Chapter 2: The Components of Matter

2.1: Elements, Compounds, and Mixtures: An Atomic Overview2.2: The Observations That Led to an Atomic View of Matter2.3: Dalton’s Atomic Theory2.4: The Observations That Led to the Nuclear Atom Model2.5: The Atomic Theory Today2.6: Elements: A First Look at the Periodic Table2.7: Compounds: Introduction to Bonding2.8: Compounds: Formulas, Names, and Masses2.9: Mixtures: Classification and Separation

Fig. 2.1

Definitions for Components of Matter

Pure Substances - Their compositions are fixed! Elements and compounds are examples of Pure Substances.Element - Is the simplest type of substance with unique physical and chemical properties. An element consists of only one type of atom. It cannot be broken down into any simpler substances by physical or chemical means.Molecule - Is a structure that is consisting of two or more atoms that are chemically bound together and thus behaves as an independent unit.Compound - Is a substance composed of two or more elements that are chemically combined.Mixture - Is a group of two or more elements and/or compounds that are physically intermingled.

Fig. 2.2

(p. 43)

Fig. 2.3

Separating

a Mixture

Fig. 2.4

Laws of Mass Conservation & Definite Composition

Law of Mass conservation: The total mass of substances does not change during a chemical reaction.

Law of Definite ( or constant ) composition: No matter what its source, a particular chemical compound is composed of the same elements in the same parts (fractions) by mass.

The Meaning of Mass Fraction and Mass Percent

Fig. 2.5

Mass Percent Composition of Na2SO4

Na2SO4 = 2 atoms of Sodium + 1 atom of Sulfur + 4 atoms of Oxygen

Elemental masses

2 x Na = 2 x 22.99 = 45.981 x S = 1 x 32.07 = 32.07

4 x O = 4 x 16.00 = 64.00

142.05

Percent of each Element

Check

% Na + % S + % O = 100%

32.37% + 22.58% + 45.05% = 100.00%

Mass Percent Composition of Na2SO4

Na2SO4 = 2 atoms of Sodium + 1 atom of Sulfur + 4 atoms of Oxygen

Elemental masses

2 x Na = 2 x 22.99 = 45.981 x S = 1 x 32.07 = 32.07

4 x O = 4 x 16.00 = 64.00

142.05

Percent of each Element

%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%

Calculating the Mass of an Element in a Compound Ammonium Nitrate

Ammonium Nitrate = NH4NO3

How much Nitrogen is in 455 kg of Ammonium Nitrate?

The Formula Mass of Cpd is:

4 x H = 4 x 1.008 = 4.032 g2 x N = 2 X 14.01 = 28.02 g3 x O = 3 x 16.00 = 48.00 g

80.052 g

Therefore gm Nitrogen/ gm Cpd

Calculating the Mass of an Element in a Compound Ammonium Nitrate

Ammonium Nitrate = NH4NO3

How much Nitrogen is in 455 kg of Ammonium Nitrate?

The Formula Mass of Cpd is:

4 x H = 4 x 1.008 = 4.032 g2 x N = 2 X 14.01 = 28.02 g3 x O = 3 x 16.00 = 48.00 g

80.052 g

Therefore gm Nitrogen/ gm Cpd

28.02 g Nitrogen

80.052 g Cpd = 0.35002249 g N / g Cpd

455 kg x 1000g / kg = 455,000 g NH4NO3

455,000 g Cpd x 0.35002249 g N / g Cpd = 1.59 x 105 g Nitrogen

28.02 kg Nitrogen

80.052 kg NH4NO4

= 159 kg Nitrogen 455 kg NH4NO3 Xor:

Fig. 2.7

Law of Multiple Proportions

If elements A and B react to form two compounds,the different masses of B that combine with a fixed mass of A can be expressed as a ratio of small wholenumbers:

Example: Nitrogen Oxides I & II

Nitrogen Oxide I : 46.68% Nitrogen and 53.32% OxygenNitrogen Oxide II : 30.45% Nitrogen and 69.55% Oxygen

in 100 g of each Compound: g O = 53.32 g & 69.55 g g N = 46.68 g & 30.45 g

g O /g N = 1.142 & 2.2842.284 2

1.142 1=

Dalton’s Atomic Theory

1. All matter consists of atoms.

2. Atoms of one element cannot be converted into atoms of another element.

3. Atoms of an element are identical in mass and other properties and are different from atoms of any other element.

4. Compounds result from the chemical combination of a specific ratio of atoms of different elements.

The Combining Volumes of Hydrogen and Oxygen in the Formation of Water Vapor

Fig. 2.8

Fig. 2.9

Fig. 2.11

Rutherford Experiment

• Alpha particles bombarding the atom.

