copyright  2011 Pearson Education, Inc. Elements and Compounds Elements combine together to make an almost limitless number of compounds The properties of the compound are totally different from the constituent elements 2 Tro: Chemistry: A Molecular Approach, 2/e

3. Copyright  2011 Pearson Education, Inc. Formation of Water from Its Elements 3 Tro: Chemistry: A Molecular Approach, 2/e

4. Copyright  2011 Pearson Education, Inc. Chemical Bonds Compounds are made of atoms held together by bonds Chemical bonds are forces of attraction between atoms The bonding attraction comes from attractions between protons and electrons 4 Tro: Chemistry: A Molecular Approach, 2/e

5. Copyright  2011 Pearson Education, Inc. Bond Types Two general types of bonding between atoms found in compounds, ionic and covalent Ionic bonds result when electrons have been transferred between atoms, resulting in oppositely charged ions that attract each other generally found when metal atoms bond to nonmetal atoms Covalent bonds result when two atoms share some of their electrons generally found when nonmetal atoms bond together 5 Tro: Chemistry: A Molecular Approach, 2/e

6. Copyright  2011 Pearson Education, Inc. 6 Tro: Chemistry: A Molecular Approach, 2/e

7. Copyright  2011 Pearson Education, Inc. Formulas Describe Compounds A compound is a distinct substance that is composed of atoms of two or more elements Describe the compound by describing the number and type of each atom in the simplest unit of the compound molecules or ions Each element is represented by its letter symbol The number of atoms of each element is written to the right of the element as a subscript if there is only one atom, the 1 subscript is not written Polyatomic ions are placed in parentheses if more than one 7 Tro: Chemistry: A Molecular Approach, 2/e

8. Copyright  2011 Pearson Education, Inc. Representing Compounds with Chemical Formula Compounds are generally represented with a chemical formula The amount of information about the structure of the compound varies with the type of formula all formula and models convey a limited amount of information – none are perfect representations All chemical formulas tell what elements are in the compound use the letter symbol of the element 8 Tro: Chemistry: A Molecular Approach, 2/e

9. Copyright  2011 Pearson Education, Inc. Types of Formula: Empirical Formula An empirical formula gives the relative number of atoms of each element in a compound it does not describe how many atoms, the order of attachment, or the shape the formulas for ionic compounds are empirical The empirical formula for the ionic compound fluorspar is CaCl 2. This means that there is 1 Ca 2+ ion for every 2 Cl − ions in the compound. The empirical formula for the molecular compound oxalic acid is CHO 2. This means that there is 1 C atom and 1 H atom for every 2 O atoms in the molecule. The actual molecular formula is C 2 H 2 O 4. 9 Tro: Chemistry: A Molecular Approach, 2/e

10. Copyright  2011 Pearson Education, Inc. Types of Formula: Molecular Formula A molecular formula gives the actual number of atoms of each element in a molecule of a compound it does not describe the order of attachment, or the shape The molecular formula is C 2 H 2 O 4. This does not tell you that the carbon atoms are attached together in the center of the molecule, and that each is attached to two oxygen atoms. 10 Tro: Chemistry: A Molecular Approach, 2/e

11. Copyright  2011 Pearson Education, Inc. Types of Formula: Structural Formula A structural formula uses lines to represent covalent bonds and shows how atoms in a molecule are connected or bonded to each other it does not directly describe the 3-dimensional shape, but an experienced chemist can make a good guess at it each line describes the number of electrons shared by the bonded atoms  single line = two shared electrons, a single covalent bond  double line = four shared electrons, a double covalent bond  triple line = six shared electrons, a triple covalent bond 11 Tro: Chemistry: A Molecular Approach, 2/e

12. Copyright  2011 Pearson Education, Inc. Structural Formula of Oxalic Acid 12 Tro: Chemistry: A Molecular Approach, 2/e

13. Copyright  2011 Pearson Education, Inc. Representing Compounds: Molecular Models Models show the 3-dimensional structure along with all the other information given in the structural formula Ball-and-stick models use balls to represent the atoms and sticks to represent the attachments between them Space-filling models use interconnected spheres to show the electron clouds of atoms connecting together 13 Tro: Chemistry: A Molecular Approach, 2/e

14. Copyright  2011 Pearson Education, Inc. Models of Oxalic Acid Ball and stick Space filling 14 Tro: Chemistry: A Molecular Approach, 2/e

15. Copyright  2011 Pearson Education, Inc. 15 Tro: Chemistry: A Molecular Approach, 2/e

16. Copyright  2011 Pearson Education, Inc. Practice — Find the empirical formula for each of the following The ionic compound that has two aluminum ions for every three oxide ions arabinose, C 5 H 10 O 5 pyrimidine ethylene glycol Al 2 O 3 CH 2 O C2H2NC2H2N CH 3 O 16 Tro: Chemistry: A Molecular Approach, 2/e

17. Copyright  2011 Pearson Education, Inc. Molecular View of Elements and Compounds 17 Tro: Chemistry: A Molecular Approach, 2/e

18. Copyright  2011 Pearson Education, Inc. Classifying Elements & Compounds 18 Tro: Chemistry: A Molecular Approach, 2/e Atomic elements = elements whose particles are single atoms Molecular elements = elements whose particles are multi-atom molecules Molecular compounds = compounds whose particles are molecules made of only nonmetals Ionic compounds = compounds whose particles are cations and anions

19. Copyright  2011 Pearson Education, Inc. Elements Most elements have single atoms as their constituent particles The atoms may be physically attracted to each other, but are not chemically bonded together A few elements have molecules as their constituent particles The molecules are made of two or more atoms chemically bonded together by covalent bonds 19 Tro: Chemistry: A Molecular Approach, 2/e

20. Copyright  2011 Pearson Education, Inc. Molecular Elements H2H2 Cl 2 Br 2 I2I2 7 7A N 2 O 2 F 2 20 Tro: Chemistry: A Molecular Approach, 2/e Certain elements occur as 2 atom molecules rule of 7’s Other elements occur as polyatomic molecules P 4, S 8, Se 8

21. Copyright  2011 Pearson Education, Inc. Molecular Elements 21 Tro: Chemistry: A Molecular Approach, 2/e

22. Copyright  2011 Pearson Education, Inc. Compounds Some compounds are composed of ions arranged in a 3-dimensional pattern – these are called ionic compounds each cation is surrounded by anions, and vice-versa Other compounds are composed of individual molecule units Each molecule contains atoms of different elements chemically attached by covalent bonds 22 Tro: Chemistry: A Molecular Approach, 2/e

23. Copyright  2011 Pearson Education, Inc. Ionic vs. Molecular Compounds Propane – contains individual C 3 H 8 molecules Table salt – contains an array of Na + ions and Cl - ions 23 Tro: Chemistry: A Molecular Approach, 2/e

24. Copyright  2011 Pearson Education, Inc. Classify Each of the Following as Either an Atomic Element, Molecular Element, Molecular Compound, or Ionic Compound Aluminum, Al Aluminum chloride, AlCl 3 Chlorine, Cl 2 Acetone, C 3 H 6 O Carbon monoxide, CO Cobalt, Co 24 atomic element ionic compound molecular element molecular compound atomic element Tro: Chemistry: A Molecular Approach, 2/e

25. Copyright  2011 Pearson Education, Inc. Ionic Compounds Compounds of metals with nonmetals are made of ions metal atoms form cations, nonmetal atoms form anions No individual molecule units, instead they have a 3-dimensional array of cations and anions made of formula units Many contain polyatomic ions several atoms attached together by covalent bonds into one ion 25 Tro: Chemistry: A Molecular Approach, 2/e

