Chapter 5 Molecules and Compounds 2 Molecules and Compounds Salt Sodium – shiny, reactive,...

Chapter 5Molecules

andCompounds

2

Molecules and Compounds• Salt

Sodium – shiny, reactive, poisonousChlorine – pale yellow gas, reactive,

poisonousSodium chloride – table salt

• SugarCarbon – pencil or diamondsHydrogen – flammable gasOxygen – a gas in airCombine to form white crystalline

sugar

http://wps.prenhall.com/wps/media/objects/165/169061/MixturesandCompounds.html

3

Law of Constant Composition• all pure substances have constant composition

all samples of a pure substance contain the same elements in the same percentages (ratios)

mixtures have variable composition

4

Compounds Display Constant Composition

If we decompose water by electrolysis, we find 16.0 grams of oxygen to every 2.00 grams of hydrogen.

Water has a constant Mass Ratio of Oxygen to Hydrogen of 8.0.

0.8g 2.0

g 0.16

hydrogen of mass

oxygen of mass Ratio Mass

5

Why do Compounds ShowConstant Composition

• smallest piece of a compound is called a molecule

• every molecule of a compound has the same number and type of atoms as determined by the electronic structures of the atoms (more on that later in the year)

• since all the molecules of a compound are identical…every sample will have the same ratio of the

elementsevery sample of the compound will have the

same properties

6

Two samples of carbon dioxide, obtained from different sources, were decomposed into their constituent elements. One sample produced 4.8 g of oxygen and 1.8 g of carbon, and the other sample produced 17.1 g of oxygen and 6.4 g of carbon. Show that these results are consistent with the law of constant composition.

EXAMPLE 5.1 Constant Composition of Compounds

To show this, compute the mass ratio of one element to the other by dividing the larger mass by the smaller one. For the first sample:

For the second sample:

Solution:

Since the ratios are the same for the two samples, these results are consistent with the law of constant composition.

7

1. 12. 0.333. 34. 45. 0.25

5.3A What is the carbon-hydrogen mass ratio for methane (CH4)?

8

1. 13.32. 1853. 228 4. 3125. 553

5.1B If the mass ratio of lead(II) sulfide is 270.0 g lead and 41.8 g sulfur, how much lead is required to completely react with 85.6 g of sulfur?

9

Formulas Describe Compounds

water = H2O two atoms of

hydrogen and 1 atom of oxygen

table sugar = C12H22O11 12 atoms

of C, 22 atoms of H and 11 atoms O

10

Order of Elements in a Formula• metals written first

NaCl

• nonmetals written in order from Table 5.1CO2

are occasional exceptions for historical or informational reasonsH2O, but NaOH

Table 5.1Order of Listing Nonmetalsin Chemical Formulas

C P N H S I Br Cl O F

11

Molecules with Polyatomic Ions

Mg(NO3)2

compound calledmagnesium nitrate

symbol of the polyatomic ion called nitrate

symbol of the polyatomic ion called sulfate

CaSO4

compound calledcalcium sulfate

implied “1” subscripton magnesium

implied “1” subscripton calcium

parentheses to group two NO3’s no parentheses for one SO4

12

Molecules with Polyatomic Ions

Mg(NO3)2

compound calledmagnesium nitrate

CaSO4

compound calledcalcium sulfate

subscript indicatingtwo NO3 groups

no subscript indicatingone SO4 group

implied “1” subscripton nitrogen, total 2 N

implied “1” subscripton sulfur, total 1 S

stated “3” subscripton oxygen, total 6 O

stated “4” subscripton oxygen, total 4 O

13

Classifying Materials• 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

14

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, Nmake a figure 7 by going over to Group 7A, then downdon’t forget to include H2

H2

Cl2

Br2

I2

7VIIA

N2 O2 F2

P4 S8

15

Molecular Compounds

• two or more nonmetals

• smallest unit is a molecule

16

Ionic Compounds

• metals + nonmetals

• no individual molecule units, instead have a 3-dimensional array of cations and anions made of formula units

