1. Chapter 3 Molecules, Ions, & their Compounds

2. Chapter goals Interpret, predict, and write formulas for ionic and molecular compounds.Interpret, predict, and write formulas for ionic and molecular compounds. Name compounds.Name compounds. Understand some properties of ionic compounds.Understand some properties of ionic compounds. Calculate and use molar mass.Calculate and use molar mass. Calculate percent composition for a compound and derive formulas from experimental data.Calculate percent composition for a compound and derive formulas from experimental data.

3. Molecule an assembly of 2 or more atoms (mostly of non-metals) bound together in a particular ratio and a particular manneran assembly of 2 or more atoms (mostly of non-metals) bound together in a particular ratio and a particular manner is the smallest identifiable unit into which a pure substance can be divided and still retain the composition and chemical properties of the substanceis the smallest identifiable unit into which a pure substance can be divided and still retain the composition and chemical properties of the substance

4. Elements that exist as molecules Atoms of most of the nonmetals form discrete molecules, except for the noble gases.Atoms of most of the nonmetals form discrete molecules, except for the noble gases. Some elements can exist in more than one form of molecule; the different forms are called allotropes. Examples are:Some elements can exist in more than one form of molecule; the different forms are called allotropes. Examples are: Diamond, graphite, and buckyballs for carbon.Diamond, graphite, and buckyballs for carbon. O 2 and O 3 (ozone) for oxygen.O 2 and O 3 (ozone) for oxygen.

5. ELEMENTS THAT EXIST AS MOLECULES Allotropes of C

6. ELEMENTS THAT EXIST AS POLYATOMIC MOLECULES White P 4 and polymeric red phosphorus Sulfur: crown- shaped rings of S 8 molecules

7. Molecular compounds models - shapes Methane, CH 4 Water, H 2 O Ammonia, NH 3

8. Molecular compounds Are made of non-metalsAre made of non-metals H 2 O, carbon dioxide (CO 2 ), ammonia (NH 3 ), nitric acid (HNO 3 ), ethanol (CH 3 CH 2 OH), sulfuric acid (H 2 SO 4 ), glucose (C 6 H 12 O 6 ), are examples among thousandsH 2 O, carbon dioxide (CO 2 ), ammonia (NH 3 ), nitric acid (HNO 3 ), ethanol (CH 3 CH 2 OH), sulfuric acid (H 2 SO 4 ), glucose (C 6 H 12 O 6 ), are examples among thousands In the molecules of theses compounds, atoms pairs of electronsIn the molecules of theses compounds, atoms share pairs of electrons

9. Molecular compounds: Formulas NAME MOLECULAR CONDENSED STRUCTURAL FORMULA FORMULA FORMULA FORMULA FORMULA H H | | Ethanol C 2 H 6 O Ethanol C 2 H 6 O CH 3 CH 2 OH H─C ─ C─O─H | | H H | | Dimethyl C 2 H 6 O Dimethyl C 2 H 6 O CH 3 OCH 3 H─C ─ O─ C─H ether ether | | H H Ethanol and dimethyl ether are said to be structural Ethanol and dimethyl ether are said to be structural isomers.

10. Ionic Compounds Ion charged particle (atom or group of atoms)charged particle (atom or group of atoms) cation: + chargecation: + charge anion: – chargeanion: – charge

11. Ionic Compounds Sodium chloride or “table salt” is an example of an ionic compound.

12. Ionic compounds consist of positive and negative ions, mostly a metal and a non-metal, respectively.consist of positive and negative ions, mostly a metal and a non-metal, respectively. have attractions called ionic bonds between positively (cations) and negatively charged ions (anions).have attractions called ionic bonds between positively (cations) and negatively charged ions (anions). have high melting and boiling points.have high melting and boiling points. T m of NaCl = 800 °C = 1472 °F T m of NaCl = 800 °C = 1472 °F are solid at room temperature.are solid at room temperature.

13. Charge Balance Group IA VIIA NaCl, sodium chloride IIA of periodic table

14. Periodic Table

15. Monatomic Cations Metal atoms of group 1A lose one electron to produce a mono-positive ion

16. Monatomic Cations Metal atoms of group 2A lose two electrons to produce a di-positive ion

17. Monatomic Anions Nonmetals often gain one or more electrons and form ions having a negative charge equal to the group number of the element minus 8: Group Atom gained e - Resulting anion 5A N 3 N 3- 5A N 3 N 3- P 6A O 2 O 2- 6A O 2 O 2- S 7A F 1 F - 7A F 1 F - Cl, Br, I Cl −, Br −, I − Cl, Br, I Cl −, Br −, I −

18. Charges of Representative Elements

19. Monatomic Cations Transition metals (B-group elements) can form a no easily predictable variety of cations: Group Atom Electrons loss Resulting cation 7B Mn 2 Mn 2+ 7B Mn 2 Mn 2+ 8B Fe 2 Fe 2+ 8B Fe 2 Fe 2+ 8B Fe 3 Fe 3+ 8B Fe 3 Fe 3+ 1B Cu 1 Cu + 1B Cu 1 Cu + 1B Cu 2 Cu 2+ 1B Cu 2 Cu 2+ 2B Zn 2 Zn 2+ 2B Zn 2 Zn 2+ 2B Cd 2 Cd 2+ 2B Cd 2 Cd 2+

20. Transition Metals form Positive Ions Most transition metals and Group 4A metals form 2 or more positive ions. Zn, Ag, and Cd form only one ion.