26. Copyright  2011 Pearson Education, Inc. Compounds that Contain Ions Compound must have no total charge, therefore we must balance the numbers of cations and anions in a compound to get 0 charge If Na + is combined with S 2−, you will need two Na + ions for every S 2− ion to balance the charges, therefore the formula must be Na 2 S 26 Tro: Chemistry: A Molecular Approach, 2/e

27. Copyright  2011 Pearson Education, Inc. Example 3.4: Write a formula for ionic compound that forms between calcium and oxygen 27 1.Write the symbol for the metal cation and its charge followed by the symbol for the nonmetal anion and its charge. Obtain charges from the element’s group number on the periodic table. Ca 2+ O 2− 2.Adjust the subscript on each cation and anion to balance the overall charge. CaO 3.Check that the sum of the charges of the cations equals the sum of the charges of the anions. cations: +2 anions: −2 The charges cancel. Tro: Chemistry: A Molecular Approach, 2/e

28. Copyright  2011 Pearson Education, Inc. Writing Formulas for Ionic Compounds 28 Tro: Chemistry: A Molecular Approach, 2/e 1. Write the symbol for the metal cation and its charge 2. Write the symbol for the nonmetal anion and its charge 3. Charge (without sign) becomes subscript for other ion 4. Reduce subscripts to smallest whole number ratio 5. Check that the sum of the charges of the cations cancels the sum of the anions

29. Copyright  2011 Pearson Education, Inc. Example 3.3: Write the formula of a compound made from aluminum ions and oxide ions 1. Write the symbol for the metal cation and its charge 2. Write the symbol for the nonmetal anion and its charge 3. Charge (without sign) becomes subscript for other ion 4. Reduce subscripts to smallest whole number ratio 5. Check that the total charge of the cations cancels the total charge of the anions Al 3+ column 3A O 2− column 6A Al +3 O 2− Al 2 O 3 Al = (2)∙(+3) = +6 O = (3)∙(−2) = −6 29 Tro: Chemistry: A Molecular Approach, 2/e

30. Copyright  2011 Pearson Education, Inc. Practice — What are the formulas for compounds made from the following ions? Potassium ion with a nitride ion Calcium ion with a bromide ion Aluminum ion with a sulfide ion 30 Tro: Chemistry: A Molecular Approach, 2/e

31. Copyright  2011 Pearson Education, Inc. Practice — What are the formulas for compounds made from the following ions? K + with N 3− K 3 N Ca 2+ with Br − CaBr 2 Al 3+ with S 2− Al 2 S 3 31 Tro: Chemistry: A Molecular Approach, 2/e

32. Copyright  2011 Pearson Education, Inc. Formula-to-Name Rules for Ionic Compounds 32 Tro: Chemistry: A Molecular Approach, 2/e Made of cation and anion Some have one or more nicknames that are only learned by experience NaCl = table salt, NaHCO 3 = baking soda Write systematic name by simply naming the ions if cation is:  metal with invariant charge = metal name  metal with variable charge = metal name(charge)  polyatomic ion = name of polyatomic ion if anion is:  nonmetal = stem of nonmetal name + ide  polyatomic ion = name of polyatomic ion

33. Copyright  2011 Pearson Education, Inc. Metals with variable Charges metals whose ions can have more than one possible charge determine charge by charge on anion and cation name = metal name with Roman numeral charge in parentheses Metals with invariant charge metals whose ions can only have one possible charge  Groups 1A 1+ & 2A 2+, Al 3+, Ag 1+, Zn 2+, Sc 3+ cation name = metal name Naming Metal Cations 33 Tro: Chemistry: A Molecular Approach, 2/e

34. Copyright  2011 Pearson Education, Inc. Naming Monatomic Nonmetal Anion 34 Tro: Chemistry: A Molecular Approach, 2/e Determine the charge from position on the Periodic Table To name anion, change ending on the element name to –ide

35. Copyright  2011 Pearson Education, Inc. Naming Binary Ionic Compounds for Metals with Invariant Charge 35 Tro: Chemistry: A Molecular Approach, 2/e Contain metal cation + nonmetal anion Metal listed first in formula and name 1.name metal cation first, name nonmetal anion second 2.cation name is the metal name 3.nonmetal anion named by changing the ending on the nonmetal name to -ide

36. Copyright  2011 Pearson Education, Inc. Example: Naming Binary Ionic with Invariant Charge Metal CsF 36 Tro: Chemistry: A Molecular Approach, 2/e 1.Identify cation and anion Cs = Cs + because it is Group 1A F = F − because it is Group 7A 2.Name the cation Cs + = cesium 3.Name the anion F − = fluoride 4.Write the cation name first, then the anion name cesium fluoride

37. Copyright  2011 Pearson Education, Inc. Practice — Name the following compounds 1.KCl 2.MgBr 2 3.Al 2 S 3 potassium chloride magnesium bromide aluminum sulfide 37 Tro: Chemistry: A Molecular Approach, 2/e

38. Copyright  2011 Pearson Education, Inc. Naming Binary Ionic Compounds for Metals with Variable Charge Tro: Chemistry: A Molecular Approach, 2/e Contain metal cation + nonmetal anion Metal listed first in formula and name 1.name metal cation first, name nonmetal anion second 2.metal cation name is the metal name followed by a Roman numeral in parentheses to indicate its charge determine charge from anion charge common ions Table 3.4 3.nonmetal anion named by changing the ending on the nonmetal name to -ide

39. Copyright  2011 Pearson Education, Inc. Determining the Charge on a Cation with Variable Charge — Au 2 S 3 39 Tro: Chemistry: A Molecular Approach, 2/e 1.Determine the charge on the anion Au 2 S 3 : the anion is S, since it is in Group 6A, its charge is 2 − 2.Determine the total negative charge since there are 3 S in the formula, the total negative charge is 6 − 3.Determine the total positive charge since the total negative charge is 6 −, the total positive charge is 6 + 4.Divide by the number of cations since there are 2 Au in the formula and the total positive charge is 6 +, each Au has a 3 + charge

40. Copyright  2011 Pearson Education, Inc. Practice — Find the charge on the cation 1.TiCl 4 2.CrO 3 3.Fe 3 N 2 4 Cl = 4−,  Ti = 4+ 3 O = 6−,  Cr = 6+ 2 N = 6−,  3 Fe = 6+, Fe = 2+ 40 Tro: Chemistry: A Molecular Approach, 2/e

41. Copyright  2011 Pearson Education, Inc. Example: Naming binary ionic with variable charge metal CuF 2 41 Tro: Chemistry: A Molecular Approach, 2/e 1.Identify cation and anion F = F − because it is Group 7 Cu = Cu 2+ to balance the two (−) charges from 2 F − 2.Name the cation Cu 2+ = copper(II) 3.Name the anion F − = fluoride 4.Write the cation name first, then the anion name copper(II) fluoride

42. Copyright  2011 Pearson Education, Inc. Name the following compounds 1.TiCl 4 2.PbBr 2 3.Fe 2 S 3 titanium(IV) chloride lead(II) bromide iron(III) sulfide 42 Tro: Chemistry: A Molecular Approach, 2/e

43. Copyright  2011 Pearson Education, Inc. Example: Writing formula for binary ionic compounds containing variable charge metal manganese(IV) sulfide 1.Write the symbol for the cation and its charge 2.Write the symbol for the anion and its charge 3.Charge (without sign) becomes subscript for other ion 4.Reduce subscripts to smallest whole number ratio 5.Check that the total charge of the cations cancels the total charge of the anions Mn 4+ S 2- Mn 4+ S 2− Mn 2 S 4 Mn = (1)∙(4+) = +4 S = (2)∙(2 − ) = −4 MnS 2 43 Tro: Chemistry: A Molecular Approach, 2/e