17

Molecular View of Elements and Compounds

18

Classify each of the following as either an atomic element, molecular element, molecular

compound or ionic compound

• aluminum, Al

• aluminum chloride, AlCl3

• chlorine, Cl2

• acetone, C3H6O

• carbon monoxide, CO

• cobalt, Co

19

Classify each of the following as either an atomic element, molecular element, molecular

compound or ionic compound

• aluminum, Al = atomic element

• aluminum chloride, AlCl3 = ionic compound

• chlorine, Cl2 = molecular element

• acetone, C3H6O = molecular compound

• carbon monoxide, CO = molecular compound

• cobalt, Co = atomic element

Formula-to-NameStep 1

Is the compound one of the exceptions to the rules?

•H2O = water, steam, ice

•NH3 = ammonia


What major class of compound is it?

Ionic or Molecular

22

Major Classes• Ionic

metal + nonmetalmetal first in formulaBinary Ionic

compounds with polyatomic ions

• Molecular2 nonmetals

Binary Molecular (or Binary Covalent)

Acids – formula starts with H though acids are molecular, they behave as ionic when

dissolved in watermay be binary or oxyacid


What major subclass of compound is it?

Binary Ionic, Ionic with Polyatomic Ions,

Binary Molecular,

Binary Acid, Oxyacid

24

Classifying Compounds• Compounds containing a metal and a nonmetal =

binary ionicType I and II

• Compounds containing a polyatomic ion = ionic with polyatomic ion

• Compounds containing two nonmetals = binary molecular compounds

• Compounds containing H and a nonmetal = binary acids

• Compounds containing H and a polyatomic ion = oxyacids


Apply Rules for the Class and Subclass

26

Formula-to-NameRules for Ionic

• Made of cation and anion

• Name by simply naming the ionsIf cation is:

Type I metal = metal nameType II metal = metal name(charge)Polyatomic ion = name of polyatomic ion

If anion is:Nonmetal = stem of nonmetal name + idePolyatomic ion = name of polyatomic ion

27

Monatomic Nonmetal Anion• determine the charge from position on the

Periodic Table

• to name anion, change ending on the element name to –ide

4A = -4 5A = -3 6A = -2 7A = -1

C = carbide N = nitride O = oxide F = fluoride

Si = silicide P = phosphide S = sulfide Cl = chloride

28

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 IImetals whose ions can have more than

one possible chargedetermine charge by charge on anion

How do you know a metal cation is Type II?

its not Type I !!!

29

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

30

Determine if the following metals are Type I or Type II. If Type I, determine the charge on

the cation it forms.

• lithium, Li Type I +1

• copper, Cu Type II

• gallium, Ga Type I +3

• tin, Sn Type II

• strontium, Sr Type I +2

31

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

32

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 Common Type II cations in Table 5.5

3. nonmetal anion named by changing the ending on the nonmetal name to -ide

33

Examples

• LiCl = lithium chloride

• AlCl3 = aluminum chloride

• PbO = lead(II) oxide

• PbO2 = lead(IV) oxide

• Mn2O3 = manganese(III) oxide

• ZnCl2 = zinc(II) chloride or zinc chloride

• AgCl = silver(I) chloride or silver chloride

34

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 anionNon-polyatomic cations named like Type I and IINon-polyatomic anions named with -ide

35

Li+1

Na+1

K+1

Rb+1

Cs+1

Be+2

Mg+2

Ca+2

Sr+2

Ba+2

Al+3

Ga+3

In+3

O-2

S-2

Se-2

Te-2

F-1

Cl-1

Br-1

I-1

N-3

P-3

As-3

IA

IIA IIIA VIIAVIAVA

Zn+2

Cd+2Ag+1

Fixed Charge Metals and Nonmetals

36

Some Common Polyatomic IonsName Formulaacetate C2H3O2

–

carbonate CO32–

hydrogen carbonate(aka bicarbonate)

