2. Unit 5 Nomenclature Slides adapted from Nivaldo Tro

3. Formulas Describe Compounds 2 water = H 2 O  two atoms of hydrogen and 1 atom of oxygen

4. Order of Elements in a Formula metals written first ▫NaCl nonmetals written in order from Table 5.1 ▫CO 2 ▫are occasional exceptions for historical or informational reasons  H 2 O, but NaOH 3 Table 5.1 Order of Listing Nonmetals in Chemical Formulas CPNHSIBrClOF

5. Molecular View of Elements and Compounds 4

6. Molecular Elements Certain elements occur as 2 atom molecules Rule of 7’s ▫there are 7 common diatomic elements ▫find the element with atomic number 7, N ▫make a figure 7 by going over to Group 7A, then down ▫don’t forget to include H 2 5 H2H2 Cl 2 Br 2 I2I2 7 VIIA N 2 O 2 F 2 P4P4 S8S8

7. Major Classes Ionic ▫metal + nonmetal  metal first in formula  Binary Ionic (Type I & II) ▫compounds with polyatomic ions Molecular ▫2 nonmetals  Binary Molecular (Type III) or Binary Covalent ▫Includes Acids – formula starts with H  though acids are molecular, they behave as ionic when dissolved in water  may be binary or oxyacid 6

8. 7 Ionic Bonds metal to nonmetal metal loses electrons to form cation nonmetal gains electrons to form anion ionic bond results from + to - attraction ▫larger charge = stronger attraction ▫smaller ion = stronger attraction Lewis Theory allow us to predict the correct formulas of ionic compounds

9. Formula-to-Name Rules for Ionic Made of cation and anion Name by simply naming the ions ▫If cation is:  Type I metal = metal name  Type II metal = metal name(charge)  Polyatomic ion = name of polyatomic ion ▫If anion is:  Nonmetal = stem of nonmetal name + ide  Polyatomic ion = name of polyatomic ion 8

10. Monatomic Nonmetal Anion determine the charge from position on the Periodic Table to name anion, change ending on the element name to –ide 9 4A = -45A = -36A = -27A = -1 C = carbideN = nitrideO = oxideF = fluoride Si = silicideP = phosphideS = sulfideCl = chloride

11. Metal Cations Type I ▫metals whose ions can only have one possible charge  IA, IIA, (Al, Ga, In) ▫determine charge by position on the Periodic Table  IA = +1, IIA = +2, (Al, Ga, In = +3) Type II ▫metals whose ions can have more than one possible charge ▫determine charge by charge on anion 10 How do you know a metal cation is Type II? its not Type I !!!

12. Determine if the following metals are Type I or Type II. If Type I, determine the charge on the cation it forms. lithium, Li copper, Cu gallium, Ga tin, Sn strontium, Sr 11

13. Determine if the following metals are Type I or Type II. If Type I, determine the charge on the cation it forms. lithium, LiType I+1 copper, CuType II gallium, GaType I+3 tin, SnType II strontium, SrType I+2 12

14. Type I Binary Ionic Compounds Contain Metal Cation + Nonmetal Anion Metal listed first in formula & 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 13

15. Examples LiCl = Lithium Chloride AlCl 3 = Aluminum Chloride NaBr = Sodium Bromide MgS = Magnesium Sulfide K 3 N = Potassium Nitride 14

16. Name to Formula Compounds that Contain Ions compounds of metals with nonmetals are made of ions ▫metal atoms form cations, nonmetal atoms for anions 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 2 Na + ions for every S 2- ion to balance the charges, therefore the formula must be Na 2 S 15

17. Writing Formulas for Ionic Compounds 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 cation cancels the sum of the anions 16

18. 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 17 Al +3 column IIIA O 2- column VIA Al +3 O 2- Al 2 O 3 Al = (2)∙(+3) = +6 O = (3)∙(-2) = -6

19. 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 18

20. Type II Binary Ionic Compounds Contain Metal Cation + Nonmetal Anion Metal listed first in formula & 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 3.nonmetal anion named by changing the ending on the nonmetal name to -ide 19

21. Example – Ionic Compounds 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 20 Mn +4 S 2- Mn +4 S 2- Mn 2 S 4 Mn = (1)∙(+4) = +4 S = (2)∙(-2) = -4 MnS 2