• Rationale - to study the internal structure of the atom, and to know more about the mass distribution in the atom!

• Bombarded a thin Gold foil with Alpha particles from Radium.

Fig. 2.12

Ernest Rutherford (1871-1937)

• Won the Nobel Prize in Chemistry in 1908

• “It was quite the most incredible event..... It was almost as if a gunner were to fire a shell at a piece of tissue and the shell bounced right back!!!!! ”

Fig. 2.13

Properties of the Subatomic Particles

Name Charge Mass Location(Symbol) Relative Absolute Relative(amu) Absolute In Atom

Proton 1 + 1.60218 x 10-19 C* 1.00727 1.67262 x 10-24 g Nucleus (P +)

Neutron 0 0 1.00866 1.67593 x 10-24 g Nucleus (n o)

Electron 1 - -1.60218 x 10-19 C* 9.10939 x 10-28g Outside (e -) 0.00054858 Nucleus

* the coulomb (C) is the SI unit of Charge

Table 2.2 (p. 53)

Atomic Definitions I: Symbols, Isotopes,Numbers

XA

Z

X = Atomic symbol of the element, or element symbol

A = The Mass number; A = Z + N

Z = The Atomic Number, the Number of Protons in the Nucleus

N = The Number of Neutrons in the Nucleus

Isotopes = atoms of an element with the same number of protons, but different numbers of Neutrons in the Nucleus

The Nuclear Symbol of the Atom, or Isotope

Depicting the Atom

Fig. 2.14

Neutral ATOMS• 51 Cr = P+ (24), e- (24), • N (27)

• 239 Pu = P+(94), e- (94), • N (145)

• 15 N = P+(7), e-(7), N(8)

• 56 Fe = P+(26), e-(26),• N (50)

• 235 U =P+(92), e-(92), • N (143)

Atomic Definitions II: AMU, Dalton, 12C Std.

Atomic mass Unit (AMU) = 1/12 the mass of a carbon - 12 atom on this scale Hydrogen has a mass of 1.008 AMU.

Dalton (D) = The new name for the Atomic Mass Unit, one dalton = one Atomic Mass Unit on this scale, 12C has a mass of 12.00 daltons.

Isotopic Mass = The relative mass of an Isotope relative to the Isotope 12C the chosen standard.

Atomic Mass = “Atomic Weight” of an element is the average of the masses of its naturally occurring isotopes weighted according to their abundances.

Isotopes of Hydrogen

• 11H 1 Proton 0 Neutrons 99.985 % 1.00782503 amu

• 21H (D) 1 Proton 1 Neutron 0.015 % 2.01410178 amu

• 31H (T) 1 Proton 2 Neutrons -------- ----------

The average mass of Hydrogen is 1.008 amu

• 3H is Radioactive with a half life of 12 years.

• H2O Normal water “light water “

• mass = 18.0 g/mole , BP = 100.000000C

• D2O Heavy water

• mass = 20.0 g/mole , BP = 101.42 0C

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

Formation of a Positively Charged Neon Particle in a Mass Spectrometer

Fig. 2.A

Fig. 2.B part A

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: Calculate the abundance of the two Bromine isotopes: 79Br = 78.918336 g/mol and 81Br = 80.91629 g/mol , given that the average mass of Bromine is 79.904 g/mol.

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

Solution: X(78.918336) + Y(80.91629) = 79.904

X + Y = 1.00 therefore X = 1.00 - Y (1.00 - Y)(78.918336) + Y(80.91629) = 79.904

78.918336 - 78.918336 Y + 80.91629 Y = 79.904

1.997954 Y = 0.985664 or Y = 0.4933

X = 1.00 - Y = 1.00 - 0.4933 = 0.5067

%X = % 79Br = 0.5067 x 100% = 50.67% = 79Br %Y = % 81Br = 0.4933 x 100% = 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 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

Fig. 2.16

Fig. 2.17

He

Ne

Ar

Kr

Xe

Rn

The Periodic Table of the Elements

Most Probable Oxidation State

CrMn Fe Co Ni

Mo

W

Tc

Re

Ru

Os

Rh

Ir

Pd

Pt

+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

Newly Discovered Elements

Atomic No.