44. Copyright  2011 Pearson Education, Inc. Practice — What are the formulas for compounds made from the following ions? copper(II) ion with a nitride ion iron(III) ion with a bromide ion Cu 2+ with N 3− Cu 3 N 2 Fe 3+ with Br − FeBr 3 44 Tro: Chemistry: A Molecular Approach, 2/e

45. Copyright  2011 Pearson Education, Inc. Compounds Containing Polyatomic Ions 45 Tro: Chemistry: A Molecular Approach, 2/e Polyatomic ions are single ions that contain more than one atom Often identified by parentheses around ion in formula Name and charge of polyatomic ion do not change Name any ionic compound by naming cation first and then anion

46. Copyright  2011 Pearson Education, Inc. Some Common Polyatomic Ions 46 Tro: Chemistry: A Molecular Approach, 2/e

47. Copyright  2011 Pearson Education, Inc. Patterns for Polyatomic Ions 47 Tro: Chemistry: A Molecular Approach, 2/e 1.Elements in the same column form similar polyatomic ions same number of O’s and same charge ClO 3 − = chlorate  BrO 3 − = bromate 2.If the polyatomic ion starts with H, add hydrogen- prefix before name and add 1 to the charge CO 3 2− = carbonate  HCO 3 − = hydrogen carbonate

48. Copyright  2011 Pearson Education, Inc. Periodic Pattern of Polyatomic Ions -ate groups 48 Tro: Chemistry: A Molecular Approach, 2/e

49. Copyright  2011 Pearson Education, Inc. Patterns for Polyatomic Ions 49 Tro: Chemistry: A Molecular Approach, 2/e -ate ion chlorate = ClO 3 − -ate ion + 1 O  same charge, per- prefix perchlorate = ClO 4 − -ate ion – 1 O  same charge, -ite suffix chlorite = ClO 2 − -ate ion – 2 O  same charge, hypo- prefix, - ite suffix hypochlorite = ClO −

50. Copyright  2011 Pearson Education, Inc. Example: Naming ionic compounds containing a polyatomic ion Na 2 SO 4 50 Tro: Chemistry: A Molecular Approach, 2/e 1.Identify the ions Na = Na + because in Group 1A SO 4 = SO 4 2− a polyatomic ion 2.Name the cation Na + = sodium, metal with invariant charge 3.Name the anion SO 4 2− = sulfate 4.Write the name of the cation followed by the name of the anion sodium sulfate

51. Copyright  2011 Pearson Education, Inc. Example: Naming ionic compounds containing a polyatomic ion Fe(NO 3 ) 3 51 Tro: Chemistry: A Molecular Approach, 2/e 1.Identify the ions NO 3 = NO 3 − a polyatomic ion Fe = Fe 3+ to balance the charge of the 3 NO 3 − 2.Name the cation Fe 3+ = iron(III), metal with variable charge 3.Name the anion NO 3 − = nitrate 4.Write the name of the cation followed by the name of the anion iron(III) nitrate

52. Copyright  2011 Pearson Education, Inc. Name the Following Compounds 1.NH 4 Cl 2.Ca(C 2 H 3 O 2 ) 2 3.Cu(NO 3 ) 2 ammonium chloride calcium acetate copper(II) nitrate 52 Tro: Chemistry: A Molecular Approach, 2/e

53. Copyright  2011 Pearson Education, Inc. Example – Writing formula for ionic compounds containing polyatomic ion Iron(III) phosphate 1.Write the symbol for the cation and its charge 2.Write the symbol for the anion and its charge 3.Charge (without sign) becomes subscript for other ion 4.Reduce subscripts to smallest whole number ratio 5.Check that the total charge of the cations cancels the total charge of the anions Fe 3+ PO 4 3− Fe 3+ PO 4 3− Fe 3 (PO 4 ) 3 Fe = (1)∙(3+) = +3 PO 4 = (1)∙(3−) = −3 FePO 4 53 Tro: Chemistry: A Molecular Approach, 2/e

54. Copyright  2011 Pearson Education, Inc. Practice — What are the formulas for compounds made from the following ions? aluminum ion with a sulfate ion chromium(II) with hydrogen carbonate Al 3+ with SO 4 2− Al 2 (SO 4 ) 3 Cr 2+ with HCO 3 − Cr(HCO 3 ) 2 54 Tro: Chemistry: A Molecular Approach, 2/e

55. Copyright  2011 Pearson Education, Inc. Hydrates 55 Tro: Chemistry: A Molecular Approach, 2/e Hydrates are ionic compounds containing a specific number of waters for each formula unit Water of hydration often “driven off” by heating In formula, attached waters follow ∙ CoCl 2 ∙6H 2 O In name attached waters indicated by prefix+hydrate after name of ionic compound CoCl 2 ∙6H 2 O = cobalt(II) chloride hexahydrate CaSO 4 ∙½H 2 O = calcium sulfate hemihydrate

56. Copyright  2011 Pearson Education, Inc. Cobalt(II) chloride hexahydrate 56 Tro: Chemistry: A Molecular Approach, 2/e

57. Copyright  2011 Pearson Education, Inc. Practice What is the formula of magnesium sulfate heptahydrate? What is the name of NiCl 2 6H 2 O? Mg 2+ + SO 4 2− MgSO 4 MgSO 4  7H 2 O Cl −  Ni 2+ nickel(II) chloride nickel(II) chloride hexahydrate 57 Tro: Chemistry: A Molecular Approach, 2/e

58. Copyright  2011 Pearson Education, Inc. Writing Names of Binary Molecular Compounds of Two Nonmetals 58 Tro: Chemistry: A Molecular Approach, 2/e 1.Write name of first element in formula a)element furthest left and down on the Periodic Table b)use the full name of the element 2.Writes name the second element in the formula with an -ide suffix a)as if it were an anion, however, remember these compounds do not contain ions! 3.Use a prefix in front of each name to indicate the number of atoms a)Never use the prefix mono- on the first element

59. Copyright  2011 Pearson Education, Inc. Subscript – Prefixes Drop last “a” if name begins with a vowel 59 Tro: Chemistry: A Molecular Approach, 2/e 1 = mono- not used on first nonmetal 2 = di- 3 = tri- 4 = tetra- 5 = penta- 6 = hexa- 7 = hepta- 8 = octa- 9 = nona- 10 = deca-

60. Copyright  2011 Pearson Education, Inc. Example: Naming binary molecular BF 3 60 Tro: Chemistry: A Molecular Approach, 2/e 1.Name the first element boron 2.Name the second element with an –ide fluorine  fluoride 3.Add a prefix to each name to indicate the subscript monoboron, trifluoride 4.Write the first element with prefix, then the second element with prefix a)drop prefix mono from first element boron trifluoride

61. Copyright  2011 Pearson Education, Inc. Name the Following NO 2 PCl 5 I 2 F 7 nitrogen dioxide phosphorus pentachloride diiodine heptafluoride 61 Tro: Chemistry: A Molecular Approach, 2/e

62. Copyright  2011 Pearson Education, Inc. Example: Binary Molecular dinitrogen pentoxide 62 Tro: Chemistry: A Molecular Approach, 2/e Identify the symbols of the elements nitrogen = N oxide = oxygen = O Write the formula using prefix number for subscript di = 2, penta = 5 N2O5N2O5

63. Copyright  2011 Pearson Education, Inc. Write Formulas for the Following dinitrogen tetroxide sulfur hexafluoride diarsenic trisulfide N 2 O 4 SF 6 As 2 S 3 63 Tro: Chemistry: A Molecular Approach, 2/e

