General Chemistry
Principles and Modern Applications
Petrucci • Harwood • Herring
8th Edition
Chapter 3: Chemical Compounds
Dr. Burak Esat
Fatih University
Chem 107Fall 2013
1
Compound:
A combination of two or more different elements.
A substance composed of two or more elements chemically
combined in fixed ratios by mass.
Compounds are represented by chemical formulas indicating
the fixed ratios of each elements chemically combined
Water - H2O
Carbon dioxide - CO2
Sodium Chloride - NaCl
Iron(II) sulfide - FeS
•Molecule
Molecule: The smallest entity having the same elemental combination as
•the compound.
•Compounds are made of individual molecules
2
1
Chemical Formulas
Empirical Formula - Shows the relative number of atoms
of each element in the compound. It is the simplest
formula, and is derived from masses of the elements.
Molecular Formula - Shows the actual number of atoms
of each element in the molecule of the compound.
Structural Formula - Shows the actual number of atoms,
and the bonds between them ; that is, the arrangement
of atoms in the molecule.
3
Molecular compounds
Structural Formula shows the order in which atoms are
bonded together
4
2
Standard color scheme
5
Some molecules
H2O2
CH3CH(OH)CH3
CH3CH2Cl
P4O10
HCO2H
6
3
Chemical Formulas
Empirical Formula - Shows the relative number of atoms
of each element in the compound. It is the simplest
formula, and is derived from masses of the elements.
Molecular Formula - Shows the actual number of atoms
of each element in the molecule of the compound.
Structural Formula - Shows the actual number of atoms,
and the bonds between them ; that is, the arrangement
of atoms in the molecule.
7
Some Examples of Compounds with the
Same Elemental Ratio’s
Empirical Formula
Molecular Formula
CH2(unsaturated Hydrocarbons)
C2H4 , C3H6 , C4H8
OH or HO
H2 O2
S
S8
P
P4
Cl
Cl2
CH2O (carbohydrates)
C6H12O6
8
4
Some Compounds with Empirical Formula
CH2O (Composition by Mass 40.0% C, 6.71% H, 53.3%O)
Molecular
Formula
M
(g/mol)
Name
CH2O
30.03
Formaldehyde
C2H4O2
60.05
Acetic acid
C3H6O3
90.08
Lactic acid
C4H8O4
120.10
Erythrose
C5H10O5
150.13
Ribose
C6H12O6
180.16
Glucose
Use or Function
Disinfectant; Biological
preservative
Acetate polymers; vinegar
( 5% solution)
Causes milk to sour; forms
in muscle during exercise
Forms during sugar
metabolism
Component of many nucleic
acids and vitamin B2
Major nutrient for energy
9
in cells
Mole
• The Mole is based upon the definition:
• The amount of substance that contains
as many elementary parts (atoms,
molecules, or other) as there are atoms
in exactly 12 grams of carbon 12.
• 1 Mole = 6.022045 x 1023 particles
10
5
One Mole of Common Substances
CaCO3
100.09 g
Oxygen Gas (O2)
32.00 g
Copper (Cu)
63.55 g
Water (H2O)
18.02 g
11
Mole - Mass Relationships of Elements
Element Atom/Molecule Mass
1 atom of H = 1.008 amu
Mole Mass
Number of Atoms
1 mole of H = 1.008 g = 6.022 x 1023 atoms
1 atom of Fe = 55.85 amu 1 mole of Fe = 55.85 g = 6.022 x 1023 atoms
1 atom of S = 32.07 amu
1 mole of S = 32.07 g = 6.022 x 1023 atoms
1 atom of O = 16.00 amu
1 mole of O = 16.00 g = 6.022 x 1023 atoms
1 molecule of O2 = 32.00 amu
1 mole of O2 = 32.00 g = 6.022 x 1023 molecule
1 molecule of S8 = 205.952 amu
12
1 mole of S8 = 205.952 g = 6.022 x 1023 molecules
6
Information Contained in the Chemical
Formula of Glucose C6H12O6