21. Names of Some Common Ions Main group metals Nonmetal: change the last element name only part of name to ide

22. Naming Cations transition metals and In, Sn, Tl, Pb, Bitransition metals and In, Sn, Tl, Pb, Bi new system: element name (charge in Roman numerals) new system: element name (charge in Roman numerals) eg. Mn 2+ manganese(II) eg. Mn 2+ manganese(II) Mn 3+ manganese(III) Mn 3+ manganese(III) Cr 2+ chromium(II) Cr 3+ chromium(III) Cr 2+ chromium(II) Cr 3+ chromium(III) Fe 2+ iron(II) Fe 3+ iron(III) Fe 2+ iron(II) Fe 3+ iron(III) Exceptions: when only one cation Exceptions: when only one cation Ag + silver Zn 2+ zinc Cd 2+ cadmium

23. old system old system Latin name-suffix Latin name-suffix suffix = -ic for higher charge, -ous for lower charge suffix = -ic for higher charge, -ous for lower charge eg. Cu + copper(I) cuprous ion eg. Cu + copper(I) cuprous ion Cu 2+ copper(II) cupric ion Cu 2+ copper(II) cupric ion Co 2+ cobalt(II) cobaltous ion Co 2+ cobalt(II) cobaltous ion Co 3+ cobalt(III) cobaltic ion Co 3+ cobalt(III) cobaltic ion Fe 2+ iron(II) ferrous ion Fe 2+ iron(II) ferrous ion Fe 3+ iron(III) ferric ion Fe 3+ iron(III) ferric ion

24. Polyatomic Ions A A polyatomic ion is a group of atoms.is a group of atoms. has an overall ionic charge, positive or negative.has an overall ionic charge, positive or negative.

25. Polyatomic Ions (memorize) Some examples of polyatomic ions are NH 4 + ammoniumH 3 O + hydronium OH − hydroxide N 3 − azide OH − hydroxide N 3 − azide CO 3 2− carbonateCN − cyanide CH 3 CO 2 − acetate C 2 O 4 2− oxalate CH 3 CO 2 − acetate C 2 O 4 2− oxalate NO 3 − nitrateNO 2 − nitrite NO 3 − nitrateNO 2 − nitrite PO 4 3− phosphatePO 3 3− phosphite SO 4 2− sulfateSO 3 2− sulfite CrO 4 2− chromateCr 2 O 7 2− dichromate MnO 4 − permanganate MnO 4 2− manganate

26. Hydrogenated Polyatomic Ions HCO 3 − hydrogen carbonate (bicarbonate) HSO 4 − hydrogen sulfate (bisulfate) HSO 3 − hydrogen sulfite (bisulfite) HPO 4 2− hydrogen phosphate H 2 PO 4 − dihydrogen phosphate HS − hydrogen sulfide (from sulfide, S 2− )

27. Systematics ClO 4 – perchlorateClO 4 – perchlorate ClO 3 – chlorateClO 3 – chlorate ClO 2 – chloriteClO 2 – chlorite ClO – hypochloriteClO – hypochlorite The same with other halogens (except for F) IO 4 – periodateIO 4 – periodate IO 3 – iodateIO 3 – iodate IO 2 – ioditeIO 2 – iodite IO – hypoioditeIO – hypoiodite

29. PO 5 3– perphosphatePO 5 3– perphosphate PO 4 3– phosphatePO 4 3– phosphate PO 3 3– phosphitePO 3 3– phosphite PO 2 3– hypophosphitePO 2 3– hypophosphite AsO 5 3– perarsenateAsO 5 3– perarsenate AsO 4 3– arsenateAsO 4 3– arsenate AsO 3 3– arseniteAsO 3 3– arsenite AsO 2 3– hypoarseniteAsO 2 3– hypoarsenite

30. Oxyanions that end in  ate Group III A Group IV A Group V A Group VI A Group VII A BO 3 3– borate CO 3 2– carbonate NO 3 – nitrate SiO 3 2– silicate PO 4 3– phosphate SO 4 2– sulfate ClO 3 – chlorate AsO 4 3– arsenate SeO 4 2– selenate BrO 3 – bromate TeO 4 2– tellurate IO 3 – iodate

31. Other oxyanions: Once ions ending in  ate are memorized, those with different number of O atoms are named with following prefixes and suffixes 2 oxygen less than the “ate” ion is hypo__ite 1 oxygen less than the “ate” ion is ___ite ___ite “ate” ion is ____ate 1 oxygen more than “ate” ion is per__ate BrO – hypobromite BrO 2 – bromite BrO 3 – bromate BrO 4 – perbromate

32. Formula Unit combination of ions in simplest whole number ratio to be electrically neutral (the simplest unit of an ionic compound)combination of ions in simplest whole number ratio to be electrically neutral (the simplest unit of an ionic compound) consists of positively and negatively charged ions.consists of positively and negatively charged ions. is neutral.is neutral. has charge balance.has charge balance. total positive charge = total negative charge total positive charge = total negative charge The symbol of the metal is written first followed by the symbol of the nonmetal. The symbol of the metal is written first followed by the symbol of the nonmetal.

33. Naming Ionic Compounds with Two Elements (binary compound) To name a compound that contains two that contains two elements, elements, identify the cation and anion.identify the cation and anion. name the cation first followed by the name of the anion.name the cation first followed by the name of the anion.