64. Copyright  2011 Pearson Education, Inc. Acids Acids are molecular compounds that form H + when dissolved in water to indicate the compound is dissolved in water (aq) is written after the formula  not named as acid if not dissolved in water Sour taste Dissolve many metals such as Zn, Fe, Mg; but not Au, Ag, Pt Formula generally starts with H e.g., HCl, H 2 SO 4 64 Tro: Chemistry: A Molecular Approach, 2/e

65. Copyright  2011 Pearson Education, Inc. Reaction of Acids with Metals H 2 gas 65 Tro: Chemistry: A Molecular Approach, 2/e

66. Copyright  2011 Pearson Education, Inc. Acids 66 Tro: Chemistry: A Molecular Approach, 2/e Contain H +1 cation and anion in aqueous solution Binary acids have H +1 cation and nonmetal anion Oxyacids have H + cation and polyatomic anion

67. Copyright  2011 Pearson Education, Inc. Naming Binary Acids Write a hydro prefix Follow with the nonmetal name Change ending on nonmetal name to –ic Write the word acid at the end of the name 67 Tro: Chemistry: A Molecular Approach, 2/e

68. Copyright  2011 Pearson Education, Inc. Example: Naming binary acids HCl(aq) 68 Tro: Chemistry: A Molecular Approach, 2/e 1.Identify the anion Cl = Cl −, chloride because Group 7A 2.Name the anion with an –ic suffix Cl − = chloride  chloric 3.Add a hydro- prefix to the anion name hydrochloric 4.Add the word acid to the end hydrochloric acid

69. Copyright  2011 Pearson Education, Inc. Naming Oxyacids If polyatomic ion name ends in –ate, then change ending to –ic suffix If polyatomic ion name ends in –ite, then change ending to –ous suffix Write word acid at end of all names 69 Tro: Chemistry: A Molecular Approach, 2/e

70. Copyright  2011 Pearson Education, Inc. Example: Naming oxyacids H 2 SO 4 (aq) 70 Tro: Chemistry: A Molecular Approach, 2/e 1.Identify the anion SO 4 = SO 4 2− = sulfate 2.If the anion has –ate suffix, change it to –ic. If the anion has –ite suffix, change it to -ous SO 4 2− = sulfate  sulfuric 3.Write the name of the anion followed by the word acid sulfuric acid (kind of an exception, to make it sound nicer!)

71. Copyright  2011 Pearson Education, Inc. Example: Naming oxyacids H 2 SO 3 (aq) 71 Tro: Chemistry: A Molecular Approach, 2/e 1.Identify the anion SO 3 = SO 3 2− = sulfite 2.If the anion has –ate suffix, change it to –ic. If the anion has –ite suffix, change it to -ous SO 3 2− = sulfite  sulfurous 3.Write the name of the anion followed by the word acid sulfurous acid

72. Copyright  2011 Pearson Education, Inc. Name the Following H 2 S HClO 3 HNO 2 hydrosulfuric acid chloric acid nitrous acid 72 Tro: Chemistry: A Molecular Approach, 2/e

73. Copyright  2011 Pearson Education, Inc. Writing Formulas for Acids When name ends in acid, formulas starts with H Write formulas as if ionic, even though it is molecular Hydro prefix means it is binary acid, no prefix means it is an oxyacid For oxyacid, if ending is –ic, polyatomic ion ends in –ate; if ending is –ous, polyatomic ion ends in –ous 73 Tro: Chemistry: A Molecular Approach, 2/e

74. Copyright  2011 Pearson Education, Inc. Example: Binary Acids hydrosulfuric acid 1.Write the symbol for the cation and its charge 2.Write the symbol for the anion and its charge 3.Charge (without sign) becomes subscript for other ion 4.Add (aq) to indicate dissolved in water 5.Check that the total charge of the cations cancels the total charge of the anions H+H+ S2−S2− H + S 2− H2SH2S H = (2)∙(1+) = +2 S = (1)∙(2−) = −2 H 2 S(aq) in all acids the cation is H + hydro means binary 74 Tro: Chemistry: A Molecular Approach, 2/e

75. Copyright  2011 Pearson Education, Inc. Example: Oxyacids carbonic acid 1.Write the symbol for the cation and its charge 2.Write the symbol for the anion and its charge 3.Charge (without sign) becomes subscript for other ion 4.Add (aq) to indicate dissolved in water 5.Check that the total charge of the cations cancels the total charge of the anions H+H+ CO 3 2− H + CO 3 2− H 2 CO 3 H = (2)∙(1+) = +2 CO 3 = (1)∙(2−) = −2 H 2 CO 3 (aq) in all acids the cation is H + no hydro means polyatomic ion -ic means -ate ion 75 Tro: Chemistry: A Molecular Approach, 2/e

76. Copyright  2011 Pearson Education, Inc. Example: Oxyacids sulfurous acid 1.Write the symbol for the cation and its charge 2.Write the symbol for the anion and its charge 3.Charge (without sign) becomes subscript for other ion 4.Add (aq) to indicate dissolved in water 5.Check that the total charge of the cations cancels the total charge of the anions H+H+ SO 3 2− H + SO 3 2− H 2 SO 3 H = (2)∙(1+) = +2 SO 3 = (1)∙(2−) = −2 H 2 SO 3 (aq) in all acids the cation is H + no hydro means polyatomic ion -ous means -ite ion 76 Tro: Chemistry: A Molecular Approach, 2/e

77. Copyright  2011 Pearson Education, Inc. Practice — What are the formulas for the following acids? H + with ClO 2 – HClO 2 H + with PO 4 3– H 3 PO 4 H + with Br – HBr chlorous acid phosphoric acid hydrobromic acid 77 Tro: Chemistry: A Molecular Approach, 2/e

78. Copyright  2011 Pearson Education, Inc. Formula Mass 78 Tro: Chemistry: A Molecular Approach, 2/e The mass of an individual molecule or formula unit also known as molecular mass or molecular weight Sum of the masses of the atoms in a single molecule or formula unit whole = sum of the parts! mass of 1 molecule of H 2 O = 2(1.01 amu H) + 16.00 amu O = 18.02 amu

79. Copyright  2011 Pearson Education, Inc. Molar Mass of Compounds 79 Tro: Chemistry: A Molecular Approach, 2/e The relative masses of molecules can be calculated from atomic masses Formula Mass = 1 molecule of H 2 O = 2(1.01 amu H) + 16.00 amu O = 18.02 amu 1 mole of H 2 O contains 2 moles of H and 1 mole of O so molar mass = 1 mole H 2 O = 2(1.01 g H) + 16.00 g O = 18.02 g so the Molar Mass of H 2 O is 18.02 g/mole

80. Copyright  2011 Pearson Education, Inc. Practice — Converting grams to moles How many moles are in 50.0 g of PbO 2 ? (Pb = 207.2, O = 16.00) 80 Tro: Chemistry: A Molecular Approach, 2/e

81. Copyright  2011 Pearson Education, Inc. Practice — How many moles are in 50.0 g of PbO 2 ? (Pb = 207.2, O = 16.00) because the given amount is less than 239.2 g, the moles being < 1 makes sense 1 mol PbO 2 = 239.2 g 50.0 g mol PbO 2 moles PbO 2 Check: Solution: Conceptual Plan: Relationships: Given: Find: g PbO 2 mol PbO 2 81 Tro: Chemistry: A Molecular Approach, 2/e

