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Chapter 5 Molecules and Compounds. Molecules and Compounds. Salt Sodium – shiny, reactive, poisonous Chlorine – pale yellow gas, reactive, poisonous Sodium chloride – table salt Sugar Carbon – pencil or diamonds Hydrogen – flammable gas Oxygen – a gas in air

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molecules and compounds
Molecules and Compounds
  • Salt
    • Sodium – shiny, reactive, poisonous
    • Chlorine – pale yellow gas, reactive, poisonous
    • Sodium chloride – table salt
  • Sugar
    • Carbon – pencil or diamonds
    • Hydrogen – flammable gas
    • Oxygen – a gas in air
    • Combine to form white crystalline sugar
law of constant composition
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
compounds display constant composition
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.

why do compounds show constant composition
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 elements
    • every sample of the compound will have the same properties
slide6

Solution:

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:

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

EXAMPLE 5.1 Constant Composition of Compounds

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.

slide8

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?

5.1B

13.3

185

228

312

553

formulas describe compounds
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

order of elements in a formula
Order of Elements in a Formula
  • metals written first
    • NaCl
  • nonmetals written in order from Table 5.1
    • CO2
    • are occasional exceptions for historical or informational reasons
      • H2O, but NaOH
molecules with polyatomic ions

symbol of the polyatomic

ion called nitrate

symbol of the polyatomic

ion called sulfate

implied “1” subscript

on magnesium

implied “1” subscript

on calcium

parentheses to group two NO3’s

no parentheses for one SO4

Molecules with Polyatomic Ions

Mg(NO3)2

compound called

magnesium nitrate

CaSO4

compound called

calcium sulfate

molecules with polyatomic ions1

subscript indicating

two NO3 groups

no subscript indicating

one SO4 group

implied “1” subscript

on nitrogen, total 2 N

implied “1” subscript

on sulfur, total 1 S

stated “4” subscript

on oxygen, total 4 O

stated “3” subscript

on oxygen, total 6 O

Molecules with Polyatomic Ions

Mg(NO3)2

compound called

magnesium nitrate

CaSO4

compound called

calcium sulfate

classifying materials
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
molecular elements

P4

S8

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 H2

VIIA

7

H2

N2 O2 F2

Cl2

Br2

I2

molecular compounds
Molecular Compounds
  • two or more nonmetals
  • smallest unit is a molecule
ionic compounds
Ionic Compounds
  • metals + nonmetals
  • no individual molecule units, instead have a 3-dimensional array of cations and anions made of formula units
slide18
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
slide19
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 name step 1

Formula-to-NameStep 1

Is the compound one of the exceptions to the rules?

  • H2O = water, steam, ice
  • NH3 = ammonia
formula to name step 2

Formula-to-NameStep 2

What major class of compound is it?

Ionic or Molecular

major classes
Major Classes
  • Ionic
    • metal + nonmetal
      • metal first in formula
      • Binary Ionic
    • compounds with polyatomic ions
  • Molecular
    • 2 nonmetals
      • Binary Molecular (or Binary Covalent)
    • Acids – formula starts with H
      • though acids are molecular, they behave as ionic when dissolved in water
      • may be binary or oxyacid
formula to name step 3

Formula-to-NameStep 3

What major subclass of compound is it?

Binary Ionic, Ionic with Polyatomic Ions,

Binary Molecular,

Binary Acid, Oxyacid

classifying compounds
Classifying Compounds
  • Compounds containing a metal and a nonmetal = binary ionic
    • Type I and II
  • Compounds containing a polyatomic ion = ionic with polyatomic ion
  • Compounds containing two nonmetals = binarymolecular compounds
  • Compounds containing H and a nonmetal = binary acids
  • Compounds containing H and a polyatomic ion = oxyacids
formula to name step 4

Formula-to-NameStep 4

Apply Rules for the Class and Subclass

formula to name rules for ionic
Formula-to-NameRules 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
monatomic nonmetal anion
Monatomic Nonmetal Anion
  • determine the charge from position on the Periodic Table
  • to name anion, change ending on the element name to –ide
metal cations
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

How do you know a metal cation is Type II?

its not Type I !!!