HCO3–

hydroxide OH–

nitrate NO3–

nitrite NO2–

chromate CrO42–

dichromate Cr2O72–

ammonium NH4+

Name Formulahypochlorite ClO–

chlorite ClO2–

chlorate ClO3–

perchlorate ClO4–

sulfate SO42–

sulfite SO32–

hydrogensulfate(aka bisulfate)

HSO4–

hydrogensulfite(aka bisulfite)

HSO3–

37

Patterns for Polyatomic Ions

1. elements in the same column form similar polyatomic ions

same number of O’s and same charge

ClO3- = chlorate BrO3

- = bromate

2. if the polyatomic ion starts with H, the name adds hydrogen- prefix before name and add 1 to the chargeCO3

2- = carbonate HCO3-1 = hydrogencarbonate

38

Periodic Pattern of Polyatomic Ions-ate groups

BO3-3 NO3

-1

SiO3-2

PO4-3

SO4-2

ClO3-1

AsO4-3

SeO4-2

BrO3-1

TeO4-2

IO3-1

CO3-2

IIIA IVA VA VIA VIIA

39

Binary Molecular Compounds of 2 Nonmetals

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

40

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-

drop last “a” if name begins with vowel

41

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

42

Formula-to-NameAcids

• 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

43

Example – Naming Binary AcidsHCl

1. Is it one of the common exceptions?

H2O, NH3, CH4, NaCl, C12H22O11 = No!

2. Identify Major Classfirst element listed is H, Acid

3. Identify the Subclass2 elements, Binary Acid

44

Sample - Naming Binary Acids – HCl

4. Identify the anionCl = Cl-, chloride because Group 7A

5. Name the anion with an –ic suffixCl- = chloride chloric

6. Add a hydro- prefix to the anion namehydrochloric

7. Add the word acid to the endhydrochloric acid

45

Example – Naming OxyacidsH2SO4




3. Identify the Subclass3 elements in the formula, Oxyacid

46

Example – Naming Oxyacids H2SO4

4. Identify the anionSO4 = SO4

2- = sulfate

5. If the anion has –ate suffix, change it to –ic. If the anion has –ite suffix, change it to -ous

SO42- = sulfate sulfuric

6. Write the name of the anion followed by the word acidsulfuric acid

(kind of an exception, to make it sound nicer!)

47





3. Identify the Subclass3 elements in the formula, Oxyacid

48


4. Identify the anionSO3 = SO3

2- = sulfite

5. If the anion has –ate suffix, change it to –ic. If the anion has –ite suffix, change it to -ous

SO32- = sulfite sulfurous

6. Write the name of the anion followed by the word acid

sulfurous acid

49

Formula-to-Name Flow Chart

Name – to – Formula

51

Writing the Formulas from the Names

• For binary molecular compounds, use the prefixes to determine the subscripts

• For Type I, Type II, Ternary Compounds and Acids

1. Determine the ions present

2. Determine the charges on the cation and anion

3. Balance the charges to get the subscripts

52

Example – Binary Moleculardinitrogen pentoxide

• Identify the symbols of the elements

nitrogen = N

oxide = oxygen = O

• Write the formula using prefix number for subscript

di = 2, penta = 5

N2O5

53

Compounds that Contain Ions• compounds of metals with nonmetals are made

of ionsmetal 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 S2-, you will need 2 Na+ ions for every S2- ion to balance the charges, therefore the formula must be Na2S