82. Copyright  2011 Pearson Education, Inc. Example: Find the number of CO 2 molecules in 10.8 g of dry ice because the given amount is much less than 1 mol CO 2, the number makes sense 1 mol CO 2 = 44.01 g, 1 mol = 6.022 x 10 23 10.8 g CO 2 molecules CO 2 Check: Solution: Conceptual Plan: Relationships: Given: Find: g CO 2 mol CO 2 molec CO 2 82 Tro: Chemistry: A Molecular Approach, 2/e

83. Copyright  2011 Pearson Education, Inc. Practice — How many formula units are in 50.0 g of PbO 2 ? (PbO 2 = 239.2) 83 Tro: Chemistry: A Molecular Approach, 2/e

84. Copyright  2011 Pearson Education, Inc. Practice — How many formula units are in 50.0 g of PbO 2 ? (PbO 2 = 239.2) because the given amount is less than 1 mol PbO 2, the number makes sense 1 mol PbO 2 = 239.2 g,1 mol = 6.022 x 10 23 50.0 g PbO 2 formula units PbO 2 Check: Solution: Conceptual Plan: Relationships: Given: Find: g PbO 2 mol PbO 2 units PbO 2 84 Tro: Chemistry: A Molecular Approach, 2/e

85. Copyright  2011 Pearson Education, Inc. Practice — What is the mass of 4.78 x 10 24 NO 2 molecules? 85 Tro: Chemistry: A Molecular Approach, 2/e

86. Copyright  2011 Pearson Education, Inc. Practice — What is the mass of 4.78 x 10 24 NO 2 molecules? because the given amount is more than Avogadro’s number, the mass > 46 g makes sense 1 mol NO 2 = 46.01 g, 1 mol = 6.022 x 10 23 4.78 x 10 24 NO 2 molecules g NO 2 Check: Solution: Conceptual Plan: Relationships: Given: Find: molecules mol NO 2 g NO 2 86 Tro: Chemistry: A Molecular Approach, 2/e

87. Copyright  2011 Pearson Education, Inc. Percent Composition 87 Tro: Chemistry: A Molecular Approach, 2/e Percentage of each element in a compound by mass Can be determined from 1. the formula of the compound 2. the experimental mass analysis of the compound The percentages may not always total to 100% due to rounding

88. Copyright  2011 Pearson Education, Inc. Example 3.13: Find the mass percent of Cl in C 2 Cl 4 F 2 because the percentage is less than 100 and Cl is much heavier than the other atoms, the number makes sense C 2 Cl 4 F 2 % Cl by mass Check: Solution: Conceptual Plan: Relationships: Given: Find: 88 Tro: Chemistry: A Molecular Approach, 2/e

89. Copyright  2011 Pearson Education, Inc. Practice — Determine the mass percent composition of the following CaCl 2 (Ca = 40.08, Cl = 35.45) 89 Tro: Chemistry: A Molecular Approach, 2/e

90. Copyright  2011 Pearson Education, Inc. Practice — Determine the percent composition of the following 90 CaCl 2 Tro: Chemistry: A Molecular Approach, 2/e

91. Copyright  2011 Pearson Education, Inc. Mass Percent as a Conversion Factor The mass percent tells you the mass of a constituent element in 100 g of the compound the fact that CCl 2 F 2 is 58.64% Cl by mass means that 100 g of CCl 2 F 2 contains 58.64 g Cl This can be used as a conversion factor 100 g CCl 2 F 2 : 58.64 g Cl 91 Tro: Chemistry: A Molecular Approach, 2/e

92. Copyright  2011 Pearson Education, Inc. Example 3.14: Find the mass of table salt containing 2.4 g of Na because the mass of NaCl is more than 2x the mass of Na, the number makes sense 100 g NaCl : 39 g Na 2.4 g Na, 39% Na g NaCl Check: Solution: Conceptual Plan: Relationships: Given: Find: g Nag NaCl 92 Tro: Chemistry: A Molecular Approach, 2/e

93. Copyright  2011 Pearson Education, Inc. Practice — Benzaldehyde is 79.2% carbon. What mass of benzaldehyde contains 19.8 g of C? 93 Tro: Chemistry: A Molecular Approach, 2/e

94. Copyright  2011 Pearson Education, Inc. Practice — Benzaldehyde is 79.2% carbon. What mass of benzaldehyde contains 19.8 g of C? because the mass of benzaldehyde is more than the mass of C, the number makes sense 100 g benzaldehyde : 79.2 g C 19.8 g C, 79.2% C g benzaldehyde Check: Solution: Conceptual Plan: Relationships: Given: Find: g Cg benzaldehyde 94 Tro: Chemistry: A Molecular Approach, 2/e

95. Copyright  2011 Pearson Education, Inc. Conversion Factors in Chemical Formulas Chemical formulas have inherent in them relationships between numbers of atoms and molecules or moles of atoms and molecules These relationships can be used to convert between amounts of constituent elements and molecules like percent composition 95 Tro: Chemistry: A Molecular Approach, 2/e

96. Copyright  2011 Pearson Education, Inc. Example 3.15: Find the mass of hydrogen in 1.00 gal of water because 1 gallon weighs about 3800 g, and H is light, the number makes sense 3.785 L = 1 gal, 1 L = 1000 mL, 1.00 g H 2 O = 1 mL, 1 mol H 2 O = 18.02 g, 1 mol H = 1.008 g, 2 mol H : 1 mol H 2 O 1.00 gal H 2 O, d H2O = 1.00 g/ml g H Check: Solution: Conceptual Plan: Relationships: Given: Find: gal H 2 OL H 2 OmL H 2 Og H 2 O mol H 2 OmoL Hg H 96 Tro: Chemistry: A Molecular Approach, 2/e

97. Copyright  2011 Pearson Education, Inc. Practice — How many grams of sodium are in 6.2 g of NaCl? (Na = 22.99; Cl = 35.45) 97 Tro: Chemistry: A Molecular Approach, 2/e

98. Copyright  2011 Pearson Education, Inc. 1 mol NaCl = 58.44 g, 1 mol Na = 22.99 g, 1 mol Na : 1 mol NaCl Practice — Find the mass of sodium in 6.2 g of NaCl because the amount of Na is less than the amount of NaCl, the answer makes sense 6.2 g NaCl g Na Check: Solution: Conceptual Plan: Relationships: Given: Find: g NaClmol NaClmol Nag Na 98 Tro: Chemistry: A Molecular Approach, 2/e

99. Copyright  2011 Pearson Education, Inc. Empirical Formula Simplest, whole-number ratio of the atoms of elements in a compound Can be determined from elemental analysis masses of elements formed when a compound is decompose, or that react together to form a compound  combustion analysis percent composition 99 Tro: Chemistry: A Molecular Approach, 2/e

100. Copyright  2011 Pearson Education, Inc. Finding an Empirical Formula 100 Tro: Chemistry: A Molecular Approach, 2/e 1.Convert the percentages to grams a)assume you start with 100 g of the compound b)skip if already grams 2.Convert grams to moles a)use molar mass of each element 3.Write a pseudoformula using moles as subscripts 4.Divide all by smallest number of moles a)if result is within 0.1 of whole number, round to whole number 5.Multiply all mole ratios by number to make all whole numbers a)if ratio ?.5, multiply all by 2; if ratio ?.33 or ?.67, multiply all by 3; if ratio 0.25 or 0.75, multiply all by 4; etc. b)skip if already whole numbers

101. Copyright  2011 Pearson Education, Inc. Example 3.17 Laboratory analysis of aspirin determined the following mass percent composition. Find the empirical formula. C = 60.00% H = 4.48% O = 35.53% 101 Tro: Chemistry: A Molecular Approach, 2/e