1994

ACS Slate IUPAC Slate

RevisedIUPAC Slate

104 Rutherfordium Dubnium Rutherfordium

105 Hahnium Joliotium Dubnium

106 Seaborgium Rutherfordium Seaborgium

107 Neilsbohrium Bohrium Bohrium

108 Hassium Hahnium Hassium

109 Meitnerium Meitnerium Meitnerium

110 ? ? 3 Labs

111 ? ? GSI

112 ? ? GSIFinal Slate 9/12/97

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

The Periodic Table of the Elements

Metals

Non Metals

Semi - metals Metalloids

B

Si

Ge As

Sb Te

C N

P

O

S

Se

F

Cl

Br

I

At

He

Ne

Ar

Kr

Xe

RnPoBi

Al

Ga

Sn

Pb

In

Tl

Zn

Cd

Hg

Cu

Ag

Au

NiCoFeMn

Pd

Pt

Rh

Ir

Ru

Os

Tc

Re

Sg

W

Mo

CrV

Nb

Ta

Ti

Zr

Hf

Rf

Sc

Y

La

Ac

Be

Mg

Ca

Sr

Ba

Ra

H

Li

Na

K

Rb

Cs

Fr

Ce Pr

Th

Nd Pm SmEu Gd Tb Dy Ho Er

Pa U Np PuAmCm Bk Cf Es Fm

Md

TmYb Lu

No Lr

Du Bo Ha 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+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!

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.

“Cations” - Metal atoms lose electrons to form “ + ” ions.

“Anions” - Nonmetal atoms gain electrons to form “ - ” ions.

Mono-atomic ions form binary ionic compounds

Formation of a Covalent Bond between Two Hydrogen Atoms

Fig. 2.21

A Polyatomic Ion

Fig. 2.22

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.

Fig. 2.23

Table 2.3 (p. 67) Common Monoatomic IonsCations Anions

Charge Formula Name Charge Formula Name

1+ H+ hydrogen 1- H - hydride Li+ lithium F - fluoride Na+ sodium Cl - chloride K+ potassium Br - bromide Cs+ cesium I - iodide Ag+ silver2+ Mg2+ magnesium 2- O 2 - oxide Ca2+ calcium S2 - sulfide Sr2+ strontium Ba2+ barium Zn2+ zinc Cd2+ cadmium3+ Al3+ aluminum 3- N 3 - nitride

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

Some Metals That Form More than One Oxidation State

Element Ion Formula Systematic Name Common Name

Chromium Cr+2 Chromium (II) Chromous Cr+3 Chromium (III) ChromicCobalt Co+2 Cobalt (II) Co+3 Cobalt (III) Copper Cu+1 Copper (I) Cuprous Cu+2 Copper (II) CupicIron Fe+2 Iron (II) Ferrous Fe+3 Iron (III) FerricLead Pb+2 Lead (II) Pb+4 Lead (IV) Manganese Mn+2 Manganese (II) Mn+3 Manganese (III)Mercury Hg2

+2 Mercury (I) Mercurous Hg+2 Mercury (II) MercuricTin Sn+2 Tin (II) Stannous Sn+4 Tin (IV) Stannic Table 2.4 (p. 69)

Some Common Polyatomic IonsFormula Name Formula Name

CationsNH4

+ AmmoniumH3O+ Hydronium

Anions

CH3COO- AcetateCN- CyanideOH- HydroxideClO- HypochloriteClO2

- ChloriteClO3

- ChlorateClO4

- PerchlorateNO2

- NitriteNO3

- Nitrate

Anions Ctd.

MnO4- Permanganate

CO3-2 Carbonate

HCO3- Hydrogen Carbonate

CrO4-2 Chromate

Cr2O7-2 Dichromate

O2-2 Peroxide

PO4-3 Phosphate

HPO4-2 Hydrogen Phosphate

H2PO4- Dihydrogen Phosphate

SO3-2 Sulfite

SO4-2 Sulfate

HSO4- Hydrogen Sulfate

Determining Names and Formulas of Ionic Com-pounds 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”.

