• 16
8
O
• 17
8
O
• 18
8
O
8 Protons 8 Neutrons
99.759% 15.99491462 amu
8 Protons 9 Neutrons
0.037% 16.9997341 amu
8 Protons 10 Neutrons
0.204 % 17.999160 amu

Fig.2.A

Fig.2.B

Problem: Calculate the average atomic mass of Magnesium!
Magnesium Has three stable isotopes, 24 Mg ( 78.7%);
25 Mg (10.2%); 26 Mg (11.1%).
24 Mg (78.7%) 23.98504 amu
25 Mg (10.2%) 24.98584 amu
26 Mg (11.1%) 25.98636 amu
Total =
With Significant Digits = amu

Problem: Calculate the average atomic mass of Magnesium!
Magnesium Has three stable isotopes, 24 Mg ( 78.7%);
25 Mg (10.2%); 26 Mg (11.1%).
24 Mg (78.7%) 23.98504 amu x 0.787 = 18.876226 amu
25 Mg (10.2%) 24.98584 amu x 0.102 = 2.548556 amu
26 Mg (11.1%) 25.98636 amu x 0.111 = 2.884486 amu
24.309268 amu
With Significant Digits = 24.3 amu

Zirconium has five stable isotopes: 90 Zr, 91 Zr, 92 Zr, 94 Zr, 96 Zr.
Isotope (% abd.) Mass (amu) (%) Fractional Mass
90 Zr (51.45%) 89.904703 amu X 0.5145 = 46.2560 amu
91 Zr (11.27%) 90.905642 amu X 0.1127 = 10.2451 amu
92 Zr (17.17%) 91.905037 amu X 0.1717 = 15.7801 amu
94 Zr (17.33%) 93.906314 amu X 0.1733 = 16.2740 amu
96 Zr (2.78%) 95.908274 amu X 0.0278 = 2.6663 amu
91.2215 amu
With Significant Digits = 91.22 amu

Problem: Estimate the abundance of the two Bromine isotopes, given that the average mass of Br is 79.904 amu. Since exact masses of isotopes not give, estimate from: mass in amu = #p + + #n:
79 Br = 79 g/mol and 81 Br = 81 g/mol (approximately).
Plan: Let the abundance of 79 Br = X and of 81 Br = Y and X + Y = 1.0
Solution:

Problem: Estimate the abundance of the two Bromine isotopes, given that the average mass of Br is 79.904 amu. Since exact masses of isotopes not give, estimate from: mass in amu = #p + + #n:
79 Br = 79 g/mol and 81 Br = 81 g/mol (approximately).
Plan: Let the abundance of 79 Br = X and of 81 Br = Y and X + Y = 1.0
Solution: X(79) + Y(81) = 79.904
X + Y = 1.00 therefore X = 1.00 - Y
(1.00 - Y)(79) + Y(81) = 79.904
79 - 79 Y + 81Y = 79.904
2 Y = 0.904 = 1 w/ sig. figs. so Y = 0.5
X = 1.00 - Y = 1.00 - 0.5 = 0.5
%X = % 79 Br = 0.5 x 100% = 50% (Actual: 50.67% = 79 Br)
%Y = % 81 Br = 0.5 x 100% = 50% (Actual: 49.33% = 81 Br)

1. All matter is composed of atoms. Although atoms are composed of smaller particles (electrons, protons, and neutrons), the atom is the smallest body that retains the unique identity of the element.
2. Atoms of one element cannot be converted into atoms of another element in a chemical reaction. Elements can only be converted into other elements in Nuclear reactions in which protons are changed.
3. All atoms of an element have the same number of protons and electrons, which determines the chemical behavior of the element.
Isotopes of an element differ in the number of neutrons, and thus in mass number, but not in chemical behavior (much). A sample of the element is treated as though its atoms have an average mass.
4. Compounds are formed by the chemical combination of two or more elements in specific ratios, as originally stated by Dalton.

• ELEMENT - A substance that cannot be separated into simpler substances by chemical means
• COMPOUND - A substance composed of atoms of two or more elements chemically united in fixed proportions
• PERIODIC TABLE “MENDELEEV TABLE” -
A tabular arrangement of the elements, vertical groups or families of elements based upon their chemical properties - actually combining ratios with oxygen

Fig2.16

Fig.2.17

Main Group Elements (Vertical Groups)
Group IA -
Group IIA -
Group IIIA -
Group IVA -
Group VA -
Group VIA -
Group VIIA -
Group VIIIA -
Other Groups ( Vertical and Horizontal Groups)
Group IB - 8B -
Period 6 Group -
Period 7 Group -

Main Group Elements (Vertical Groups)
Group IA - Alkali Metals
Group IIA - Alkaline Earth Metals
Group IIIA - Boron Family
Group IVA - Carbon Family
Group VA - Nitrogen Family
Group VIA - Oxygen Family (Calcogens)
Group VIIA - Halogens
Group VIIIA - Noble Gases
Other Groups ( Vertical and Horizontal Groups)
Group IB - 8B Transition Metals
Period 6 Group Lanthanides (Rare Earth Elements)
Period 7 Group Actinides

The Periodic Table of the Elements
H
Li Be
Na Mg
B C
Al Si
N
P
O
S
F
He
Ne
Cl Ar
K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe
Cs Ba La Hf Ta W Re Os Ir
Fr Ra Ac Rf Du Sg Bo HaMe
Pt Au Hg Tl Pb Bi Po At Rn
Ce Pr Nd PmSm Eu Gd Tb Dy Ho Er TmYb Lu
Th Pa U Np PuAmCmBk Cf Es FmMd No Lr
The Halogens The Alkali Metals
The Alkaline
Earth Metals
The Noble Gases