22. Examples PbO = lead(II) oxide PbO 2 = lead(IV) oxide Mn 2 O 3 = manganese(III) oxide Exceptions to the Rule ZnCl 2 = zinc(II) chloride -> zinc chloride AgCl = silver(I) chloride -> silver chloride 21

23. Compounds Containing Polyatomic Ions Polyatomic ions are single ions that contain more than one atom Name any ionic compound by naming cation first and then anion ▫Non-polyatomic cations named like Type I and II ▫Non-polyatomic anions named with -ide 22

24. Molecules with Polyatomic Ions 23 Mg(NO 3 ) 2 compound called magnesium nitrate symbol of the polyatomic ion called nitrate symbol of the polyatomic ion called sulfate CaSO 4 compound called calcium sulfate implied “1” subscript on magnesium implied “1” subscript on calcium parentheses to group two NO 3 ’sno parentheses for one SO 4

25. Molecules with Polyatomic Ions 24 Mg(NO 3 ) 2 compound called magnesium nitrate CaSO 4 compound called calcium sulfate subscript indicating two NO 3 groups no subscript indicating one SO 4 group implied “1” subscript on nitrogen, total 2 N implied “1” subscript on sulfur, total 1 S stated “3” subscript on oxygen, total 6 O stated “4” subscript on oxygen, total 4 O

26. Some Common Polyatomic Ions 25 NameFormula acetateC2H3O2–C2H3O2– carbonateCO 3 2– hydrogen carbonate (aka bicarbonate) HCO 3 – hydroxideOH – nitrateNO 3 – nitriteNO 2 – chromateCrO 4 2– dichromateCr 2 O 7 2– ammoniumNH 4 + NameFormula hypochloriteClO – chloriteClO 2 – chlorateClO 3 – perchlorateClO 4 – sulfateSO 4 2– sulfiteSO 3 2– hydrogensulfate (aka bisulfate) HSO 4 – hydrogensulfite (aka bisulfite) HSO 3 –

27. Patterns for Polyatomic Ions 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, the name adds hydrogen- prefix before name and add 1 to the charge CO 3 2- = carbonate  HCO 3 -1 = hydrogencarbonate 26

28. Example – Ionic Compounds 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 27 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

29. Example – Ionic Compounds ammonium carbonate 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 28 NH 4 + CO 3 2- NH 4 + CO 3 2- (NH 4 ) 2 CO 3 NH 4 = (2)∙(+1) = +2 CO 3 = (1)∙(-2) = -2 (NH 4 ) 2 CO 3

30. 29 Covalent Bonds often found between two nonmetals typical of molecular species atoms bonded together to form molecules ▫strong attraction sharing pairs of electrons to attain octets molecules generally weakly attracted to each other ▫observed physical properties of molecular substance due to these attractions

31. Binary Molecular Compounds of 2 Nonmetals (Type III) 1.Name first element in formula first ▫use the full name of the element 2.Name the second element in the formula with an -ide ▫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 30

32. Subscript - Prefixes 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- 31 drop last “a” if name begins with vowel

33. Examples H 2 O = Water SO 2 = Sulfur Dioxide BF 3 = Boron Trifluoride CO 2 = Carbon Dioxide CO = Carbon Monoxide O 3 = Ozone NH 3 = Ammonia 32

34. Writing the Formulas from the NamesExample – Binary Molecular dinitrogen pentoxide Identify the symbols of the elements nitrogen = N oxide = oxygen = O Write the formula using prefix number for subscript di = 2, penta = 5 N 2 O 5 33

35. Acids Contain H +1 cation and anion ▫in aqueous solution Binary acids have H +1 cation and nonmetal anion Oxyacids have H +1 cation and polyatomic anion 34

36. Formula-to-Name Acids acids are molecular compounds that often behave like they are made of ions All names have acid at end Binary Acids = hydro prefix + stem of the name of the nonmetal + ic suffix Oxyacids ▫if polyatomic ion ends in –ate = name of polyatomic ion with –ic suffix ▫if polyatomic ion ends in –ite = name of polyatomic ion with –ous suffix 35

37. Sample - Naming Binary Acids – HCl 4.Identify the anion Cl = Cl -, chloride because Group 7A 5.Name the anion with an –ic suffix Cl - = chloride  chloric 6.Add a hydro- prefix to the anion name hydrochloric 7.Add the word acid to the end hydrochloric acid 36