slide29
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
slide30
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
type i binary ionic compounds
Type I Binary Ionic Compounds
  • Contain Metal Cation + Nonmetal Anion
  • Metal listed first in formula & name
  • name metal cation first, name nonmetal anion second
  • cation name is the metal name
  • nonmetal anion named by changing the ending on the nonmetal name to -ide
type ii binary ionic compounds
Type II Binary Ionic Compounds
  • Contain Metal Cation + Nonmetal Anion
  • Metal listed first in formula & name
  • name metal cation first, name nonmetal anion second
  • 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
  • nonmetal anion named by changing the ending on the nonmetal name to -ide
examples
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
compounds containing polyatomic ions
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
fixed charge metals and nonmetals
Fixed Charge Metals and Nonmetals

IA

VA

VIIA

IIA

IIIA

VIA

Be+2

Li+1

N-3

O-2

F-1

Mg+2

Na+1

P-3

S-2

Cl-1

Al+3

Ca+2

K+1

As-3

Se-2

Br-1

Zn+2

Ga+3

Sr+2

Rb+1

Te-2

I-1

Ag+1

Cd+2

In+3

Ba+2

Cs+1

patterns for polyatomic ions
Patterns for Polyatomic Ions
  • elements in the same column form similar polyatomic ions
    • same number of O’s and same charge

ClO3- = chlorate \ BrO3- = bromate

  • if the polyatomic ion starts with H, the name adds hydrogen- prefix before name and add 1 to the charge

CO32- = carbonate \ HCO3-1 = hydrogencarbonate

periodic pattern of polyatomic ions ate groups

-3

-2

-1

BO

CO

NO

3

3

3

-2

-3

-2

-1

SiO

PO

SO

ClO

3

4

4

3

-3

-2

-1

AsO

SeO

BrO

4

4

3

-2

-1

TeO

IO

4

3

Periodic Pattern of Polyatomic Ions-ate groups

IIIA IVA VA VIA VIIA

binary molecular compounds of 2 nonmetals
Binary Molecular Compounds of 2 Nonmetals
  • Name first element in formula first
    • use the full name of the element
  • Name the second element in the formula with an -ide
    • as if it were an anion, however, remember these compounds do not contain ions!
  • Use a prefix in front of each name to indicate the number of atoms
    • Never use the prefix mono- on the first element
subscript prefixes
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

acids
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
formula to name acids
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
example naming binary acids hcl
Example – Naming Binary AcidsHCl
  • Is it one of the common exceptions?

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

  • Identify Major Class

first element listed is H,  Acid

  • Identify the Subclass

2 elements,  Binary Acid

sample naming binary acids hcl
Sample - Naming Binary Acids – HCl
  • Identify the anion

Cl = Cl-, chloride because Group 7A

  • Name the anion with an –ic suffix

Cl- = chloride  chloric

  • Add a hydro- prefix to the anion name

hydrochloric

  • Add the word acid to the end

hydrochloric acid

example naming oxyacids h 2 so 4
Example – Naming OxyacidsH2SO4
  • Is it one of the common exceptions?

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

  • Identify Major Class

first element listed is H,  Acid

  • Identify the Subclass

3 elements in the formula,  Oxyacid

example naming oxyacids h 2 so 41
Example – Naming Oxyacids H2SO4
  • Identify the anion

SO4 = SO42- = sulfate

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

SO42- = sulfate  sulfuric

  • Write the name of the anion followed by the word acid

sulfuric acid

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

example naming oxyacids h 2 so 3
Example – Naming Oxyacids H2SO3
  • Is it one of the common exceptions?