Carbon (C)
Hydrogen (H)
Oxygen (O)
Atoms/molecule
of compound
6 atoms
12 atoms
6 atoms
Moles of atoms/
mole of compound
6 moles of
atoms
12 moles of
atoms
6 moles of
atoms
Atoms/mole of
compound
6(6.022 x 1023) 12(6.022 x 1023) 6(6.022 x 1023)
atoms
atoms
atoms
Mass/molecule
of compound
6(12.01 amu)
=72.06 amu
Mass/mole of
compound
72.06 g
12(1.008 amu)
=12.10 amu
12.10 g
6(16.00 amu)
=96.00 amu
96.00 g
13
Calculate the Molecular Mass of
Glucose: C6H12O6
• Carbon
6 x 12.011 g/mol = 72.066 g
• Hydrogen 12 x 1.008 g/mol = 12.096 g
• Oxygen
6 x 15.999 g/mol =
95.994 g
180.156 g
14
7
% Composition
H OH
Glucose
H O
HO
HO
H
H
H
OH
OH
Molecular formula C6H12O6
Empirical formula CH2O
Molecular Mass: Use the naturally occurring mixture of isotopes,
6 x 12.01 + 12 x 1.01 + 6 x 16.00 = 180.18
Exact Mass:
Use the most abundant isotopes,
6 x 12.000000 + 12 x 1.007825 + 6 x 15.994915
= 180.06339
15
Chemical Composition
Halothane
C2HBrClF3
Mole ratio
nC/nhalothane
Mass ratio
mC/mhalothane
M(C2HBrClF3) = 2MC + MH + MBr + MCl + 3MF
= (2 x 12.01) + 1.01 + 79.90 + 35.45 + (3 x 19.00)
= 197.38 g/mol
16
8
Example 3.4
Calculating the Mass Percent Composition of a Compound
Calculate the molecular mass
M(C2HBrClF3) = 197.38 g/mol
For one mole of compound, formulate the mass
ratio and convert to percent:
%C =
(2 ×12.01) g
×100% = 12.17%
197.38 g
17
Example 3-4
(2 ×12.01) g
×100% = 12.17%
197.38 g
1.01g
×100% = 0.51%
%H =
197.38 g
79.90 g
% Br =
×100% = 40.48%
197.38 g
35.45 g
%Cl =
×100% = 17.96%
197.38 g
(3 ×19.00) g
%F =
×100% = 28.88%
197.38 g
%C =
18
9
Mass Percent Composition of Na2SO4
Na2SO4 = 2 atoms of Sodium + 1 atom of Sulfur + 4 atoms of Oxygen
Elemental masses
Percent of each Element
2 x Na = 2 x 22.99 = 45.98
1 x S = 1 x 32.07 = 32.07
4 x O = 4 x 16.00 = 64.00
142.05
Atomic Masses
from Periodic Table
% Na = Mass Na / Total mass x 100%
% Na = (45.98 / 142.05) x 100% =32.37%
% S = Mass S / Total mass x 100%
% S = (32.07 / 142.05) x 100% = 22.58%
% O = Mass O / Total mass x 100%
% O = (64.00 / 142.05) x 100% = 45.05%
Check
% Na + % S + % O = 100%
32.37% + 22.58% + 45.05% = 100.00%
Steps to Determine Empirical Formulas
Masses (g) of Individual Elements
xA+yB AxBy
Molar Mass (g/mol )
Moles of Each Element
use no. of moles as subscripts
Preliminary Formula
change to integer subscripts
Empirical Formula
20
10
Determining Empirical Formulas from
Masses of Elements - I
Problem: The elemental analysis of a sample compound gave the
following results: 5.677g Na, 6.420 g Cr, and 7.902 g O. What is the
empirical formula and name of the compound?
Plan: First we have to convert mass of the elements to moles of the
elements using the molar masses. Then we construct a preliminary
formula and name of the compound.
Solution: Finding the moles of the elements:
1 mol Na
Moles of Na = 5.678 g Na x 22.99 g Na = 0.2469 mol Na
1 mol Cr
Moles of Cr = 6.420 g Cr x 52.00 g Cr = 0.12347 mol Cr
Moles of O = 7.902 g O x 1 mol O = 0.4939 mol O
16.00 g O
21
Determining Empirical Formulas from
Masses of Elements - II
Constructing the preliminary formula:
Na0.2469 Cr0.1235 O0.4939
Converting to integer subscripts (dividing all by smallest subscript):
Na1.99 Cr1.00 O4.02
Rounding off to whole numbers:
Na2CrO4
Sodium Chromate
22
11
Establishing Formulas from
Experimentally Determined Percent
Composition
5 Step approach:
1.
2.
3.
4.
5.