34. Examples, name each below KBrKBr K+ K+ K+ K+ potassium potassium Br – Br – bromide bromide potassium bromide potassium bromide

35. AlF 3AlF 3 Al 3+ Al 3+ aluminum aluminum F– F– F– F– fluoride fluoride aluminum fluoride aluminum fluoride

36. Sr 3 P 2 Sr PSr 3 P 2 Sr P Sr 2+ 3  (+2) = +6 2  (−3) = −6 Sr 2+ 3  (+2) = +6 2  (−3) = −6 strontium strontium P 3– P 3– phosphide phosphide strontium phosphide strontium phosphide

37. CuCl 2 Cu ClCuCl 2 Cu Cl Cl – 1  (+2) = +2 2  (−1) = −2 Cl – 1  (+2) = +2 2  (−1) = −2 chloride chloride Cu 2+ Cu 2+ copper(II) copper(II) cupric cupric copper(II) chloride copper(II) chloride cupric chloride cupric chloride CuCl: copper(I) chloride CuCl: copper(I) chloride

38. WF 6 W FWF 6 W F F – w + 6x(−1) = 0 F – w + 6x(−1) = 0 fluoride w = 6+ fluoride w = 6+ W 6+ W 6+ tungsten(VI) tungsten(VI) tungsten(VI) fluoride tungsten(VI) fluoride

39. Naming Compounds with Polyatomic Ions The positive ion is named first followed by the name of the polyatomic ion. NaNO 3 sodium nitrate K 2 SO 4 potassium sulfate Fe(HCO 3 ) 3 iron(III) bicarbonate or iron(III) hydrogen carbonate or iron(III) hydrogen carbonate fe + 3x(−1) = 0 fe = 3+ iron(III) fe + 3x(−1) = 0 fe = 3+ iron(III) (NH 4 ) 3 PO 3 ammonium phosphite (NH 4 ) 3 PO 3 ammonium phosphite

40. Writing Formulas with Polyatomic Ions The formula of an ionic compound containing a polyatomic ion must have a charge balance that equals zero (0).containing a polyatomic ion must have a charge balance that equals zero (0). Na + and NO 3 −  NaNO 3 with two or more polyatomic ions has the polyatomic ions in parentheses.with two or more polyatomic ions has the polyatomic ions in parentheses. Mg 2+ and 2NO 3 −  Mg(NO 3 ) 2 subscript 2 for charge balance Aluminum sulfateAluminum sulfate 2Al 3+ and 3SO 4 2−  Al 2 (SO 4 ) 3 2Al 3+ and 3SO 4 2−  Al 2 (SO 4 ) 3

41. Learning Check Match each formula with the correct name. A. MgS1) magnesium sulfite MgSO 3 2) magnesium sulfate MgSO 3 2) magnesium sulfate MgSO 4 3) magnesium sulfide MgSO 4 3) magnesium sulfide B. Ca(ClO 3 ) 2 1) calcium chlorate CaCl 2 2) calcium chlorite CaCl 2 2) calcium chlorite Ca(ClO 2 ) 2 3) calcium chloride Ca(ClO 2 ) 2 3) calcium chloride Name each of the following compounds: A.Mg(NO 3 ) 2 magnesium nitrate Mg 2+ and 2 NO 3 − B.Cu(ClO 3 ) 2 copper(II) chlorate Cu 2+ and 2 ClO 3 − C.PbO 2 lead(IV) oxide Pb 4+ and 2 O 2− D.Fe 2 (SO 4 ) 3 iron(III) sulfate 2 Fe 3+ and 3 SO 4 2 − E.Ba 3 (PO 3 ) 2 barium phosphite 3 Ba 2+ and 2 PO 3 3 −

42. Learning Check Select the correct formula for each. A. aluminum nitrate 1) AlNO 3 2) Al(NO) 3 3) Al(NO 3 ) 3 1) AlNO 3 2) Al(NO) 3 3) Al(NO 3 ) 3 B. copper(II) nitrate 1) CuNO 3 2) Cu(NO 3 ) 2 3) Cu 2 (NO 3 ) 1) CuNO 3 2) Cu(NO 3 ) 2 3) Cu 2 (NO 3 ) C. iron(III) hydroxide 1) FeOH2) Fe 3 OH 3) Fe(OH) 3 1) FeOH2) Fe 3 OH 3) Fe(OH) 3 D. tin(IV) hydroxide 1) Sn(OH) 4 2) Sn(OH) 2 3) Sn 4 (OH) 1) Sn(OH) 4 2) Sn(OH) 2 3) Sn 4 (OH)

43. CaSO 4CaSO 4 Ca 2+ 1  (+2) = +2 1  (−2) = −2 Ca 2+ 1  (+2) = +2 1  (−2) = −2 calcium calcium SO 4 2– SO 4 2– sulfate sulfate calcium sulfate calcium sulfate

44. (NH 4 ) 2 S(NH 4 ) 2 S NH 4 + NH 4 + ammonium ammonium S 2– S 2– sulfide sulfide ammonium sulfide ammonium sulfide

45. (NH 4 ) 3 PO 4(NH 4 ) 3 PO 4 NH 4 + NH 4 + ammonium ammonium PO 4 3– PO 4 3– phosphate phosphate ammonium phosphate ammonium phosphate

46. Mo(BrO) 6Mo(BrO) 6 BrO – mo + 6x(−1) = 0 mo = 6+ BrO – mo + 6x(−1) = 0 mo = 6+ hypobromite hypobromite Mo 6+ Mo 6+ molybdenum(VI) molybdenum(VI) molybdenum(VI) hypobromite molybdenum(VI) hypobromite