54

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

55

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



5. Check that the total charge of the cations cancels the total charge of the anions

Al+3 column IIIA

O2- column VIA

Al+3 O2-

Al2 O3

Al = (2)∙(+3) = +6O = (3)∙(-2) = -6

56

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

57


• K+ with N3- K3N

• Ca+2 with Br- CaBr2

• Al+3 with S2- Al2S3

58

Example – Ionic Compoundsmanganese(IV) sulfide

1. Write the symbol for the cation and its charge

2. Write the symbol for the anion and its charge




Mn+4

S2-

Mn+4 S2- Mn2S4

Mn = (1)∙(+4) = +4S = (2)∙(-2) = -4

MnS2

59

Example – Ionic CompoundsIron(III) phosphate






Fe+3

PO43-

Fe+3 PO43- Fe3(PO4)3

Fe = (1)∙(+3) = +3PO4 = (1)∙(-3) = -3

FePO4

60

Example – Ionic Compoundsammonium carbonate






NH4+

CO32-

NH4+ CO3

2- (NH4)2CO3

NH4 = (2)∙(+1) = +2CO3 = (1)∙(-2) = -2

(NH4)2CO3

61


• copper(II) ion with a nitride ion

• iron(III) ion with a bromide ion

• aluminum ion with a sulfate ion

62


• Cu2+ with N3- Cu3N2

• Fe+3 with Br- FeBr3

• Al+3 with SO42- Al2(SO4)3

63

Example – Binary Acidshydrosulfuric acid






H+

S2-

H+ S2- H2S

H = (2)∙(+1) = +2S = (1)∙(-2) = -2

H2S

in all acids the cation is H+

hydro meansbinary

64

Example – Oxyacidscarbonic acid






H+

CO32-

H+ CO32- H2CO3

H = (2)∙(+1) = +2CO3 = (1)∙(-2) = -2

H2CO3

in all acids the cation is H+

no hydro meanspolyatomic ion

-ic means -ate ion

65

Practice - What are the formulas for the following acids?

• chlorous acid

• phosphoric acid

• hydrobromic acid

66

Practice - What are the formulas for the following acids?

• H+ with ClO2– HClO2

• H+ with PO43– H3PO4

• H+ with Br– HBr

67

Formula Mass• 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 unitwhole = sum of the parts!

mass of 1 molecule of H2O = 2(1.01 amu H) + 16.00 amu O = 18.02 amu

68

Two samples of carbon dioxide, obtained from different sources, were decomposed into their constituent elements. One sample produced 4.8 g of oxygen and 1.8 g of carbon, and the other sample produced 17.1 g of oxygen and 6.4 g of carbon. Show that these results are consistent with the law of constant composition.


To show this, compute the mass ratio of one element to the other by dividing the larger mass by the smaller one. For the first sample:

For the second sample:

Solution:

Since the ratios are the same for the two samples, these results are consistent with the law of constant composition.

69

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) Al2O3

(b) SO3

(c) CCl4

70

Determine the number of each type of atom in Mg3(PO4)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).

71

Classify each of the following substances as an atomic element, molecular element, molecular compound, or ionic compound.

(a) krypton(b) CoCl2

(c) nitrogen(d) SO2

(e) KNO3

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) CoCl2 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) SO2 is a compound composed of two nonmetals; therefore, it is a molecular compound.(e) KNO3 is a compound composed of a metal and two nonmetals; therefore, it is an ionic

compound.

72

Write a formula for the ionic compound that forms from aluminum and oxygen.

EXAMPLE 5.5 Writing Formulas for Ionic Compounds

Solution:

Al3+ O2–

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 Al2O3.

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.

73

Write a formula for the ionic compound that forms from magnesium and oxygen.


Solution:Mg2+ O2–

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.

Mg2O2 ÷ 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.

74

Write a formula for the compound that forms from potassium and oxygen.


Solution:We first write the symbol for each ion along with its appropriate charge from its group number in the periodic table.

K+ O2–

We then make the magnitude of each ion’s charge become the subscript for the other ion.

K+ O2– becomes K2O

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 K2O.

75

Give the name for the compound MgF2.

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.

76

EXAMPLE 5.9 Naming Type II Ionic Compounds

Give the name for the compound PbCl4.

Solution:The name for PbCl4 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 Pb4+.

PbCl4 lead (IV) chloride

77

EXAMPLE 5.10 Naming Ionic Compounds That Contain a Polyatomic Ion

Give the name for the compound K2CrO4.

The name for K2CrO4 consists of the name of the cation, potassium, followed by the name of the polyatomic ion, chromate.

Solution:

K2CrO4 potassium chromate

78

EXAMPLE 5.11 Naming Molecular Compounds

Name each of the following: CCl4 BCl3 SF6.

Solution:

CCl4

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.