38. Example – Naming Oxyacids H 2 SO 4 4.Identify the anion SO 4 = SO 4 2- = sulfate 5.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 6.Write the name of the anion followed by the word acid sulfuric acid (kind of an exception, to make it sound nicer!) 37

39. Example – Naming Oxyacids H 2 SO 3 1.Is it one of the common exceptions? H 2 O, NH 3, CH 4, NaCl, C 12 H 22 O 11 = No! 2.Identify Major Class first element listed is H,  Acid 3.Identify the Subclass 3 elements in the formula,  Oxyacid 38

40. Example – Naming Oxyacids H 2 SO 3 4.Identify the anion SO 3 = SO 3 2- = sulfite 5.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 6.Write the name of the anion followed by the word acid sulfurous acid 39

41. Name to Formula for Acids 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.Reduce subscripts to smallest whole number ratio 5.Check that the total charge of the cations cancels the total charge of the anions 40 H+H+ S 2- H + S 2- H2SH2S H = (2)∙(+1) = +2 S = (1)∙(-2) = -2 H2SH2S in all acids the cation is H + hydro means binary

42. 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.Reduce subscripts to smallest whole number ratio 5.Check that the total charge of the cations cancels the total charge of the anions 41 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 in all acids the cation is H + no hydro means polyatomic ion -ic means -ate ion

43. Know the following strong acids: HBr – Hydrobromic Acid HCl – Hydrochloric Acid HI – Hydroiodic Acid H 2 SO 4 – Sulfuric Acid HNO 3 – Nitric Acid HClO 4 – Perchloric Acid Other important acids: HC 2 H 3 O 2 – Acetic Acid (vinegar) H 3 PO 4 – Phosphoric Acid H 2 CO 3 – Carbonic Acid (carbon dioxide in water) 42

44. Practice - What are the formulas for the following acids? chlorous acid- phosphoric acid- hydrobromic acid- 43

45. Practice - What are the formulas for the following acids? chlorous acid-HClO 2 phosphoric acid-H 3 PO 4 hydrobromic acid-HBr 44

46. Formula-to-Name Flow Chart 45

47. Alkanes… 1Carbon and 2C+2 hydrogens Methane1 carbon4 hydrogens Ethane2 carbons6 hydrogens Propane3 carbons8 hydrogens Butane4 carbons10 hydrogens Pentane5 carbons12 hydrogens Hexane6 carbons14 hydrogens Heptane7 carbons16 hydrogens Octane8 carbons18 hydrogens Nonane9 carbons20 hydrogens Decane10 carbons22 hydrogens 46

48. 47

49. More Examples and Help with Nomenclature 48

50. 49 Write a chemical formula for each of the following: (a) the compound containing two aluminum atoms to every three oxygen atoms (b) the compound containing three oxygen atoms to every sulfur atom (c) the compound containing four chlorine atoms to every carbon atom EXAMPLE 5.2 Writing Chemical Formulas Since aluminum is the metal, it is listed first. Since sulfur is below oxygen on the periodic table and since it occurs before oxygen in Table 5.1, it is listed first. Since carbon is to the left of chlorine on the periodic table and since it occurs before chlorine in Table 5.1, it is listed first. Solution: (a)Al 2 O 3 (b)SO 3 (c)CCl 4

51. 50 Determine the number of each type of atom in Mg 3 (PO 4 ) 2. EXAMPLE 5.3 Total Number of Each Type of Atom in a Chemical Formula Solution: Mg: There are three Mg atoms, as indicated by the subscript 3. P: There are two P atoms, as we see by multiplying the subscript outside the parentheses (2) by the subscript for P inside the parentheses, which is 1 (implied). O:There are eight O atoms, as we see by multiplying the subscript outside the parentheses (2) by the subscript for O inside the parentheses (4).

52. 51 Classify each of the following substances as an atomic element, molecular element, molecular compound, or ionic compound. (a) krypton (b) CoCl 2 (c) nitrogen (d) SO 2 (e) KNO 3 EXAMPLE 5.4 Classifying Substances as Atomic Elements, Molecular Elements, Molecular Compounds, or Ionic Compounds Solution: (a) Krypton is an element that is not listed as diatomic in Table 5.2; therefore, it is an atomic element. (b) CoCl 2 is a compound composed of a metal (left side of periodic table) and nonmetal (right side of the periodic table); therefore, it is an ionic compound. (c) Nitrogen is an element that is listed as diatomic in Table 5.2; therefore, it is a molecular element. (d) SO 2 is a compound composed of two nonmetals; therefore, it is a molecular compound. (e) KNO 3 is a compound composed of a metal and two nonmetals; therefore, it is an ionic compound.