47. Write formula for rubidium bromiderubidium bromide Rb Rb Rb + Rb + Br Br Br – Br – RbBr RbBr

48. calcium phosphidecalcium phosphide Ca Ca Ca 2+ Ca 2+ P For neutrality we need P For neutrality we need P 3– 3 Ca 2+ and 2 P 3– P 3– 3 Ca 2+ and 2 P 3– Ca 3 P 2 that is, +6 – 6 = 0 Ca 3 P 2 that is, +6 – 6 = 0

49. niobium(IV) sulfiteniobium(IV) sulfite Nb 4+ In order to have the same Nb 4+ In order to have the same SO 3 2– total + and – charge, we need SO 3 2– total + and – charge, we need one Nb 4+ ion and two SO 3 2–, that is one Nb 4+ ion and two SO 3 2–, that is 1  (+4) + 2  (–2) = 4 – 4 =0 (neutrality) 1  (+4) + 2  (–2) = 4 – 4 =0 (neutrality) Nb(SO 3 ) 2 Nb(SO 3 ) 2 barium phosphite: Ba 2+ and PO 3 3–barium phosphite: Ba 2+ and PO 3 3– for the compound to be neutral, we need for the compound to be neutral, we need 3 Ba 2+ and 2 PO 3 3–, that is, +6 – 6 = 0 3 Ba 2+ and 2 PO 3 3–, that is, +6 – 6 = 0 Ba 3 (PO 3 ) 2 Ba 3 (PO 3 ) 2

50. Electrostatic Forces The oppositely charged ions in ionic compounds are attracted to one another by ELECTROSTATIC FORCES. These forces are governed by COULOMB’S LAW.

51. Electrostatic Forces COULOMB’S LAW As ion charge increases, the attractive force increases. As the distance between ions (d) increases, the attractive force decreases. This idea is important and will come up many times in future discussions!

52. F QcQa d2d2d2d2 NaCl Na + Cl – CaCl 2 Ca 2+ Cl – CaS Ca 2+ S –2 Al 2 S 3 Al 3+ S 2– Product of charges 1 6 strongest force 2 4

53. F QcQa d2d2d2d2 F–F–F–F– Na + Cl – Br – I–I–I–I– d d d d NaF has the strongest force and NaI the weakest

54. Acids H + is only cation in acidsH + is only cation in acids Binary Acids hydrogen bonded to one other element hydrogen bonded to one other element hydro-anion-ic acid hydro-anion-ic acid

55. Examples HCl in aqueous solution (water is the solvent)HCl in aqueous solution (water is the solvent) H + and Cl – H + and Cl – anion is chloride anion is chloride hydro-chlor-ic acid Same for F, Br, and I hydro-chlor-ic acid Same for F, Br, and I hydrochloric acid hydrochloric acid H 2 SH 2 S hydrosulfuric acid hydrosulfuric acid H 3 PH 3 P hydrophosphoric acid hydrophosphoric acid H 2 TeH 2 Te hydrotelluric acid hydrotelluric acid

56. Oxoacids hydrogen, oxygen, and another element hydrogen, oxygen, and another element anion-suffix acid anion-suffix acid anion anion suffix: -ate  -ic, -ite  -ous suffix: -ate  -ic, -ite  -ous acid acid

57. Examples, name H 3 PO 4 H 2 SO 4H 3 PO 4 H 2 SO 4 phosphoric acid sulfuric acid phosphoric acid sulfuric acid PO 4 3– SO 4 2– PO 4 3– SO 4 2– phosphate sulfate phosphate sulfate H 3 PO 3 H 2 SO 3H 3 PO 3 H 2 SO 3 phosphorous acid sulfurous acid phosphorous acid sulfurous acid PO 3 3– SO 3 2– PO 3 3– SO 3 2– phosphite sulfite phosphite sulfite

58. If more than two oxoacids, that with less O than the –ous will be hypo– –ous; that with more O than the –ic will be per– –ic. HBrO HBrO 2HBrO HBrO 2 hypobromous acid bromous acid hypobromous acid bromous acid BrO – BrO 2 – BrO – BrO 2 – hypobromite bromite hypobromite bromite HBrO 3 HBrO 4HBrO 3 HBrO 4 bromic acid perbromic acid bromic acid perbromic acid BrO 3 – BrO 4 – BrO 3 – BrO 4 – bromate perbromate bromate perbromate bromic acid perbromic acid bromic acid perbromic acid The same applies to Cl and I, not to F

59. Molecular Compounds contain only nonmetals bound by covalent bonds methane, CH 4 Covalent bonds form when atoms share electrons to complete octets. There are no ions.

60. Naming Molecular (Covalent) Compounds To name covalent compounds STEP 1: Name the first nonmetal as an element.STEP 1: Name the first nonmetal as an element. STEP 2: Name the second nonmetal with an ending.STEP 2: Name the second nonmetal with an ide ending. STEP 3: Use to indicate the number of atoms (subscript) of each element.STEP 3: Use prefixes to indicate the number of atoms (subscript) of each element. Prefixes Used in Naming Covalent Compounds # of Atoms Prefix 1 Mono 2 Di 3 Tri 4 Tetra 5 Penta 6 Hexa 7 Hepta 8 Octa 9Nona 10Deca

61. Exception prefix mono- omitted from name of first elementprefix mono- omitted from name of first element eg. CO 2eg. CO 2 carbon dioxide, not monocarbon dioxide carbon dioxide, not monocarbon dioxide