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

  • Identify Major Class

first element listed is H,  Acid

  • Identify the Subclass

3 elements in the formula,  Oxyacid

example naming oxyacids h 2 so 31
Example – Naming Oxyacids H2SO3
  • Identify the anion

SO3 = SO32- = sulfite

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

SO32- = sulfite  sulfurous

  • Write the name of the anion followed by the word acid

sulfurous acid

writing the formulas from the names
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
    • Determine the ions present
    • Determine the charges on the cation and anion
    • Balance the charges to get the subscripts
example binary molecular dinitrogen pentoxide
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

compounds that contain ions
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 S2-, you will need 2 Na+ ions for every S2- ion to balance the charges, therefore the formula must be Na2S
writing formulas for ionic compounds
Writing Formulas for Ionic Compounds
  • Write the symbol for the metal cation and its charge
  • Write the symbol for the nonmetal anion and its charge
  • Charge (without sign) becomes subscript for other ion
  • Reduce subscripts to smallest whole number ratio
  • Check that the sum of the charges of the cation cancels the sum of the anions
write the formula of a compound made from aluminum ions and oxide ions
Write the formula of a compound made from aluminum ions and oxide ions

Al+3 column IIIA

  • Write the symbol for the metal cation and its charge
  • Write the symbol for the nonmetal anion and its charge
  • Charge (without sign) becomes subscript for other ion
  • Reduce subscripts to smallest whole number ratio
  • Check that the total charge of the cations cancels the total charge of the anions

O2- column VIA

Al+3 O2-

Al2 O3

Al = (2)∙(+3) = +6

O = (3)∙(-2) = -6

practice what are the formulas for compounds made from the following ions
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
practice what are the formulas for compounds made from the following ions1
Practice - What are the formulas for compounds made from the following ions?
  • K+ with N3- K3N
  • Ca+2 with Br- CaBr2
  • Al+3 with S2-Al2S3
example ionic compounds manganese iv sulfide
Example – Ionic Compoundsmanganese(IV) sulfide

Mn+4

  • Write the symbol for the cation and its charge
  • Write the symbol for the anion and its charge
  • Charge (without sign) becomes subscript for other ion
  • Reduce subscripts to smallest whole number ratio
  • Check that the total charge of the cations cancels the total charge of the anions

S2-

Mn+4 S2-

Mn2S4

MnS2

Mn = (1)∙(+4) = +4

S = (2)∙(-2) = -4

example ionic compounds iron iii phosphate
Example – Ionic CompoundsIron(III) phosphate

Fe+3

  • Write the symbol for the cation and its charge
  • Write the symbol for the anion and its charge
  • Charge (without sign) becomes subscript for other ion
  • Reduce subscripts to smallest whole number ratio
  • Check that the total charge of the cations cancels the total charge of the anions

PO43-

Fe+3 PO43-

Fe3(PO4)3

FePO4

Fe = (1)∙(+3) = +3

PO4 = (1)∙(-3) = -3

example ionic compounds ammonium carbonate
Example – Ionic Compoundsammonium carbonate

NH4+

  • Write the symbol for the cation and its charge
  • Write the symbol for the anion and its charge
  • Charge (without sign) becomes subscript for other ion
  • Reduce subscripts to smallest whole number ratio
  • Check that the total charge of the cations cancels the total charge of the anions

CO32-

NH4+ CO32-

(NH4)2CO3

(NH4)2CO3

NH4 = (2)∙(+1) = +2

CO3 = (1)∙(-2) = -2

practice what are the formulas for compounds made from the following ions2
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
  • aluminum ion with a sulfate ion
practice what are the formulas for compounds made from the following ions3
Practice - What are the formulas for compounds made from the following ions?
  • Cu2+ with N3- Cu3N2
  • Fe+3 with Br- FeBr3
  • Al+3 with SO42-Al2(SO4)3
example binary acids hydrosulfuric acid
Example – Binary Acidshydrosulfuric acid

in all acids the

cation is H+

H+

  • Write the symbol for the cation and its charge
  • Write the symbol for the anion and its charge
  • Charge (without sign) becomes subscript for other ion
  • Reduce subscripts to smallest whole number ratio
  • Check that the total charge of the cations cancels the total charge of the anions

hydro means

binary

S2-

H+ S2-

H2S

H2S

H = (2)∙(+1) = +2

S = (1)∙(-2) = -2

example oxyacids carbonic acid
Example – Oxyacidscarbonic acid

in all acids the

cation is H+

H+

  • Write the symbol for the cation and its charge
  • Write the symbol for the anion and its charge
  • Charge (without sign) becomes subscript for other ion
  • Reduce subscripts to smallest whole number ratio
  • Check that the total charge of the cations cancels the total charge of the anions

no hydro means

polyatomicion

CO32-

-ic means -ate ion

H+ CO32-

H2CO3

H2CO3

H = (2)∙(+1) = +2

CO3 = (1)∙(-2) = -2

practice what are the formulas for the following acids
Practice - What are the formulas for the following acids?
  • chlorous acid
  • phosphoric acid
  • hydrobromic acid
practice what are the formulas for the following acids1
Practice - What are the formulas for the following acids?
  • H+ with ClO2–HClO2
  • H+ with PO43–H3PO4
  • H+ with Br–HBr
formula mass
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 unit
    • whole = sum of the parts!