53. 52 Write a formula for the ionic compound that forms from aluminum and oxygen. EXAMPLE 5.5 Writing Formulas for Ionic Compounds Solution: Al 3 + O 2– 1. Write the symbol for the metal and its charge followed by the symbol of the nonmetal and its charge. For many elements, these charges can be determined from their group number in the periodic table (refer to Figure 4.14). 2. Make the magnitude of the charge on each ion (without the sign) become the subscript for the other ion. In this case, the numbers cannot be reduced any further; the correct formula is Al 2 O 3. 3. Reduce the subscripts to give a ratio with the smallest whole numbers. 4. Check that the sum of the charges of the cations exactly cancels the sum of the charges of the anions. Cations: 2(3 + ) = 6 + Anions: 3(2–) = 6– The charges cancel.

54. 53 Write a formula for the ionic compound that forms from magnesium and oxygen. EXAMPLE 5.6 Writing Formulas for Ionic Compounds Solution: Mg 2 + O 2– 1. Write the symbol for the metal and its charge followed by the symbol of the nonmetal and its charge. For many elements, these charges can be determined from their group number in the periodic table (refer to Figure 4.14). 2. Make the magnitude of the charge on each ion (without the sign) become the subscript for the other ion. To reduce the subscripts, divide both subscripts by 2. Mg 2 O 2 ÷ 2 = MgO 3. Reduce the subscripts to give a ratio with the smallest whole numbers. 4. Check that the sum of the charges of the cations exactly cancels the sum of the charges of the anions. Cations: 2 + Anions: 2– The charges cancel.

55. 54 Write a formula for the compound that forms from potassium and oxygen. EXAMPLE 5.7 Writing Formulas for Ionic Compounds Solution: We first write the symbol for each ion along with its appropriate charge from its group number in the periodic table. K + O 2– We then make the magnitude of each ion’s charge become the subscript for the other ion. K + O 2– becomes K 2 O No reducing of subscripts is necessary in this case. Finally, we check to see that the sum of the charges of the cations [2(1 + ) = 2 + ] exactly cancels the sum of the charges of the anion (2–). The correct formula is K 2 O.

56. 55 Give the name for the compound MgF 2. EXAMPLE 5.8 Naming Type I Ionic Compounds Solution: The cation is magnesium. The anion is fluorine, which becomes fluoride. The correct name is magnesium fluoride.

57. 56 EXAMPLE 5.9 Naming Type II Ionic Compounds Give the name for the compound PbCl 4. Solution: The name for PbCl 4 consists of the name of the cation, lead, followed by the charge of the cation in parenthesis (IV), followed by the base name of the anion, chlor-, with the ending -ide. The full name is lead(IV) chloride. We know the charge on Pb is 4 + because the charge on Cl is 1 –. Since there are 4 Cl – anions, the Pb cation must be Pb 4 +. PbCl 4 lead (IV) chloride

58. 57 EXAMPLE 5.10 Naming Ionic Compounds That Contain a Polyatomic Ion Give the name for the compound K 2 CrO 4. The name for K 2 CrO 4 consists of the name of the cation, potassium, followed by the name of the polyatomic ion, chromate. Solution: K 2 CrO 4 potassium chromate

59. 58 EXAMPLE 5.11 Naming Molecular Compounds Name each of the following: CCl 4 BCl 3 SF 6. Solution: CCl 4 The name of the compound is the name of the first element, carbon, followed by the base name of the second element, chlor, prefixed by tetra- to indicate four, and the suffix -ide. The entire name is carbon tetrachloride. BCl 3 The name of the compound is the name of the first element, boron, followed by the base name of the second element, chlor, prefixed by tri- to indicate three, and the suffix -ide. The entire name is boron trichloride. SF 4 The name of the compound is the name of the first element, sulfur, followed by the base name of the second element, fluor, prefixed by hexa- to indicate six, and the suffix -ide. The entire name is sulfur hexafluoride.