Choose an arbitrary sample size (100g).
Convert masses to amounts in moles.
Write a formula.
Convert formula to small whole numbers.
Multiply all subscripts by a small whole number to make the
subscripts integral.
23
Determining the Molecular Formula from
Elemental Composition and Molar Mass - I
Problem: The sugar burned for energy in cells of the body is Glucose
(M = 180.16 g/mol), elemental analysis shows that it contains
40.00 mass % C, 6.719 mass % H, and 53.27 mass % O.
(a) Determine the empirical formula of glucose.
(b) Determine the molecular formula.
Plan: We are only given mass %, and no weight of the compound so we
will assume 100g of the compound, and % becomes grams, and
we can do as done previously with masses of the elements.
Solution:
Mass Carbon = 40.00% x 100g/100% = 40.00 g C
Mass Hydrogen = 6.719% x 100g/100% = 6.719g H
Mass Oxygen = 53.27% x 100g/100% = 53.27 g O
99.989 g Cmpd
24
12
Determining the Molecular Formula from
Elemental Composition and Molar Mass - II
Converting from Grams of Elements to moles:
Moles of C = Mass of C x 1 mole C = 3.3306 moles C
12.01 g C
1 mol H
Moles of H = Mass of H x
= 6.6657 moles H
1.008 g H
Moles of O = Mass of O x 1 mol O = 3.3294 moles O
16.00 g O
Constructing the preliminary formula C 3.33 H 6.67 O 3.33
Converting to integer subscripts, divide all subscripts by the smallest:
C 3.33/3.33 H 6.667 / 3.33 O3.33 / 3.33 = CH2O
25
Determining the Molecular Formula from
Elemental Composition and Molar Mass - III
(b) Determining the Molecular Formula:
The formula weight of the empirical formula is:
1 x C + 2 x H + 1 x O = 1 x 12.01 + 2 x 1.008 + 1 x 16.00 = 30.03
Whole-number multiple =
=
M of Glucose
empirical formula mass
=
180.16
= 6.00 = 6
30.03
Therefore the Molecular Formula is:
C1x6H2x6O1x6 =
C6H12O6
26
13
Adrenaline Is a Very Important
Compound in the Body - I
• Analysis gives :
•
C = 56.8 %
•
H = 6.50 %
•
O = 28.4 %
•
N = 8.28 %
• Calculate the
Empirical Formula
27
Adrenaline - II
•
•
•
•
•
•
•
•
•
•
Assume 100g!
C = 56.8 g C/(12.01 g C/ mol C) = 4.73 mol C
H = 6.50 g H/( 1.008 g H / mol H) = 6.45 mol H
O = 28.4 g O/(16.00 g O/ mol O) = 1.78 mol O
N = 8.28 g N/(14.01 g N/ mol N) = 0.591 mol N
Divide by 0.591 =
C = 8.00 mol C = 8.0 mol C
or
H = 10.9 mol H = 11.0 mol H
O = 3.01 mol O = 3.0 mol O C8H11O3N
N = 1.00 mol N = 1.0 mol N
28
14
Combustion Train for the Determination of the
Chemical Composition of Organic Compounds.
CnHm + (n+ m ) O2 = n CO2(g) +m H2O(g)
2
2
29
Fig. 3.4
Combustion Analysis
30
15
Ascorbic Acid ( Vitamin C ) - I
Contains C , H , and O
• Upon combustion in excess oxygen, a 6.49 mg
sample yielded 9.74 mg CO2 and 2.64 mg H2O
• Calculate it’s Empirical formula!
• C: 9.74 x10-3g CO2 x(12.01 g C/44.01 g CO2)
= 2.65 x 10-3 g C
• H: 2.64 x10-3g H2O x (2.016 g H2/18.02 gH2O)
= 2.92 x 10-4 g H
• Mass Oxygen = 6.49 mg - 2.65 mg - 0.30 mg
= 3.54 mg O
31
Vitamin C Combustion - II
• C = 2.65 x 10-3 g C / ( 12.01 g C / mol C ) =
= 2.21 x 10-4 mol C
• H = 0.295 x 10-3 g H / ( 1.008 g H / mol H ) =
= 2.92 x 10-4 mol H
• O = 3.54 x 10-3 g O / ( 16.00 g O / mol O ) =
= 2.21 x 10-4 mol O
• Divide each by 2.21 x 10-4
• C = 1.00
• H = 1.32
• O = 1.00
Multiply each by 3
C3H4O3
= 3.00 = 3.0
= 3.96 = 4.0
= 3.00 = 3.0
32
16
Determining a Chemical Formula from
Combustion Analysis - I
Problem: Erythrose (M = 120 g/mol) is an important chemical
compound as a starting material in chemical synthesis, and
contains Carbon Hydrogen, and Oxygen. Combustion
analysis of a 700.0 mg sample yielded 1.027 g CO2 and
0.4194 g H2O.
