Element #8 : Oxygen, Isotopes

• 168O 8 Protons 8 Neutrons

99.759% 15.99491462 amu

• 178O 8 Protons 9 Neutrons

0.037% 16.9997341 amu

• 188O 8 Protons 10 Neutrons

0.204 % 17.999160 amu

Fig.2.A

Fig.2.B

Calculating the “Average” Atomic Mass of an Element

24Mg (78.7%) 23.98504 amu 25Mg (10.2%) 24.98584 amu

26Mg (11.1%) 25.98636 amu

Total =

With Significant Digits = amu

Problem: Calculate the average atomic mass of Magnesium! Magnesium Has three stable isotopes, 24Mg ( 78.7%); 25Mg (10.2%); 26Mg (11.1%).

Calculating the “Average” Atomic Mass of an Element

24Mg (78.7%) 23.98504 amu x 0.787 = 18.876226 amu 25Mg (10.2%) 24.98584 amu x 0.102 = 2.548556 amu26Mg (11.1%) 25.98636 amu x 0.111 = 2.884486 amu

24.309268 amu

With Significant Digits = 24.3 amu

Problem: Calculate the average atomic mass of Magnesium! Magnesium Has three stable isotopes, 24Mg ( 78.7%); 25Mg (10.2%); 26Mg (11.1%).

Calculate the Average Atomic Mass of Zirconium, Element #40

Zirconium has five stable isotopes: 90Zr, 91Zr, 92Zr, 94Zr, 96Zr.

Isotope (% abd.) Mass (amu) (%) Fractional Mass

90Zr (51.45%) 89.904703 amu X 0.5145 = 46.2560 amu91Zr (11.27%) 90.905642 amu X 0.1127 = 10.2451 amu92Zr (17.17%) 91.905037 amu X 0.1717 = 15.7801 amu94Zr (17.33%) 93.906314 amu X 0.1733 = 16.2740 amu96Zr (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:

79Br = 79 g/mol and 81Br = 81 g/mol (approximately).

Plan: Let the abundance of 79Br = X and of 81Br = Y and X + Y = 1.0Solution:

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:

79Br = 79 g/mol and 81Br = 81 g/mol (approximately).

Plan: Let the abundance of 79Br = X and of 81Br = Y and X + Y = 1.0Solution: 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 = % 79Br = 0.5 x 100% = 50% (Actual: 50.67% = 79Br)%Y = % 81Br = 0.5 x 100% = 50% (Actual: 49.33% = 81Br)

Modern Reassessment of the Atomic Theory

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.

Definitions• 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

Groups in the Periodic Table

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 -

Groups in the Periodic Table

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 GasesOther Groups ( Vertical and Horizontal Groups)Group IB - 8B - Transition MetalsPeriod 6 Group - Lanthanides (Rare Earth Elements)Period 7 Group - Actinides