102. Copyright  2011 Pearson Education, Inc. Example: Find the empirical formula of aspirin with the given mass percent composition Write down the given quantity and its units Given:C = 60.00% H = 4.48% O = 35.53% Therefore, in 100 g of aspirin there are 60.00 g C, 4.48 g H, and 35.53 g O 102 Tro: Chemistry: A Molecular Approach, 2/e

103. Copyright  2011 Pearson Education, Inc. Example: Find the empirical formula of aspirin with the given mass percent composition Write down the quantity to find and/or its units Find: empirical formula, C x H y O z Information Given:60.00 g C, 4.48 g H, 35.53 g O 103 Tro: Chemistry: A Molecular Approach, 2/e

104. Copyright  2011 Pearson Education, Inc. Example: Find the empirical formula of aspirin with the given mass percent composition Write a conceptual plan Information Given:60.00 g C, 4.48 g H, 35.53 g O Find: empirical formula, C x H y O z g C mol C g H mol H pseudo- formula pseudo- formula empirical formula empirical formula mole ratio whole number ratio g O mol O 104 Tro: Chemistry: A Molecular Approach, 2/e

105. Copyright  2011 Pearson Education, Inc. Example: Find the empirical formula of aspirin with the given mass percent composition Collect needed relationships 1 mole C = 12.01 g C 1 mole H = 1.008 g H 1 mole O = 16.00 g O Information Given:60.00 g C, 4.48 g H, 35.53 g O Find: empirical formula, C x H y O z CP: g C,H,O  mol C,H,O  pseudo form.  mol ratio  emp. form. 105 Tro: Chemistry: A Molecular Approach, 2/e

106. Copyright  2011 Pearson Education, Inc. Example: Find the empirical formula of aspirin with the given mass percent composition Apply the conceptual plan calculate the moles of each element 106 Information Given:60.00 g C, 4.48 g H, 35.53 g O Find: empirical Formula, C x H y O z CP: g C,H,O  mol C,H,O  pseudo form.  mol ratio  emp. form. Rel: 1 mol C = 12.01 g; 1 mol H = 1.008 g; 1 mol O = 16.00 g Tro: Chemistry: A Molecular Approach, 2/e

107. Copyright  2011 Pearson Education, Inc. Example: Find the empirical formula of aspirin with the given mass percent composition Apply the conceptual plan write a pseudoformula C 4.996 H 4.44 O 2.220 107 Information Given:60.00 g C, 4.48 g H, 35.53 g O Find: empirical formula, C x H y O z CP: g C,H,O  mol C,H,O  pseudo form.  mol ratio  emp. form. Rel: 1 mol C = 12.01 g; 1 mol H = 1.008 g; 1 mol O = 16.00 g Tro: Chemistry: A Molecular Approach, 2/e

108. Copyright  2011 Pearson Education, Inc. Example: Find the empirical formula of aspirin with the given mass percent composition Apply the concept plan find the mole ratio by dividing by the smallest number of moles Information Given:60.00 g C, 4.48 g H, 35.53 g O Find: empirical formula, C x H y O z CP: g C,H,O  mol C,H,O  pseudo form.  mol ratio  emp. form. Rel: 1 mol C = 12.01 g; 1 mol H = 1.008 g; 1 mol O = 16.00 g 108 Tro: Chemistry: A Molecular Approach, 2/e

109. Copyright  2011 Pearson Education, Inc. Example: Find the empirical formula of aspirin with the given mass percent composition Apply the conceptual plan multiply subscripts by factor to give whole number {C 2.25 H 2 O 1 } x 4 C9H8O4C9H8O4 Information Given:60.00 g C, 4.48 g H, 35.53 g O Find: empirical formula, C x H y O z CP: g C,H,O  mol C,H,O  pseudo form.  mol ratio  emp. form. Rel: 1 mol C = 12.01 g; 1 mol H = 1.008 g; 1 mol O = 16.00 g 109 Tro: Chemistry: A Molecular Approach, 2/e

110. Copyright  2011 Pearson Education, Inc. Practice — Determine the empirical formula of stannous fluoride, which contains 75.7% Sn (118.70) and the rest fluorine (19.00) 110 Tro: Chemistry: A Molecular Approach, 2/e

111. Copyright  2011 Pearson Education, Inc. Practice — Determine the empirical formula of stannous fluoride, which contains 75.7% Sn (118.70) and the rest fluorine (19.00) Given: 75.7% Sn, (100 – 75.3) = 24.3% F  in 100 g stannous fluoride there are 75.7 g Sn and 24.3 g F Find: Sn x F y Rel:1 mol Sn = 118.70 g; 1 mol F = 19.00 g Conceptual plan: g Sn mol Sn g F mol F pseudo- formula pseudo- formula empirical formula empirical formula mole ratio whole number ratio 111 Tro: Chemistry: A Molecular Approach, 2/e

112. Copyright  2011 Pearson Education, Inc. Practice — Determine the empirical formula of stannous fluoride, which contains 75.7% Sn (118.70) and the rest fluorine (19.00) Apply the conceptual plan Sn 0.638 F 1.28 SnF 2 112 Tro: Chemistry: A Molecular Approach, 2/e

113. Copyright  2011 Pearson Education, Inc. Practice — Determine the empirical formula of magnetite, which contains 72.4% Fe (55.85) and the rest oxygen (16.00) 113 Tro: Chemistry: A Molecular Approach, 2/e

114. Copyright  2011 Pearson Education, Inc. Practice — Determine the empirical formula of magnetite, which contains 72.4% Fe (55.85) and the rest oxygen (16.00) Given: 72.4% Fe, (100 – 72.4) = 27.6% O  in 100 g magnetite there are 72.4 g Fe and 27.6 g O Find: Fe x O y Rel:1 mol Fe = 55.85 g; 1 mol O = 16.00 g Conceptual plan: g Fe mol Fe g O mol O pseudo- formula pseudo- formula empirical formula empirical formula mole ratio whole number ratio 114 Tro: Chemistry: A Molecular Approach, 2/e

115. Copyright  2011 Pearson Education, Inc. Practice — Determine the empirical formula of magnetite, which contains 72.4% Fe (55.85) and the rest oxygen (16.00) Apply the conceptual plan Fe 1.30 O 1.73 115 Tro: Chemistry: A Molecular Approach, 2/e

116. Copyright  2011 Pearson Education, Inc. Molecular Formulas 116 Tro: Chemistry: A Molecular Approach, 2/e The molecular formula is a multiple of the empirical formula To determine the molecular formula you need to know the empirical formula and the molar mass of the compound

117. Copyright  2011 Pearson Education, Inc. Example 3.18: Find the molecular formula of butanedione the molar mass of the calculated formula is in agreement with the given molar mass emp. form. = C 2 H 3 O; MM = 86.03 g/mol molecular formula Check: Solution: Conceptual Plan: and Relationships: Given: Find: 117 Tro: Chemistry: A Molecular Approach, 2/e

118. Copyright  2011 Pearson Education, Inc. Practice — Benzopyrene has a molar mass of 252 g/mol and an empirical formula of C 5 H 3. What is its molecular formula? (C = 12.01, H=1.01) 118 Tro: Chemistry: A Molecular Approach, 2/e

119. Copyright  2011 Pearson Education, Inc. Practice – Benzopyrene has a molar mass of 252 g and an empirical formula of C 5 H 3. What is its molecular formula? (C = 12.01, H=1.01) C 5 =5(12.01 g) = 60.05 g H 3 =3(1.01 g) = 3.03 g C 5 H 3 = 63.08 g Molecular formula = {C 5 H 3 } x 4 = C 20 H 12 119 Tro: Chemistry: A Molecular Approach, 2/e