Plan: 1) We find the masses of Hydrogen and Carbon using the mass
fractions of H in H2O, and C in CO2.
2) The mass of Carbon and Hydrogen are subtracted from the
sample mass to get the mass of Oxygen.
3) We then calculate moles, and
4) Construct the empirical formula, and
5) From the given molar mass we can calculate the molecular33
formula.
Determining a Chemical Formula from
Combustion Analysis - II
Calculating the mass fractions of the elements:
Mass fraction of C in CO2 =
mol C x M of C
=
mass of 1 mol CO2
= 1 mol C x 12.01 g C/ 1 mol C =
44.01 g CO2
0.2729 g C / 1 g CO2
mol H x M of H
=
mass of 1 mol H2O
2 mol H x 1.008 g H / 1 mol H
=
= 0.1119 g H / 1 g H2O
18.02 g H2O
Mass fraction of H in H2O =
Calculating masses of C and H
Mass of Element = mass of compound x mass fraction of element
34
17
Determining a Chemical Formula from
Combustion Analysis - III
0.2729 g C
= 0.2803 g C
1 g CO2
0.1119 g H
Mass (g) of H = 0.4194 g H2O x
= 0.04693 g H
1 g H2O
Mass (g) of C = 1.027 g CO2 x
Calculating the mass of O:
Mass (g) of O = Sample mass -( mass of C + mass of H )
= 0.700 g - 0.2803 g C - 0.04693 g H = 0.37277 g O
Calculating moles of each element:
C = 0.2803 g C / 12.01 g C/ mol C = 0.02334 mol C
H = 0.04693 g H / 1.008 g H / mol H = 0.04656 mol H
O = 0.37277 g O / 16.00 g O / mol O = 0.02330 mol O
C0.02334H0.04656O0.02330 = CH2O formula weight = 30 g / formula
120 g /mol / 30 g / formula = 4 formula units / cpd = C4H8O4
35
Oxidation States
Metals tend to
lose electrons.
Non-metals tend
to gain electrons.
Na -> Na+ + e-
Cl + e- -> Cl-
Reducing agents
Oxidizing agents
We use the Oxidation State to keep track of the number of
electrons that have been gained or lost by an element.
36
18
Ionic Bonding - An ionic bond is a chemical
bond that results from an electrostatic attraction
among oppositely charged ions in a compound.
They form when electrons are transferred from
one atom to another to form ions.
Mono-atomic ions form binary ionic compounds
Na [Ne] 3s1
Na [Ne]+
Na+
Cl [Ne] 3s2 3p5
+
+
Cl [Ar]-
Cl-
“Cations” - Metal atoms lose electrons to form “ + ” ions.
37
“Anions” - Nonmetal atoms gain electrons to form “ - ” ions.
Fig. 2.18
38
19
Rules for Oxidation States
1.
The oxidation state (OS) of an individual atom in a
free element is 0.
2.
The total of the OS in all atoms in:
i. Neutral species is 0.
ii. Ionic species is equal to the charge on the
ion.
3.
In their compounds, the alkali metals and the
alkaline earths have OS of +1 and +2 respectively.
4.
In compounds the OS of fluorine is always –1
39
Rules for Oxidation States
5.
In compounds, the OS of hydrogen is usually +1
6.
In compounds, the OS of oxygen is usually –2.
7.
In binary (two-element) compounds with metals:
i. Halogens have OS of –1,
ii. Group 16 have OS of –2 and
iii. Group 15 have OS of –3.
40
20
Example 3-7
Assigning Oxidation States.
What is the oxidation state of the underlined element in each
of the following? a) P4; b) Al2O3; c) MnO4-; d) NaH
a) P4 is an element. P OS = 0
b) Al2O3: O is –2. O3 is –6. Since (+6)/2=(+3), Al OS = +3.
c) MnO4-: net OS = -1, O4 is –8. Mn OS = +7.
d) NaH: net OS = 0, rule 3 beats rule 5, Na OS = +1 and
H OS = -1.