O

S

Se

Te

Po

N

P

As

Sb

Bi

C

Si

Ge

Sn

Pb

B

Al

Ga

In

Tl

ZnCu

Cd

Hg

Ag

Au

Ni

Pd

Pt

Co

Rh

Ir

Fe

Ru

Os

Mn

Tc

Re

Cr

Mo

W

V

Nb

Ta

Ti

Zr

Hf

Sc

Y

La

Ac

The Periodic Table of the Elements

The Alkali Metals

The Alkaline Earth Metals

Ce Pr Nd PmSmEu Gd Tb Dy Ho Er TmYb Lu

Th Pa Np PuAmCmBk Cf Es FmMd No LrU

H

Li

Na

K

Rb

Cs

Fr

Be

Mg

Ca

Sr

Ba

Ra Rf Sg

The Halogens

The Noble Gases

He

Ne

Ar

Kr

Xe

Rn

F

Cl

Br

I

At

Du Bo HaMe

The Periodic Table of the ElementsH

Li Be

NaMg

K Ca Sc

Rb

Cs

Fr

Sr

Ba

Ra

Ti V CrMn Fe

Y

La

Ac

Co Ni Cu Zn

Zr

Hf

Nb

Ta

Rf

Mo

W

Tc

Re

Ru

Os

Rh

Ir

Pd

Pt

Ag

Au

Cd

Hg

F

He

Ne

ArCl

Br Kr

Xe

Rn

I

At

Ce Pr Nd Pm

Th

SmEu Gd Tb Dy Ho Er Tm Yb Lu

Pa U Np PuAmCm Bk Cf Es FmMd No Lr

Boron family

B

Al

Ga

In

Tl

Carbon Family

C

Si

Ge

Sn

Pb

Nitrogen family

N

P

As

Sb

Bi

Oxygen Family

O

S

Se

Te

Po

Du Sg Bo Ha Me

O

S

Se

Te

Po

N

P

As

Sb

Bi

C

Si

Ge

Sn

Pb

B

Al

Ga

In

Tl

The Periodic Table of the Elements

Lanthanides: The

Rare Earth ElementsThe Actinides

F

Cl

Br

I

At

H He

Ne

Ar

Kr

Xe

Rn

Li

Na

K

Rb

Cs

Fr

Be

Mg

Ca

Sr

Ba

Ra

Ce

The Transition Metals

Pr Nd PmSmEu Gd Tb DyHo Er TmYb Lu

Th Pa U Np PuAmCmBk Cf Es FmMd No Lr

Sc Ti V CrMn

Y

La

Fe Co Ni Cu Zn

Zr NbMo Tc Ru Rh PdAg Cd

Hf Ta W Re Os Ir Pt Au Hg

Ac Rf Sg HaDu Bo Me

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!

He

Ne

Ar

Kr

Xe

Rn

The Periodic Table of the Elements

CrMn Fe Co Ni

Mo

W

Tc

Re

Ru

Os

Rh

Ir

Pd

Pt

Most Probable Oxidation State

+1

+2

+3 +4

+3 +_4 - 3 - 2 - 1

0

H

Li

Na

K

Rb

Cs

Fr

Sc

Y

Be

Mg

Ca

Sr

Ba

Ra

La

Ac

B

Al

Ga

In

Tl

Ti

Rf

Hf

Zr

C

Si

Ge

Sn

Pb

F

Cl

Br

I

At

O

S

Se

Te

Po

N

P

As

Sb

Bi

Zn

CdHg

+ 2+1

Cu

Ag

Au

+5

V

Nb

Ta

CeTh

Pr Nd PmSmEu Gd Tb Dy Ho Er TmYb LuPa U Np Pu AmCmBk Cf Es FmMd No Lr

+3

+3

Du Sg Bo Ha Me

(same concept as Fig. 2.20)

Chemical Compounds and Bonds

Chemical Bonds - The electrostatic forces that hold the atoms of elements together in the compound.

Ionic Compounds - Electrons are transferred from one atom to another to form Ionic Cpds.

Covalent Compounds - Electrons are shared between atoms of different elements to form Covalent 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)

Thepolyatomicion

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.

Empirical and Molecular Formulas Name Molecular Empiricalwater H2O

hydrogen H2O2 peroxide

ethane C2H6

sulfur S8

acetic acid CH3COOH

Empirical and Molecular Formulas Name Molecular Empiricalwater H2O H2O

hydrogen H2O2 HOperoxide

ethane C2H6 CH3

sulfur S8 S

acetic acid CH3COOH COH2

Fig.2.23

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

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 c) Calcium and Fluorine

d) Strontium and Nitrogen

e) Hydrogen and Iodine

f) Scandium and Sulfur

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

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: Fe2S3

b) CoF2 -

c) Stannic Oxide -

d) NiCl3 -

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

b) CoF2 - 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) Fluoridec) 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: Nickel (III) Chloride

Rules for Families of OxoanionsFamilies 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-” and thesuffix “-ite”.

Fig.2.24

NAMING OXOANIONS - EXAMPLES

Prefixes Root Suffixes Chlorine Bromine Iodine

per “ ” ate perchlorate perbromate periodate [ ClO4

-] [ BrO4-] [ IO4

-] “ ” ate chlorate bromate iodate [ ClO3

-] [BrO3-] [ IO3

-]

“ ” ite chlorite bromite iodite [ ClO2

-] [ BrO2-] [ IO2

-]

hypo “ ” ite hypochlorite hypobromite hypoiodite [ ClO -] [ BrO -] [ IO -]

No.

of

O a

tom

s