mass of 1 molecule of H2O

= 2(1.01 amu H) + 16.00 amu O = 18.02 amu

slide68

Solution:

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:

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

EXAMPLE 5.1 Constant Composition of Compounds

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.

slide69

Solution:

(a) Al2O3

(b) SO3

(c) CCl4

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.

EXAMPLE 5.2 Writing Chemical Formulas

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

slide70

EXAMPLE 5.3 Total Number of Each Type of Atom in a Chemical Formula

Determine the number of each type of atom in Mg3(PO4)2.

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).

slide71

EXAMPLE 5.4 Classifying Substances as Atomic Elements, Molecular Elements, Molecular Compounds, or Ionic Compounds

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

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.

slide72

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.

EXAMPLE 5.5 Writing Formulas for Ionic Compounds

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

slide73

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.

3.Reduce the subscripts to give a ratio with the smallest whole numbers.

To reduce the subscripts, divide both subscripts by 2.Mg2O2÷ 2 = MgO

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.

EXAMPLE 5.6 Writing Formulas for Ionic Compounds

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

slide74

EXAMPLE 5.7 Writing Formulas for Ionic Compounds

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.

slide75

EXAMPLE 5.8 Naming Type I Ionic Compounds

Give the name for the compound MgF2.

Solution:The cation is magnesium. The anion is fluorine, which becomes fluoride. The correct name is magnesium fluoride.

slide76

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

slide77

Solution:K2CrO4 potassium chromate

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

EXAMPLE 5.10 Naming Ionic Compounds That Contain a Polyatomic Ion

Give the name for the compound K2CrO4.

slide78

EXAMPLE 5.11 Naming Molecular Compounds

Name each of the following: CCl4 BCl3 SF6.

Solution:CCl4The 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.BCl3The 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. SF4The 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.

slide79

Solution:H2Shydrosulfuric acid

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

EXAMPLE 5.12 Naming Binary Acids

Give the name of H2S.

slide80

Solution:HC2H3O2 acetic acid

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

EXAMPLE 5.13 Naming Oxyacids

Give the name of HC2H3O2.

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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 monoxide

CaF2 calcium fluoride

HF hydrofluoric acid

Fe(NO3)3 iron(III) nitrate

HClO4 perchloric acid

H2SO3 sulfurous acid

Ionic Type I

Acid Binary

Ionic Type II

Acid Oxyacid -ate

Acid Oxyacid -ite

EXAMPLE 5.14 Nomenclature Using Figure 5.17

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

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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

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Solution: Sample 1

These results are not consistent with the law of constant composition and the data given must therefore be in error.

EXAMPLE 5.16 Constant Composition of Compounds

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?

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EXAMPLE 5.17 Writing Chemical Formulas

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

Solution:

NO2

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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 atom Two Cl atoms

Six O atoms

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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 (I2)

N: molecular (N2)

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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

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Solution:

Li+SO 2–

Li2(SO4)

In this case, the subscripts cannot be further reduced.

Li2SO4

Cations: Anions:

2(1+) = 2+ 2–

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EXAMPLE 5.21 Writing Formulas for Ionic Compounds

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

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EXAMPLE 5.22 Naming Type I Binary Ionic Compounds

Give the name for the compound Al2O3

Solution:

aluminum oxide

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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

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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

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EXAMPLE 5.25 Naming Molecular Compounds

Name the compound NO2.

Solution:

nitrogen dioxide

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EXAMPLE 5.26 Naming Binary Acids

Name the acid HI.

Solution:

hydroiodic acid

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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.

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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.

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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