41
Fig. 2.23
42
21
Fig. 2.20
43
Predicting the Ion an Element Will Form in Chemical Reactions
Problem: What mono-atomic ions will each of the elements form?
(a) Barium(z=56) (b) Sulfur(z=16) (c) Titanium(z =22) (d) Fluorine(z=9)
Plan: We use the “z” value to find the element in the periodic
table and the nearest noble gas.
Elements that lie after a noble gas will loose electrons,
and those before a noble gas will gain electrons.
Solution:
(a) Ba+2, Barium is an alkaline earth element, Group 2A, and is
expected to loose two electrons to attain the same number of electrons
as the noble gas Xenon!
(b) S -2, Sulfur is in the Oxygen family, Group 6A, and is expected to
gain two electrons to attain the same number of electrons as the noble
gas Argon!
(c) Ti+4, Titanium is in Group 4B, and is expected to loose 4 electrons
to attain the same number of electrons as the noble gas Argon!
(d) F -, Fluorine is in a halogen, Group 7A, and is expected to gain one
electron, to attain the same number of electrons as the noble gas Neon!
44
22
3-5 Naming Compounds:
Organic and Inorganic Compounds
Trivial names are used for common compounds.
A systematic method of naming compounds is known as a
system of nomenclature.
Organic compounds
Inorganic compounds
Lead (IV) oxide
Lead (II) oxide
45
3-6 Names and Formulas of
Inorganic Compounds
Binary Compounds of Metals and Nonmetals
46
23
Table 2.3 (p. 67)
Common Mono-atomic Ions
Cations
Charge Formula
1+
2+
3+
H+
Li+
Na+
K+
Cs+
Ag+
Mg2+
Ca2+
Sr2+
Ba2+
Zn2+
Cd2+
Al3+
Name
hydrogen
lithium
sodium
potassium
cesium
silver
magnesium
calcium
strontium
barium
zinc
cadmium
aluminum
Anions
Charge
Formula
HFCl Br I-
1-
Name
hydride
fluoride
chloride
bromide
iodide
2-
O2S2 -
oxide
sulfide
3-
N 3-
nitride
47
Listed by charge; those in boldface are most common
Give the Name and Chemical Formulas of
the Compounds Formed from the
Following Pairs of Elements
a) Sodium and Oxygen
Na2O
Sodium Oxide
b) Zinc and Chlorine
ZnCl2
Zinc Chloride
c) Calcium and Fluorine
CaF2
Calcium Fluoride
d) Strontium and Nitrogen
Sr3N2
Strontium Nitride
e) Hydrogen and Iodine
HI
Hydrogen Iodide
f) Scandium and Sulfur
Sc2S3
Scandium Sulfide
48
24
Some Metals That Form More
than One Oxidation State
Element
Chromium
Cobalt
Copper
Iron
Lead
Manganese
Mercury
Tin
Ion Formula
Cr+2
Cr+3
Co+2
Co+3
Cu+1
Cu+2
Fe+2
Fe+3
Pb+2
Pb+4
Mn+2
Mn+3
Hg2+2
Hg+2
Sn+2
Sn+4
Table 2.4 (p. 69)
Systematic Name
Chromium (II)
Chromium (III)
Cobalt (II)
Cobalt (III)
Copper (I)
Copper (II)
Iron (II)
Iron (III)
Lead (II)
Lead (IV)
Manganese (II)
Manganese (III)
Mercury (I)
Mercury (II)
Tin (II)
Tin (IV)
Common Name
Chromous
Chromic
Cuprous
Cupric
Ferrous
Ferric
Mercurous
Mercuric
Stannous
Stannic
49
Determining Names and Formulas of Ionic Compounds of Elements That Form More than One Ion
Give the systematic names for the formulas or the formulas
for the names of the following compounds.
a) Iron III Sulfide - Fe is +3, and S is -2 therefore the compound is:
Fe2S3
-1
b) CoF2 - the anion is Fluoride (F ) and there are two F -1, the
cation is Cobalt and it must be Co+2 therefore the compound is:
Cobalt (II) Fluoride
c) Stannic Oxide - Stannic is the common name for Tin (IV), Sn+4, the
Oxide ion is O-2, therefore the formula of the compound is:
SnO2
d) NiCl3 - The anion is chloride (Cl-1), there are three anions, so the
Nickel cation is Ni+3, therefore the name of the compound is:
50
Nickel (III) Chloride
25
51
Rules for Families of Oxoanions
Families with Two Oxoanions
The ion with more O atoms takes the nonmetal root and the
suffix “-ate”.