BCl3

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.

SF4

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.

79

EXAMPLE 5.12 Naming Binary Acids

Give the name of H2S.

The base name of S is sulfur, so the name is hydrosulfuric acid.

Solution:

H2S hydrosulfuric acid

80

EXAMPLE 5.13 Naming Oxyacids

Give the name of HC2H3O2.

The oxyanion is acetate, which ends in -ate; therefore, the name of the acid is acetic acid.

Solution:

HC2H3O2 acetic acid

81

EXAMPLE 5.14 Nomenclature Using Figure 5.17

Name each of the following: CO, CaF2, HF, Fe(NO3)3, HClO4, H2SO3

Solution:For each compound, the following table shows how to use Figure 5.17 to arrive at a name for the compound.

Formula Flow Chart Path Name

CO Molecular carbon monoxideCaF2 calcium fluorideHF hydrofluoric acidFe(NO3)3 iron(III) nitrateHClO4 perchloric acidH2SO3 sulfurous acid

Ionic Type IAcid BinaryIonic Type IIAcid Oxyacid -ateAcid Oxyacid -ite

82

EXAMPLE 5.15 Calculating Formula Mass

Calculate the formula mass of carbon tetrachloride, CCl4.

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

83


Two samples said to be carbon disulfide (CS2) 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.

84

EXAMPLE 5.17 Writing Chemical Formulas

Write a chemical formula for the compound containing one nitrogen atom for every two oxygen atoms.

Solution:

NO2

85

EXAMPLE 5.18 Total Number of Each Type of Atom in a

Chemical Formula

Determine the number of each type of atom in PB(ClO3)2.

Solution:

One Pb atomTwo Cl atomsSix O atoms

86

EXAMPLE 5.19 Classifying Elements as Atomic or Molecular

Classify each of the following elements as atomic or molecular: Na, I, N.

Solution:

Na: atomicI: molecular (I2)N: molecular (N2)

87

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:

FeCl3, K2SO4, CCl4

Solution:

FeCl3: ionic, Type II K2SO4: ionic, Type I CCl4: molecular

88


Write a formula for the compound that forms from lithium and sulfate ions.

Solution:

Li+ SO 2–

Li2(SO4)

In this case, the subscripts cannot be further reduced.

Li2SO4

Cations: Anions:

2(1+) = 2+ 2–

4

89

EXAMPLE 5.22 Naming Type I Binary Ionic Compounds

Give the name for the compound Al2O3

Solution:

aluminum oxide

90

EXAMPLE 5.23 Naming Type II Binary Ionic Compounds

Give the name for the compound Fe2S3

Solution:

3 sulfide ions x (2–) = 6–

2 iron ions x (?) = 6+? = 3+

Charge of each iron ion = 3+

iron(III) sulfide

91

EXAMPLE 5.24 Naming Compounds Containing a Polyatomic Ion

Give the name for the compound Co(ClO4)2.

Solution:

3 perchlorate x (1–) = 2–

Charge of cobalt ion = 2+

cobalt(II) perchlorate

92

EXAMPLE 5.25 Naming Molecular Compounds

Name the compound NO2.

Solution:nitrogen dioxide

93

EXAMPLE 5.26 Naming Binary Acids

Name the acid HI.

Solution:hydroiodic acid

94

EXAMPLE 5.27 Naming Oxyacids with an Oxyanion Ending in -ate

Name the acid H2SO4.

Solution:The oxyanion is sulfate. The name of the acid is sulfuric acid.

95

EXAMPLE 5.28 Naming Oxyacids with an Oxyanion Ending in -ite

Name the acid HClO2.

Solution:The oxyanion is chlorite. The name of the acid is chlorous acid.

96

97

98

99

100

EXAMPLE 5.29 Calculating Formula Mass

Calculate the formula mass of Mg(NO3)2

Solution:

Formula mass = 24.31 + 2(14.01) + 6(16.00)= 148.33 amu

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Chapter 5 Molecules and Compounds 2 Molecules and Compounds Salt Sodium – shiny, reactive,...

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