60. 59 EXAMPLE 5.12 Naming Binary Acids Give the name of H 2 S. The base name of S is sulfur, so the name is hydrosulfuric acid. Solution: H 2 S hydrosulfuric acid

61. 60 EXAMPLE 5.13 Naming Oxyacids Give the name of HC 2 H 3 O 2. The oxyanion is acetate, which ends in -ate; therefore, the name of the acid is acetic acid. Solution: HC 2 H 3 O 2 acetic acid

62. 61 EXAMPLE 5.14 Nomenclature Using Figure 5.17 Name each of the following: CO, CaF 2, HF, Fe(NO 3 ) 3, HClO 4, H 2 SO 3 Solution: For each compound, the following table shows how to use Figure 5.17 to arrive at a name for the compound. FormulaFlow Chart PathName COMolecularcarbon monoxide CaF 2 calcium fluoride HFhydrofluoric acid Fe(NO 3 ) 3 iron(III) nitrate HClO 4 perchloric acid H 2 SO 3 sulfurous acid Ionic Type I Acid Binary Ionic Type II Acid Oxyacid -ate Acid Oxyacid -ite

63. 62 EXAMPLE 5.15 Calculating Formula Mass Calculate the formula mass of carbon tetrachloride, CCl 4. Solution: To find the formula mass, we sum the atomic masses of each atom in the chemical formula. Formula mass = 1 x (Formula mass C) + 4 x (Formula mass Cl) = 12.01 amu + 4(35.45 amu) = 153.81 amu

64. 63 EXAMPLE 5.16 Constant Composition of Compounds Two samples said to be carbon disulfide (CS 2 ) are decomposed into their constituent elements. One sample produced 8.08 g S and 1.51 g C, while the other produced 31.3 g S and 3.85 g C. Are these results consistent with the law of constant composition? Solution: Sample 1 These results are not consistent with the law of constant composition and the data given must therefore be in error.

65. 64 EXAMPLE 5.17 Writing Chemical Formulas Write a chemical formula for the compound containing one nitrogen atom for every two oxygen atoms. Solution: NO 2

66. 65 EXAMPLE 5.18 Total Number of Each Type of Atom in a Chemical Formula Determine the number of each type of atom in PB(ClO 3 ) 2. Solution: One Pb atom Two Cl atoms Six O atoms

67. 66 EXAMPLE 5.19 Classifying Elements as Atomic or Molecular Classify each of the following elements as atomic or molecular: Na, I, N. Solution: Na: atomic I: molecular (I 2 ) N: molecular (N 2 )

68. 67 EXAMPLE 5.20 Classifying Compounds as Ionic or Molecular Classify each of the following compounds as ionic or molecular. If they are ionic, classify them as Type I or Type II ionic compounds: FeCl 3, K 2 SO 4, CCl 4 Solution: FeCl 3 : ionic, Type II K 2 SO 4 : ionic, Type I CCl 4 : molecular

69. 68 EXAMPLE 5.21 Writing Formulas for Ionic Compounds Write a formula for the compound that forms from lithium and sulfate ions. Solution: Li + SO 2– Li 2 (SO 4 ) In this case, the subscripts cannot be further reduced. Li 2 SO 4 Cations:Anions: 2(1 + ) = 2 + 2– 4

70. 69 EXAMPLE 5.22 Naming Type I Binary Ionic Compounds Give the name for the compound Al 2 O 3 Solution: aluminum oxide

71. 70 EXAMPLE 5.23 Naming Type II Binary Ionic Compounds Give the name for the compound Fe 2 S 3 Solution: 3 sulfide ions x (2–) = 6– 2 iron ions x (?) = 6 + ? = 3 + Charge of each iron ion = 3 + iron(III) sulfide

72. 71 EXAMPLE 5.24 Naming Compounds Containing a Polyatomic Ion Give the name for the compound Co(ClO 4 ) 2. Solution: 3 perchlorate x (1–) = 2– Charge of cobalt ion = 2 + cobalt(II) perchlorate

73. 72 EXAMPLE 5.25 Naming Molecular Compounds Name the compound NO 2. Solution: nitrogen dioxide

74. 73 EXAMPLE 5.26 Naming Binary Acids Name the acid HI. Solution: hydroiodic acid

75. 74 EXAMPLE 5.27 Naming Oxyacids with an Oxyanion Ending in -ate Name the acid H 2 SO 4. Solution: The oxyanion is sulfate. The name of the acid is sulfuric acid.

76. 75 EXAMPLE 5.28 Naming Oxyacids with an Oxyanion Ending in -ite Name the acid HClO 2. Solution: The oxyanion is chlorite. The name of the acid is chlorous acid.

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