62. Examples, name What is the name of SO 3 ? 1. The first nonmetal is S sulfur. 2. The second nonmetal is O named oxide. 3.The subscript 3 of O is shown as the prefix. prefix tri. SO 3  sulfur oxide SO 3  sulfur trioxide The subscript 1 (for S) or mono is understood The subscript 1 (for S) or mono is understood

63. Examples, name P 2 O 5P 2 O 5 phosphorus phosphorus Diphosphorus (two P atoms) Diphosphorus (two P atoms) oxygen oxygen oxide oxide Pentoxide(5 O atoms) (not pentaoxide: a of penta- is dropped) Pentoxide(5 O atoms) (not pentaoxide: a of penta- is dropped) diphosphorus pentoxide diphosphorus pentoxide

64. P 3 Br 6P 3 Br 6 phosphorus phosphorus triphosphorus triphosphorus bromine bromine bromide bromide hexabromide hexabromide triphosphorus hexabromide triphosphorus hexabromide

65. Learning Check Select the correct name for each compound. A.SiCl 4 1) silicon chloride 2) tetrasilicon chloride 3) silicon tetrachloride B. P 2 O 5 1) phosphorus oxide 2) phosphorus pentoxide 3) diphosphorus pentoxide C.Cl 2 O 7 1) dichlorine heptoxide 2) dichlorine oxide 3) chlorine heptoxide

66. Learning Check Write the name of each molecular (covalent) compound. CO_____________________ CO 2 _____________________ PCl 3 _____________________ CCl 4 _____________________ N 2 O_____________________

67. Learning Check Write the correct formula for each of the following. A. phosphorus pentachloride B. dinitrogen trioxide C. sulfur hexafluoride

68. Common Names H 2 O waterH 2 O water PH 3 phosphinePH 3 phosphine CaCO 3 limestoneCaCO 3 limestone CaO limeCaO lime Ca(OH) 2 slaked limeCa(OH) 2 slaked lime CH 4 methaneCH 4 methane NaCl table saltNaCl table salt

69. N 2 O laughing gasN 2 O laughing gas NaHCO 3 baking sodaNaHCO 3 baking soda Na 2 CO 3 10H 2 O washing sodaNa 2 CO 3 10H 2 O washing soda Sodium carbonate decahydrate MgSO 4 7H 2 O epsom saltMgSO 4 7H 2 O epsom salt Mg(OH) 2 milk of magnesiaMg(OH) 2 milk of magnesia Ca(SO 4 )2H 2 O gypsumCa(SO 4 )2H 2 O gypsum

70. Nomenclature Details acid (aqueous solution) acid (aqueous solution) HBr (g) hydrogen bromide (gas phase) HBr (g) hydrogen bromide (gas phase) HBr (aq) hydrobromic acid (aqueous HBr (aq) hydrobromic acid (aqueous solution) solution) Same for HF, HCl, and HI. Same for HF, HCl, and HI.

71. Hydrogen belongs in group all its ownbelongs in group all its own generally considered a nonmetalgenerally considered a nonmetal forms both H + and H – in ionic compoundsforms both H + and H – in ionic compounds HCl (in aqueous solution) HCl (in aqueous solution) H + and Cl – H + and Cl – NaH (sodium hydride, same with Li, K, Rb, Ca, Ba…) NaH (sodium hydride, same with Li, K, Rb, Ca, Ba…) Na + and H – Na + and H – mostly forms molecular compoundsmostly forms molecular compounds e.g.. CH 4, NH 3, C 2 H 6 SO,... e.g.. CH 4, NH 3, C 2 H 6 SO,...

72. Chemical Formulas empirical formulaempirical formula –indicates the elements present and their simplest, whole-number ratio in a compound molecular formulamolecular formula –indicates elements present and the exact number of atoms of each in a unit of a compound structural formulastructural formula –a molecular formula that includes structural information

73. Benzene structural formula molecular formula C 6 H 6 dividing by smallest subscript (6) dividing by smallest subscript (6) empirical formula CH C C C C C C H H H H H H

74. Dimethylether condensed formula H 3 COCH 3condensed formula H 3 COCH 3 molecular formula C 2 H 6 O (cannot be divided to get whole numbers; by diving by 2 we get O 0.5 )molecular formula C 2 H 6 O (cannot be divided to get whole numbers; by diving by 2 we get O 0.5 ) empirical formula C 2 H 6 Oempirical formula C 2 H 6 O H H | | Dimethyl C 2 H 6 O Dimethyl C 2 H 6 O CH 3 OCH 3 H─C ─ O─ C─H ether ether | | H H

75. Calcium Chloride (ionic) structural formula none (it is ionic)structural formula none (it is ionic) molecular formula none (it is ionic)molecular formula none (it is ionic) formula unit CaCl 2formula unit CaCl 2 empirical formula CaCl 2empirical formula CaCl 2

76. Other examples NAME Molecular Empirical formula formula formula formula butanoic acid C 4 H 8 O 2 C 4 H 8 O 2 C 2 H 4 O 2 2 2 2 2 2diboron hexahydride B 2 H 6 B 2 H 6 BH 3 2 2 2 2 sodium ditionateNa 2 S 2 O 4 NaSO 2 hexane C 6 H 14 CH 2.333 C 3 H 7