The ion with fewer O atoms takes the nonmetal root and the
suffix “-ite”.
Families with Four Oxoanions (usually a Halogen)
The ion with most O atoms has the prefix “per-”, the nonmetal
root and the suffix “-ate”.
The ion with one less O atom has just the suffix “-ate”.
The ion with two less O atoms has the just the suffix “-ite”.
The ion with three less O atoms has the prefix “hypo-” and52the
suffix “-ite”.
26
Examples of Names and Formulas of
Oxoanions and Their Compounds - I
•
•
•
•
•
•
•
•
•
•
•
•
KNO2
Potassium Nitrite
BaSO3 Barium Sulfite
Mg(NO3)2 Magnesium Nitrate
Na2SO4 Sodium Sulfate
LiClO4 Lithium Perchlorate
Ca(BrO)2 Calcium Hypobromite
NaClO3 Sodium Chlorate
Al(IO2)3 Aluminum Iodite
RbClO2 Rubidium Chlorite
KBrO3 Potassium Bromate
CsClO Cesium Hypochlorite
LiIO4 Lithium Periodate
53
Examples of Names and Formulas of
Oxoanions and Their Compounds - II
•
•
•
•
•
•
•
•
•
•
•
•
Calcium Nitrate Ca(NO3)2
Ammonium Sulfite (NH4)2SO3
Strontium Sulfate SrSO4
Lithium Nitrite
LiNO2
Potassium Hypochlorite KClO
Lithium Perbromate
LiBrO4
Rubidium Chlorate RbClO3
Calcium Iodite
Ca(IO2)2
Ammonium Chlorite NH4ClO2
Boron Bromate
B(BrO3)3
Sodium Perchlorate NaClO4
Magnesium Hypoiodite Mg(IO)2
54
27
55
Naming Oxoanions - Examples
per
hypo
Root
Suffixes
“
”
ate
“
”
ate
“
”
ite
“
”
ite
Chlorine
Bromine
perchlorate perbromate
[ ClO4-]
[ BrO4-]
No. of O atoms
Prefixes
Iodine
periodate
[ IO4-]
chlorate
[ ClO3-]
bromate
[BrO3-]
iodate
[ IO3-]
chlorite
[ ClO2-]
bromite
[ BrO2-]
iodite
[ IO2-]
hypochlorite hypobromite hypoiodite
[ ClO -]
[ BrO -]
[ IO -]
56
28
Naming Acids
1) Binary acids solutions form when certain gaseous compounds
dissolve in water. For example, when gaseous hydrogen chloride
(HCl) dissolves in water, it forms a solution called hydrochloric acid.
Prefix hydro- + anion nonmetal root + suffix -ic + the word acid
hydrochloric acid
2) Oxoacid names are similar to those of the oxoanions,
except for two suffix changes:
Anion “-ate” suffix becomes an “-ic” suffix in the acid. Anion “-ite”
suffix becomes an “-ous” suffix in the acid.
The oxoanion prefixes “hypo-” and “per-” are retained. Thus, BrO4is perbromate, and HBrO4 is perbromic acid; IO2- is iodite, and
HIO2 is iodous acid.
57
Binary Acids
Acids produce H+ when dissolved in water.
They are compounds that ionize in water.
Emphasize the fact that a molecule is an acid by altering the name.