120. Copyright  2011 Pearson Education, Inc. Combustion Analysis A common technique for analyzing compounds is to burn a known mass of compound and weigh the amounts of product made generally used for organic compounds containing C, H, O By knowing the mass of the product and composition of constituent element in the product, the original amount of constituent element can be determined all the original C forms CO 2, the original H forms H 2 O, the original mass of O is found by subtraction Once the masses of all the constituent elements in the original compound have been determined, the empirical formula can be found 120 Tro: Chemistry: A Molecular Approach, 2/e

121. Copyright  2011 Pearson Education, Inc. Combustion Analysis 121 Tro: Chemistry: A Molecular Approach, 2/e

122. Copyright  2011 Pearson Education, Inc. Example 3.20 Combustion of a 0.8233 g sample of a compound containing only carbon, hydrogen, and oxygen produced the following: CO 2 = 2.445 g H 2 O = 0.6003 g Determine the empirical formula of the compound 122 Tro: Chemistry: A Molecular Approach, 2/e

123. Copyright  2011 Pearson Education, Inc. Example 3.19: Find the empirical formula of compound with the given amounts of combustion products Write down the given quantity and its units Given:compound = 0.8233 g CO 2 = 2.445 g H 2 O = 0.6003 g 123 Tro: Chemistry: A Molecular Approach, 2/e

124. Copyright  2011 Pearson Education, Inc. Example: Find the empirical formula of compound with the given amounts of combustion products Write down the quantity to find and/or its units Find: empirical formula, C x H y O z Information Given: 0.8233 g compound, 2.445 g CO 2, 0.6003 g H 124 Tro: Chemistry: A Molecular Approach, 2/e

125. Copyright  2011 Pearson Education, Inc. Write a conceptual plan g CO 2, H 2 O mol ratio empirical formula Example: Find the empirical formula of compound with the given amounts of combustion products mol CO 2, H 2 O mol C, H g C, H gOgO mol O mol C, H, O Information Given: 0.8233 g compound, 2.445 g CO 2, 0.6003 g H Find: empirical formula, C x H y O z pseudo formula 125 Tro: Chemistry: A Molecular Approach, 2/e

126. Copyright  2011 Pearson Education, Inc. Example: Find the empirical formula of compound with the given amounts of combustion products Collect needed relationships 1 mole CO 2 = 44.01 g CO 2 1 mole H 2 O = 18.02 g H 2 O 1 mole C = 12.01 g C 1 mole H = 1.008 g H 1 mole O = 16.00 g O 1 mole CO 2 = 1 mole C 1 mole H 2 O = 2 mole H Information Given: 0.8233 g compound, 2.445 g CO 2, 0.6003 g H 2 O Find: empirical formula, C x H y O z CP: g CO 2 & H 2 O  mol CO 2 & H 2 O  mol C & H  g C & H  g O  mol O  mol ratio  empirical formula 126 Tro: Chemistry: A Molecular Approach, 2/e

127. Copyright  2011 Pearson Education, Inc. Example: Find the empirical formula of compound with the given amounts of combustion products Apply the conceptual plan calculate the moles of C and H Information Given: 0.8233 g compound, 2.445 g CO 2, 0.6003 g H 2 O Find: empirical formula, C x H y O z CP: g CO 2 & H 2 O  mol CO 2 & H 2 O  mol C & H  g C & H  g O  mol O  mol ratio  empirical formula Rel: MM of CO 2, H 2 O, C, H, O; mol element : 1 mol compound 127 Tro: Chemistry: A Molecular Approach, 2/e

128. Copyright  2011 Pearson Education, Inc. Example: Find the empirical formula of compound with the given amounts of combustion products Apply the conceptual plan calculate the grams of C and H Information Given: 0.8233 g compound, 2.445 g CO 2, 0.6003 g H 2 O Find: empirical formula, C x H y O z CP: g CO 2 & H 2 O  mol CO 2 & H 2 O  mol C & H  g C & H  g O  mol O  mol ratio  empirical formula Rel: MM of CO 2, H 2 O, C, H, O; mol element : 1 mol compound 128 Tro: Chemistry: A Molecular Approach, 2/e

129. Copyright  2011 Pearson Education, Inc. Apply the conceptual plan calculate the grams and moles of O Example: Find the empirical formula of compound with the given amounts of combustion products Information Given: 0.8233 g compound, 2.445 g CO 2, 0.6003 g H 2 O Find: empirical formula, C x H y O z CP: g CO 2 & H 2 O  mol CO 2 & H 2 O  mol C & H  g C & H  g O  mol O  mol ratio  empirical formula Rel: MM of CO 2, H 2 O, C, H, O; mol element : 1 mol compound 129 Tro: Chemistry: A Molecular Approach, 2/e

130. Copyright  2011 Pearson Education, Inc. Apply the conceptual plan write a pseudoformula C 0.05556 H 0.06662 O 0.00556 Example: Find the empirical formula of compound with the given amounts of combustion products Information Given: 0.8233 g compound, 2.445 g CO 2, 0.6003 g H 2 O 0.05556 mol C, 0.6673 g C, 0.06662 mol H, 0.06715 g H, 0.0889 g O, 0.00556 mol O Find: empirical formula, C x H y O z CP: g CO 2 & H 2 O  mol CO 2 & H 2 O  mol C & H  g C & H  g O  mol O  mol ratio  empirical formula Rel: MM of CO 2, H 2 O, C, H, O; mol element : 1 mol compound 130 Tro: Chemistry: A Molecular Approach, 2/e

131. Copyright  2011 Pearson Education, Inc. Apply the conceptual plan find the mole ratio by dividing by the smallest number of moles Example: Find the empirical formula of compound with the given amounts of combustion products Information Given: 0.8233 g compound, 2.445 g CO 2, 0.6003 g H 2 O 0.05556 mol C, 0.6673 g C, 0.06662 mol H, 0.06715 g H, 0.0889 g O, 0.00556 mol O Find: empirical formula, C x H y O z CP: g CO 2 & H 2 O  mol CO 2 & H 2 O  mol C & H  g C & H  g O  mol O  mol ratio  empirical formula Rel: MM of CO 2, H 2 O, C, H, O; mol element : 1 mol compound 131 Tro: Chemistry: A Molecular Approach, 2/e

132. Copyright  2011 Pearson Education, Inc. Apply the conceptual plan multiply subscripts by factor to give whole number, if necessary write the empirical formula 11210 OHC Example: Find the empirical formula of compound with the given amounts of combustion products Information Given: 0.8233 g compound, 2.445 g CO 2, 0.6003 g H 2 O 0.05556 mol C, 0.6673 g C, 0.06662 mol H, 0.06715 g H, 0.0889 g O, 0.00556 mol O Find: empirical formula, C x H y O z CP: g CO 2 & H 2 O  mol CO 2 & H 2 O  mol C & H  g C & H  g O  mol O  mol ratio  empirical formula Rel: MM of CO 2, H 2 O, C, H, O; mol element : 1 mol compound 132 Tro: Chemistry: A Molecular Approach, 2/e

133. Copyright  2011 Pearson Education, Inc. Practice — The smell of dirty gym socks is caused by the compound caproic acid. Combustion of 0.844 g of caproic acid produced 0.784 g of H 2 O and 1.92 g of CO 2. If the molar mass of caproic acid is 116.2 g/mol, what is the molecular formula of caproic acid? (MM C = 12.01, H = 1.008, O = 16.00) 133 Tro: Chemistry: A Molecular Approach, 2/e