77. Molar Mass of a Compound is the mass (g) of one mole (6.022 x 10 23 ) of formula units for ionic compounds molecules for molecular compounds The molar mass of a compound is the sum of the molar masses of the elements in the Formula (times respective coefficients.) molecular or formula weight =molecular or formula weight = amu/molecule amu/molecule molar mass = g/mol of moleculesmolar mass = g/mol of molecules = g/mol of formula units (for = g/mol of formula units (for ionic compounds) ionic compounds)

78. Molar Mass of a Compound Example: Calculate the molar mass of CaCl 2 (ionic) Element Number of Moles Atomic Mass Total Mass Ca140.1 g/mole 40.1 g Cl235.5 g/mole 71.0 g CaCl 2 111.1 g/mole CaCl 2 111.1 g/mole For glucose: C 6 H 12 O 6 (molecular, covalent) C612.0 g/mole 72.0 g H 12 1.0 g/mole 12.0 g O6 16.0 g/mole 96.0 g C 6 H 12 O 6 180.0 g/mol

79. Molar Mass (M), Avogadro’s number, and atoms of elements 1 mole = 6.022 x 10 23 particles = molar mass(g) molecules or formula units for compounds From the molecular formula of glucose, C 6 H 12 O 6, 1 glucose molecule contains 6 C, 12 H, and 6 O atoms 1 mol glucose contains 6 mol C,12 mol H, and 6mol O x 6.022 x 10 23 particles (molecules and atoms) M (g/mol) = 6x AW C + 12 AW H + 6 AW O (from periodic table)

80. Molar Mass and Avogadro’s number Avogadro’s number (6.022 x 10 23 ) can be written as equalities and conversion factors. Equality: (atoms, molecules, ions, electrons, protons) 1 mole = 6.022 x 10 23 particles = molar mass (g) Particles here are molecules or Formula Units (ionic) 6.022 x 10 23 particles and 1 mole 1 mole 6.022 x 10 23 particles 1 mole 6.022 x 10 23 particles 6.022 x 10 23 particles and molar mass (g) molar mass (g) 6.022 x 10 23 particles molar mass (g) 6.022 x 10 23 particles 1 mole and molar mass (g) 1 mole and molar mass (g) molar mass (g) 1 mole molar mass (g) 1 mole

81. The mass (in grams) of a single molecule The molar mass of water, H 2 O, is 2 x 1.008 + 16.00 = 18.02 g/mol (about 18 mL) Does a water molecule have a mass of 18.02 g? No, it doesn’t. 18.02 g (the molar mass) 18.02 g (the molar mass) 1 mol H 2 O 1 mol H 2 O —————————— 6.022 x 10 23 H 2 O molecules 6.022 x 10 23 H 2 O molecules by diving, we get 2.99x10 −23 g per H 2 O molecule

82. What is the mass of 12.0 million benzene (C 6 H 6 ) molecules?What is the mass of 12.0 million benzene (C 6 H 6 ) molecules? Molar mass = (6 mol C)(12.01 g/mol C) + (6 mol H)(1.008 g/mol H) = 78.11 g/mol C 6 H 6 Molar mass = (6 mol C)(12.01 g/mol C) + (6 mol H)(1.008 g/mol H) = 78.11 g/mol C 6 H 6 12.0 million= 12,000,000 = 1.20 x 10 7 molecules 1.20 x 10 7 molec.x 1 mol x 78.11 g 6.022 x 10 23 molecules 1 mol 78.11 g 78.11 g = 1.99 x 10 –17 mol C 6 H 6  ────── = mol C 6 H 6 mol C 6 H 6 1.56 x 10 –15 C 6 H 6 1.56 x 10 –15 g C 6 H 6

83. Consider 4.49 g Ca 3 Consider 4.49 g Ca 3 (PO 4 ) 2. Its formula weight FW = 3x40.08 + 2x(30.97 + 4x16.00)=310.18 g/mol How many mol of Ca 3 is this?How many mol of Ca 3 (PO 4 ) 2 is this? 1 mol 1 mol 4.49 g x ───── = 0.0145 moles Ca 3 4.49 g x ───── = 0.0145 moles Ca 3 (PO 4 ) 2 310.18 g 310.18 g How many mol of Ca 2+ ions does this contain? 3 moles Ca 2+ 3 moles Ca 2+ 0.0145 mol Ca 3 x─────── = 0.0435 mol Ca 2+ 0.0145 mol Ca 3 (PO 4 ) 2 x─────── = 0.0435 mol Ca 2+ Ca 3 1 mol Ca 3 (PO 4 ) 2 How many Ca 2+ ions are there? Ca 2+ 6.022 x10 23 Ca 2+ 0.0435 mol Ca 2+ x ───────────= 2.61x10 22 ions Ca 2+ 1 mol Ca 2+

84. Consider 4.49 g Ca 3 (PO 4 ) 2 FW = 310.18 g/mol How many mol of P does that amount of Ca 3 contain?How many mol of P does that amount of Ca 3 (PO 4 ) 2 contain? 2 moles P 2 moles P 0.0145 mol Ca 3 x─────── = 0.0290 mol P 0.0145 mol Ca 3 (PO 4 ) 2 x─────── = 0.0290 mol P Ca 3 1 mol Ca 3 (PO 4 ) 2 How many mol of O does that amount contain?How many mol of O does that amount contain? 8 moles O 8 moles O 0.0145 mol Ca 3 x─────── = 0.116 mol O 0.0145 mol Ca 3 (PO 4 ) 2 x─────── = 0.116 mol O Ca 3 1 mol Ca 3 (PO 4 ) 2 How many grams of O are in that amount of salt?How many grams of O are in that amount of salt? 16.0 g O 16.0 g O 0.116 mol O x─────── = 1.86 g O 1 mol O