HCl
hydrogen chloride
hydrochloric acid
HF
hydrogen fluoride
hydrofluoric acid
58
29
Determining Names and Formulas
of Anions and Acids
Problem: Name the following anions and give the names and
formulas of the acid solutions derived from them:
a) I
b) BrOc) SO3-2 d) NO3e) CN Solution:
a) The anion is Iodide; and the acid is Hydroiodic acid, HI
b) The anion is hypobromite; and the acid is hypobromous acid,
HBrO
c) The anion is Sulfite; and the acid is Sulfurous acid, H2SO3
d) The anion is Nitrate; and the acid is Nitric acid, HNO3
e) The anion is Cyanide; and the acid is Hydrocyanic acid, HCN
59
Some Compounds of Greater
Complexity
• Effect of Moisture
– Blue anhydrous
• CoCl2
– Pink hexahydrate
• CoCl2• 6 H2O
6 mol H2O ×
%H2O =
18.02 g H2O
1 mol H2O
237.9 g CoCl2• 6 H2O
× 100%
= 45.45% H2O
60
30
Hydrates
Compounds Containing Water Molecules
MgSO4
7H2O
Magnesium Sulfate heptahydrate
CaSO4 2H2O
Calcium Sulfate dihydrate
Ba(OH)2
Barium Hydroxide octahydrate
8H2O
CuSO4 5H2O
Copper II Sulfate pentahydrate
Na2CO3 10H2O
Sodium Carbonate decahydrate
61
Determining Names and Formulas of Ionic
Compounds Containing Polyatomic Ions
a) BaCl2 5 H2O
Ba+2 is the cation Barium, Cl- is the Chloride
anion. There are five water molecules therefore
the name is: Barium Chloride Pentahydrate
b) Magnesium Perchlorate Magnesium is the Mg+2 cation, and
perchlorate is the ClO4- anion, therefore we need
two perchlorate anions for each Mg cation
therefore the formula is: Mg( ClO4)2
c) (NH4)2SO3
NH4+ is the ammonium ion, and SO3-2 is the
sulfite anion, therefore the name is:
Ammonium Sulfite
d) Calcium Nitrate Calcium is the Ca+2 cation, and nitrate is the
NO3- anion, therefore the formula is:
62
Ca(NO3)2
31
Binary Compounds of Two Non-Metals
Molecular compounds
usually write the positive OS element first.
HCl hydrogen chloride
Some pairs form more than one compound
mono 1
penta
5
di
2
hexa
6
tri
3
hepta
7
tetra
4
octa
8
63
Names and Formulas of Binary
Molecular (Covalent) Compounds
1) The element with the lower group number in the periodic table is
the first word in the name; the element with the higher group number
is the second word. (Important exception: When the compound
contains oxygen and a halogen, the halogen is named first.)
2) If both elements are in the same group, the one with the higher
period number is named first.
3) The second element is named with its root and the suffix “-ide.”
4) Covalent compounds have Greek numerical prefixes to
indicate the number of atoms of each element in the compound. The
first word has a prefix only when more than one atom of the element
64
is present; the second word always has a numerical prefix.
32
65
66
33
67
68
34
Naming Alkanes
Alkanes are hydrocarbons that are called “saturated” hydrocarbons,
they contain only single bonds, no multiple bonds !
Alkanes have the general formula --- C n H 2n+2
Each carbon atom has four bonds to others atoms !
The names for alkanes all end in -ane
Alkanes are found in three distinct groups:
a) Straight chain hydrocarbons
b) Branched chain hydrocarbons
c) Cyclic hydrocarbons
69
70
35
Structural Isomers
Isomers have the same molecular formula but have different
arrangements of atoms in space. Are the following pairs isomers?
(c)
H
71
The First 10 Straight-Chain Alkanes
Name
Formula
Structural Formulas
Methane
Ethane
CH4
C2H6
Propane
Butane
C3H8
C4H10
CH3-CH2-CH3
Pentane
Hexane
C5H12
C6H14
CH3-(CH2)3-CH3
Heptane
Octane
C7H16
C8H18
Nonane
Decane
C9H20
C10H22
H
H
C
H
H
H
H
H
C C
H
H
H
CH3-(CH2)2-CH3
CH3-(CH2)4-CH3
CH3-(CH2)5-CH3
CH3-(CH2)6-CH3
CH3-(CH2)7-CH3
Table 2.7
CH3-(CH2)8-CH72
3
36
Functional Groups – Alcohols
73
Functional Groups – Carboxylic
Acid
74
37
Two Compounds with Molecular
Formula C2H6O
Property
Ethanol
Dimethyl Ether
M (g/mol)
46.07
Color
Colorless
Melting point
- 117oC
Boiling point
78.5oC
Density (at 20oC)
0.789 g/mL
Use
Intoxicant in
H H alcoholic beverages
H
C
C
H
H
O
H
H
46.07
Colorless
- 138.5oC
- 25oC
0.00195 g/mL
In refrigeration
H
H
C
O
C
H
75
Table 3.4
H
H
76
38
77
78
39