134. Copyright  2011 Pearson Education, Inc. Practice — Combustion of 0.844 g of caproic acid produced 0.784 g of H 2 O and 1.92 g of CO 2. If the molar mass of caproic acid is 116.2 g/mol, what is the molecular formula of caproic acid? 134 Tro: Chemistry: A Molecular Approach, 2/e

135. Copyright  2011 Pearson Education, Inc. moles g 0.01450.08700.0436 0.2320.08770.524 OHC C 0.0436 H 0.0870 O 0.0145 135 Tro: Chemistry: A Molecular Approach, 2/e

136. Copyright  2011 Pearson Education, Inc. Molecular formula = {C 3 H 6 O} x 2 = C 6 H 12 O 2 136 Tro: Chemistry: A Molecular Approach, 2/e

137. Copyright  2011 Pearson Education, Inc. Chemical Reactions Reactions involve chemical changes in matter resulting in new substances Reactions involve rearrangement and exchange of atoms to produce new molecules elements are not transmuted during a reaction Reactants  Products 137 Tro: Chemistry: A Molecular Approach, 2/e

138. Copyright  2011 Pearson Education, Inc. Chemical Equations Shorthand way of describing a reaction Provides information about the reaction formulas of reactants and products states of reactants and products relative numbers of reactant and product molecules that are required can be used to determine weights of reactants used and products that can be made 138 Tro: Chemistry: A Molecular Approach, 2/e

139. Copyright  2011 Pearson Education, Inc. This equation reads “1 molecule of CH 4 gas combines with 1 molecule of O 2 gas to make 1 molecule of CO 2 gas and 1 molecule of H 2 O gas.” Combustion of Methane Methane gas burns to produce carbon dioxide gas and gaseous water. whenever something burns it combines with O 2 (g). CH 4 (g) + O 2 (g)  CO 2 (g) + H 2 O(g) H H C H H OO + O O C + O HH 1 C + 4 H + 2 O1 C + 2 O + 2 H + O 1 C + 2 H + 3 O What incorrect assumption was made when writing this equation? We are assuming that all reactants combine 1 molecule : 1 molecule; and that 1 molecule of each product is made – an incorrect assumption 139 Tro: Chemistry: A Molecular Approach, 2/e

140. Copyright  2011 Pearson Education, Inc. Combustion of Methane, Balanced To show the reaction obeys the Law of Conservation of Mass the equation must be balanced we adjust the numbers of molecules so there are equal numbers of atoms of each element on both sides of the arrow CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O(g) H H C H H + O O C + OO OO + O HH O HH + 1 C + 4 H + 4 O 140 Tro: Chemistry: A Molecular Approach, 2/e

141. Copyright  2011 Pearson Education, Inc. Chemical Equations CH 4 and O 2 are the reactants, and CO 2 and H 2 O are the products The (g) after the formulas tells us the state of the chemical The number in front of each substance tells us the numbers of those molecules in the reaction called the coefficients CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O(g) 141 Tro: Chemistry: A Molecular Approach, 2/e

142. Copyright  2011 Pearson Education, Inc. Chemical Equations This equation is balanced, meaning that there are equal numbers of atoms of each element on the reactant and product sides to obtain the number of atoms of an element, multiply the subscript by the coefficient 1  C  1 4  H  4 4  O  2 + 2 CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O(g) 142 Tro: Chemistry: A Molecular Approach, 2/e

143. Copyright  2011 Pearson Education, Inc. Symbols Used in Equations Symbols used to indicate state after chemical (g) = gas; (l) = liquid; (s) = solid (aq) = aqueous = dissolved in water Energy symbols used above the arrow for decomposition reactions  = heat h = light shock = mechanical elec = electrical 143 Tro: Chemistry: A Molecular Approach, 2/e

144. Copyright  2011 Pearson Education, Inc. Example 3.22: Write a balanced equation for the combustion of butane, C 4 H 10 8  C  8; 20  H  20; 26  O  26 {C 4 H 10 (l) + 13/2 O 2 (g)  4 CO 2 (g) + 5 H 2 O(g)}x 2 2 C 4 H 10 (l) + 13 O 2 (g)  8 CO 2 (g) + 10 H 2 O(g) 13/2 x 2  O  13 C 4 H 10 (l) + 13/2 O 2 (g)  4 CO 2 (g) + 5 H 2 O(g) 4  C  1 x 4 C 4 H 10 (l) + O 2 (g)  4 CO 2 (g) + H 2 O(g) 10  H  2 x 5 C 4 H 10 (l) + O 2 (g)  4 CO 2 (g) + 5 H 2 O(g) C 4 H 10 (l) + O 2 (g)  CO 2 (g) + H 2 O(g) Check If fractional coefficients, multiply thru by denominator Balance free elements by adjusting coefficient in front of free element Balance atoms in complex substances first Write a skeletal equation 144 Tro: Chemistry: A Molecular Approach, 2/e

145. Copyright  2011 Pearson Education, Inc. Practice when aluminum metal reacts with air, it produces a white, powdery compound, aluminum oxide reacting with air means reacting with O 2 aluminum(s) + oxygen(g)  aluminum oxide(s) Al(s) + O 2 (g)  Al 2 O 3 (s) 4 Al(s) + 3 O 2 (g)  2 Al 2 O 3 (s) 145 Tro: Chemistry: A Molecular Approach, 2/e

146. Copyright  2011 Pearson Education, Inc. Practice Acetic acid reacts with the metal aluminum to make aqueous aluminum acetate and gaseous hydrogen acids are always aqueous metals are solid except for mercury Al(s) + HC 2 H 3 O 2 (aq)  Al(C 2 H 3 O 2 ) 3 (aq) + H 2 (g) 2 Al(s) + 6 HC 2 H 3 O 2 (aq)  2 Al(C 2 H 3 O 2 ) 3 (aq) + 3 H 2 (g) 146 Tro: Chemistry: A Molecular Approach, 2/e

147. Copyright  2011 Pearson Education, Inc. Classifying Compounds Organic vs. Inorganic In the18 th century, compounds from living things were called organic; compounds from the nonliving environment were called inorganic Organic compounds easily decomposed and could not be made in the 18 th -century lab Inorganic compounds very difficult to decompose, but able to be synthesized 147 Tro: Chemistry: A Molecular Approach, 2/e

148. Copyright  2011 Pearson Education, Inc. Modern Classifying Compounds Organic vs. Inorganic Today we commonly make organic compounds in the lab and find them all around us Organic compounds are mainly made of C and H, sometimes with O, N, P, S, and trace amounts of other elements The main element that is the focus of organic chemistry is carbon 148 Tro: Chemistry: A Molecular Approach, 2/e

149. Copyright  2011 Pearson Education, Inc. Carbon Bonding Carbon atoms bond almost exclusively covalently compounds with ionic bonding C are generally inorganic When C bonds, it forms four covalent bonds 4 single bonds, 2 double bonds, 1 triple + 1 single, etc. Carbon is unique in that it can form limitless chains of C atoms, both straight and branched, and rings of C atoms 149 Tro: Chemistry: A Molecular Approach, 2/e

150. Copyright  2011 Pearson Education, Inc. 150 Carbon Bonding Tro: Chemistry: A Molecular Approach, 2/e

151. Copyright  2011 Pearson Education, Inc. Classifying Organic Compounds There are two main categories of organic compounds, hydrocarbons and functionalized hydrocarbons Hydrocarbons contain only C and H Most fuels are mixtures of hydrocarbons 151 Tro: Chemistry: A Molecular Approach, 2/e

152. Copyright  2011 Pearson Education, Inc. 152 Tro: Chemistry: A Molecular Approach, 2/e