85. Consider 4.49 g Ca 3 (PO 4 ) 2 FW = 310.18 g/mol How many phosphate ions does that amount of Ca 3 contain? Ca 3 (PO 4 ) 2 contain? First we calculate how many moles of phosphate 2 moles PO 4 3− 2 moles PO 4 3− 0.0145 mol Ca 3 x───────── = 0.0290 mol PO 4 3− 0.0145 mol Ca 3 (PO 4 ) 2 x───────── = 0.0290 mol PO 4 3− Ca 3 1 mol Ca 3 (PO 4 ) 2 Second, using Avogadro’s number, 6.02 x 10 23 PO 4 3− 6.02 x 10 23 PO 4 3− 0.0290 mol PO 4 3− x ─────────── = 1.75 x 10 22 PO 4 3− PO 4 3− ions 1 mol PO 4 3− ions

86. What mass of C 4 H 8 O is required to supply 1.43 x 10 16 C atoms? Molar mass = 72.10 g/molMolar mass = 72.10 g/mol 1 mol C atoms 1 mol C 4 H 8 O 1 mol C atoms 1 mol C 4 H 8 O 1.43 x 10 16 C atoms x──────────── x ───────.022x10 23 C atoms 4 moles C 6.022x10 23 C atoms 4 moles C 72.10 g C 4 H 8 O 72.10 g C 4 H 8 O x──────────── = 4.28 x 10 −7 g C 4 H 8 O 1 mol C 4 H 8 O 1 mol C 4 H 8 O

87. Empirical Formula from Analysis It will be shown by example. We will use The Percent Composition, i.e, the amount (grams) of each element forming one compound in 100 g of that compound. (An example of calculating %s) For the compound A x B y C z (x, y, z are unknown) unknown) % A + % B + %C = 100

88. A compound was found to be 40.92 % C, 4.58 % H, and 54.50 % O by mass.A compound was found to be 40.92 % C, 4.58 % H, and 54.50 % O by mass. Determine its empirical formula. Determine its empirical formula. CxCxHyOzCxCxHyOz x, y, and z are the simplest whole number ratios of atoms in compound. Assume exactly 100 g of compound then 40.92 g C, 4.58 g H, and 54.50 g O are contained in those 100 g Firstly, calculate the number of moles of C, H, and O in that amount of compound… next

89. 1 mole C 1 mole C 40.92 g C x──────── = 3.407 moles C 12.01 g C 12.01 g C 1 mole H 1 mole H 4.58 g H x ──────── = 4.54 moles H 1.008 g H 1.008 g H 1 mole O 1 mole O 54.50 g O x──────── = 3.406 moles O 16.00 g O 16.00 g O C 3.407 H 4.54 O 3.406 Now, divide all # by the smallest C 3.407 H 4.54 O 3.406 Now, divide all # by the smallest C 3.407 H 4.54 O 3.406 The result is C 1.0 H 1.33 O 1.0 3.406 3.406 3.406 Finally, 1.33 x 3 = 4 Empirical formula is And the same for 1.0 x 3 C 3 H 4 O 3

90. Determine the empirical formula of a compound that is 36.4% Mn, 21.2% S, and 42.4% O by mass. assuming exactly 100 g compound 1 mole Mn 1 mole Mn 36.4 g Mn x──────── = 0.663 moles Mn 54.9 g Mn 54.9 g Mn 1 mole S 1 mole S 21.2 g S x ──────── = 0.660 moles S 32.1 g S 32.1 g S 1 mole O 1 mole O 42.4 g O x──────── = 2.65 moles O 16.0 g O 16.0 g O Mn 0.663 S 0.660 O 2.65 The result is Mn 1.0 S 1.0 O 4.02 0.660 0.660 0.660 empirical formula: MnSO 4 empirical formula: MnSO 4

91. Homework: Determine the empirical formula of a compound that is 27.6% Mn, 24.2% S, & 48.2% O answer Mn 2 S 3 O 12

92. molecular formula = n x (empirical formula) n = whole numbern = whole number molar mass (or formula weight) molar mass (or formula weight) n = ─────────────────────────n = ───────────────────────── weight of the empirical formula weight of the empirical formula Example: benzene, molecular formula: C 6 H 6 empirical formula: CH empirical formula: CH n = 6 6 x (CH) = C 6 H 6 n = 6 6 x (CH) = C 6 H 6

93. Example: A 4.99 g sample of a compound was found to contain 1.52 g N and oxygen. Determine its empirical and molecular formula if its MW is 92.04 determine empirical formula g O = 4.99 −1.52 1 mol N = 3.47 g 1.52 g N  ─────── = 0.108 mol N 14.01 g N 1 mol O 3.47 g O  ─────── = = 0.217 mol O 16.00 g O N 0.108 O 0.217 NO 2 N 0.108 O 0.217 NO 2 0.108 0.108

94. determine molecular formuladetermine molecular formula empirical formula NO 2 empirical formula NO 2 empirical weight = 14.01+ 2  16.00 = 46.01 empirical weight = 14.01+ 2  16.00 = 46.01 molecular weight = 92.04 (given in previous slide) molecular weight = 92.04 (given in previous slide) 92.04 92.04 n = ───── = 2 n = ───── = 2 46.01 46.01 Molecular formula = 2 x (emp. form.) Molecular formula = 2 x (emp. form.) = 2(NO 2 ) = 2(NO 2 ) molecular formula = N 2 O 4 molecular formula = N 2 O 4