The Periodic Table of the Elements
H
Li Be
NaMg
B
Al
C
Si
N
P
O
S
F
He
Ne
Cl Ar
K Ca Sc Ti V CrMn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe
Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn
Fr Ra Ac Rf Du Sg Bo Ha Me
Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu
Th Pa U Np PuAmCm Bk Cf Es FmMd No Lr
Boron family
Carbon Family
Nitrogen family
Oxygen Family

The Periodic Table of the Elements
H
Li Be B C
Al Si
N
P
O F
S Cl
He
Ne
Ar Na Mg
K Ca Sc Ti V CrMn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
Rb Sr Y Zr NbMo Tc Ru Rh PdAg Cd In Sn Sb Te I Xe
Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn
Fr Ra Ac Rf Du Sg Bo Ha Me
The Transition Metals
Ce Pr Nd PmSm Eu Gd Tb DyHo Er TmYb Lu
Th Pa U Np PuAmCmBk Cf Es FmMd No Lr
Lanthanides: The
Rare Earth Elements
The Actinides

Fig.2.18

Fig.2.19

Fig.2.20

Predicting the Ion an Element will form in Chemical Reactions
Problem: What monoatomic 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 which is 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
(b) S
(c) Ti
(d) F

Predicting the Ion an Element will form in Chemical Reactions
Problem: What monoatomic 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 which is 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!

The Periodic Table of the Elements
Most Probable Oxidation State
(same concept as Fig. 2.20)
+1 0
H +2 +3 +_4 - 3 - 2 - 1 He
Li Be B C N O F Ne
Na
K
Rb
Cs
Mg
Ca
+3 +4
Sc
Sr
Y
Ba
La
+5 +1 + 2 Al Si P S
Ti V Cr Mn Fe Co Ni Cu
Zr Nb
Hf Ta
Mo
W
Tc
Re
Ru Rh Pd
Ag
Os Ir Pt
Au
Zn Ga Ge As Se
Cd In
Hg
Tl
Sn
Pb
Sb
Bi
Te
Po
Cl Ar
Br
I
Kr
Xe
At Rn
Fr Ra Ac Rf Du Sg Bo Ha Me
+3
+3
Ce
Th
Pr Nd PmSm Eu Gd Tb Dy Ho Er TmYb Lu
Pa U Np Pu AmCmBk Cf Es FmMd No Lr

Chemical Bonds - The electrostatic forces that hold the atoms of elements together in the compound.
Covalent Compounds Electrons are shared between atoms of different elements to form Covalent Cpds.
Ionic Compounds - Electrons are transferred from one atom to another to form Ionic Cpds.
“Cation” - An atom that has lost electron(s) to form “ + ” ions.
May be 1 or more e s. Common with metal elements.
“Anion”
- An atom which has gained electron(s), to form “ - ” ions. Common w/ nonmetal elements.
Later we’ll learn that group of atoms can also be anion or cation
Mono-atomic (monatomic) ions form binary ionic compounds.

Fig.2.20

Fig 2.22
(P 65)
The polyatomic ion

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.

water hydrogen peroxide
H
2
O
H
2
O
2 ethane sulfur
C
2
H
6
S
8 acetic acid CH
3
COOH

water hydrogen peroxide
H
2
O H
2
O
H
2
O
2
HO ethane sulfur acetic acid
C
2
H
6
S
8
CH
3
COOH
CH
3
S
COH
2

Fig.2.23

Start trying to learn those in bold . Best done by looking at name w/ use.

Na
2
O Sodium Oxide a) Sodium and Oxygen b) Zinc and Chlorine c) Calcium and Fluorine d) Strontium and Nitrogen e) Hydrogen and Iodine f) Scandium and Sulfur

a) Sodium and Oxygen Na
2
O Sodium Oxide b) Zinc and Chlorine ZnCl
2
Zinc Chloride
Calcium Fluoride c) Calcium and Fluorine CaF
2 d) Strontium and Nitrogen Sr
3
N
2
Strontium Nitride e) Hydrogen and Iodine HI Hydrogen Iodide f) Scandium and Sulfur Sc
2
S
3
Scandium Sulfide

Start learning these boldface ones .

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:
Fe
2
S
3 b) CoF
2
c) Stannic Oxide d) NiCl
3
-

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: b) CoF
2
Fe
2
S
3
- the anion is Fluoride ( F -1 ) 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: d) NiCl
3
SnO
2
- The anion is chloride (Cl -1 ), there are three anions, so the
Nickel cation is Ni +3 , therefore the name of the compound is:
Nickel (III) Chloride

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”.
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-” and the suffix
“-ite”.

Fig.2.24

Prefixes Root Suffixes Chlorine Bromine Iodine per “ ” ate perchlorate perbromate periodate
[ ClO
4
] [ BrO
4
] [ IO
4
]
“ ” ate chlorate bromate iodate
[ ClO
3
] [BrO
3
] [ IO
3
]
“ ” ite chlorite bromite iodite
[ ClO
2
] [ BrO
2
] [ IO
2
] hypo “ ” ite hypochlorite hypobromite hypoiodite
[ ClO ] [ BrO ] [ IO ]