Chapter 5 Molecules and Compounds 2 Molecules and Compounds Salt Sodium – shiny, reactive,...
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Transcript of 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
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
Formula-to-NameStep 2
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
Formula-to-NameStep 3
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
Formula-to-NameStep 4
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
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 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
Example – Naming Oxyacids H2SO3
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 Subclass3 elements in the formula, Oxyacid
48
Example – Naming Oxyacids H2SO3
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
3. Charge (without sign) becomes subscript for other ion
4. Reduce subscripts to smallest whole number ratio
5. Check that the total charge of the cations cancels the total charge of the anions
Al+3 column 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
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
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
3. Charge (without sign) becomes subscript for other ion
4. Reduce subscripts to smallest whole number ratio
5. Check that the total charge of the cations cancels the total charge of the anions
Mn+4
S2-
Mn+4 S2- Mn2S4
Mn = (1)∙(+4) = +4S = (2)∙(-2) = -4
MnS2
59
Example – Ionic CompoundsIron(III) phosphate
1. Write the symbol for the cation and its charge
2. Write the symbol for the anion and its charge
3. Charge (without sign) becomes subscript for other ion
4. Reduce subscripts to smallest whole number ratio
5. Check that the total charge of the cations cancels the total charge of the anions
Fe+3
PO43-
Fe+3 PO43- Fe3(PO4)3
Fe = (1)∙(+3) = +3PO4 = (1)∙(-3) = -3
FePO4
60
Example – Ionic Compoundsammonium 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
NH4+
CO32-
NH4+ CO3
2- (NH4)2CO3
NH4 = (2)∙(+1) = +2CO3 = (1)∙(-2) = -2
(NH4)2CO3
61
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
62
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
63
Example – Binary Acidshydrosulfuric 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
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
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
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.
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.
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.
EXAMPLE 5.6 Writing Formulas for Ionic Compounds
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.
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+ 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
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?
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
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–
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