95. A compound was found to contain only B and H. Analysis of an 8.247 g sample indicated that it contained 1.803 g H and had a MW of ~30. Determine its molecular formula and its MW to 4 SF. determine empirical formuladetermine empirical formula total mass = g B + g H total mass = g B + g H 8.247 g = g B + 1.803 g H 8.247 g = g B + 1.803 g H g B = 8.247 g compound −1.803 g H =6.444 g 1 mol B 1 mol B 6.444 g B  ───────= 0.5961 mol of boron 10.811 g B 10.811 g B

96. 1 mol H 1 mol H 1.803 g H  ─────── = 1.789 mol H 1.803 g H  ─────── = 1.789 mol H 1.008 g H 1.008 g H B 0.5961 H 1.789 BH 3 is the Empirical formula B 0.5961 H 1.789 BH 3 is the Empirical formula 0.5961 0.5961 0.5961 0.5961 determine molecular formuladetermine molecular formula MW 30 MW 30 n = ───────  ──── = 2 n = ───────  ──── = 2 emp. wt. 13.83 emp. wt. 13.83 mol. form. = 2(BH 3 ) = B 2 H 6 mol. form. = 2(BH 3 ) = B 2 H 6 B 2 H 6 MW = 2  10.811 + 6  1.008 B 2 H 6 MW = 2  10.811 + 6  1.008 = 27.67 g/mol

97. Hydrates compounds that contain intact H 2 O compounds that contain intact H 2 O eg. CaSO 4 2H 2 O eg. CaSO 4 2H 2 O Two H 2 O moles per mole of CaSO 4 Two H 2 O moles per mole of CaSO 4 compound name prefix-hydrate compound name prefix-hydrate (prefix indicates number of H 2 O molecules) (prefix indicates number of H 2 O molecules) calcium sulfate dihydrate calcium sulfate dihydrate

98. LiCl8H 2 OLiCl8H 2 O lithium chloride octahydrate lithium chloride octahydrate tungsten(VI) sulfate heptahydratetungsten(VI) sulfate heptahydrate W 6+ W 6+ SO 4 2– SO 4 2– W(SO 4 ) 3 7H 2 O W(SO 4 ) 3 7H 2 O

99. Hydrated nickel(II) chloride is a beautiful green color, crystalline compound. When heated strongly, the compound is dehydrated. If 0.235 g of NiCl 2 xH 2 O gives 0.128 g of NiCl 2 on heating, what is the value of x?Hydrated nickel(II) chloride is a beautiful green color, crystalline compound. When heated strongly, the compound is dehydrated. If 0.235 g of NiCl 2 xH 2 O gives 0.128 g of NiCl 2 on heating, what is the value of x? g of NiCl 2 xH 2 O = g NiCl 2 + g H 2 O Then, hydrated dehydrated hydrated dehydrated g H 2 O = 0.235 g – 0.128 g = 0.107 g 1 mol NiCl 2 1 mol NiCl 2 0.128 g NiCl 2 x ──────── = 0.000987 moles NiCl 2 0.128 g NiCl 2 x ──────── = 0.000987 moles NiCl 2 129.7 g NiCl 2 129.7 g NiCl 2 1 mol H 2 O 1 mol H 2 O 0.107 g H 2 O x ──────── = 0.00594 moles H 2 O 18.0 g H 2 O 18.0 g H 2 O NiCl 2 0.000987 H 2 O 0.00594 NiCl 2 6H 2 O x = 6 moles of water 0.000987 0.000987 per mole of NiCl 2 0.000987 0.000987 per mole of NiCl 2

100. Percent Composition The percent of the total mass of a substance represented by each element within that substance In other words, how many grams of every element are in 100 g of the compound

101. Determine the % composition of aluminum bromite Al 3+Al 3+ BrO 2 – Al(BrO 2 ) 3 Al Br O 26.982FW = 26.982 + 3 x(79.904 + 2 x 16.000) or = 26.982 + 3x79.904 + 6x16.000 = 362.68 g/mol

102. g Al g Al % Al = ──────── x 100 % Al = ──────── x 100 g compound g compound 26.982 g 26.982 g % Al = ──────── x 100 = 7.440 % % Al = ──────── x 100 = 7.440 % 362.68 g 362.68 g g Br g Br % Br = ──────── x 100 % Br = ──────── x 100 g compound g compound 3 x 79.904 g 3 x 79.904 g % Br = ──────── x 100 = 66.095 % % Br = ──────── x 100 = 66.095 % 362.68 g 362.68 g % O = 100% – 7.440% – 66.095 = 26.465%

103. 3.69 Silver chloride, often used in silver plating, contains 75.27% Ag. What mass of silver chloride is required to plate 155 mg of pure silver? AgCl The 75.27% means there is 75.27 g of Ag in 100 g AgCl. That can be used as a conversion factor (ratio) 1  10 −3 g Ag 1  10 −3 g Ag 155 mg Ag  ──────── = 0.155 g Ag 155 mg Ag  ──────── = 0.155 g Ag 1 mg 1 mg 100 g AgCl 100 g AgCl 0.155 g Ag  ──────── = 0.206 g AgCl 75.27 g Ag 75.27 g Ag or, with mg, or, with mg, 100 mg AgCl 100 mg AgCl 155 mg Ag  ──────── = 206 mg AgCl (with mg) 75.27 mg Ag 75.27 mg Ag