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

Numerical Prefixes for Hydrates and Binary Covalent Compounds

NUMBER PREFIX NUMBER PREFIX

1 mono- 6 hexa-

2 di- 7 hepta-

3 tri- 8 octa-

4 tetra- 9 nona-

5 penta- 10 deca-

Table 2.6 (p. 70)

Hydrates Compounds Containing Water Molecules

MgSO4 7H2O Magnesium Sulfate heptahydrate

CaSO4 2H2O Calcium Sulfate dihydrate

Ba(OH)2 8H2O Barium Hydroxide octahydrate

CuSO4 5H2O Copper II Sulfate pentahydrate

Na2CO3 10H2O Sodium Carbonate decahydrate

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 Hypochlotite LiIO4 Lithium Periodate

Examples of Names and Formulas ofOxoanions 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

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 thereforethe 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: Ca(NO3)2

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, BrO4

-

is perbromate, and HBrO4 is perbromic acid; IO2- is iodite, and

HIO2 is iodous acid.

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) BrO3- c) SO3

-2 d) NO3- e) CN -

Solution:

a) The anion is Iodide; and the acid is Hydroiodic acid, HI

b) The anion is Bromite; and the acid is Bromous acid, HBrO2

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

Names and Formulas of Binary 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 numberis 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 (table 2.6) toindicate the number of atoms of each element in the compound. Thefirst word has a prefix only when more than one atom of the elementis present; the second word always has a numerical prefix.

Determining Names and Formulas of Binary Covalent Compounds

Problem: What are the name or Chemical formulas of the followingChemical compounds:

a) Carbon dioxide b) PCl3 c) Give the name and chemical formula of the compound formed from two P atoms and five O atoms.

Solution:

a) The prefix “di-” means “two.” The formula is CO2

b) P is the symbol for phosphorous; there are three chlorine atoms which require the prefix “tri-.” The name of the compound is: phosphorous trichloride

c) P comes first in the name (lower group number). The compound is diphosphorous pentaoxide ( commonly called “phosphorous pentaoxide”)

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

The First 10 Straight-Chain AlkanesName Formula Structural Formulas

Methane CH4

Ethane C2H6

Propane C3H8

Butane C4H10

Pentane C5H12

Hexane C6H14

Heptane C7H16

Octane C8H18

Nonane C9H20

Decane C10H22

C H

H H

H H

H

H

H H

H

C C

CH3-CH2-CH3 CH3-(CH2)2-CH3

CH3-(CH2)3-CH3

CH3-(CH2)4-CH3

CH3-(CH2)5-CH3

CH3-(CH2)6-CH3

CH3-(CH2)7-CH3

CH3-(CH2)8-CH3Table 2.7

Calculating the Molecular Mass of a Compound

Problem: Using the data in the periodic table, calculate the molecular mass of the following compounds: a) Tetraphosphorous decoxide b) Ammonium sulfatePlan: We first write the formula, then multiply the number of atoms (or ions) of each element by its atomic mass, and find the sum.Solution:

a) The formula is P4O10. Molecular mass = (4 x atomic mass of P ) +(10 x atomic mass of O ) = ( 4 x 30.97 amu) + ( 10 x 16.00 amu) = 283.88 = 283.9 amub) The formula is (NH4)2SO4

Molecular mass = ( 2 x atomic mass of N ) + ( 8 x atomic mass of H) + ( 1 x atomic mass of S ) + ( 4 x atomic mass of O) = ( 2 x 14.01 amu) + ( 8 x 1.008 amu) + ( 1 x 32.07 amu) + ( 4 x 16.00 amu) = 132.154 amu = 132.15 amu

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

Fig. 2.25

Mixtures

Heterogeneous mixtures : has one or more visible boundaries between the components.

Homogeneous mixtures : has no visible boundaries because the components are mixed as individual atoms, ions, and molecules.

Solutions : A homogeneous mixture is also called a solution. Solutions in water are called aqueous solutions, and are very important in chemistry. Although we normally think of solutions as liquids, they can exist in all three physical states.

Separating MixturesFiltration : Separates components of a mixture based upon differences in particle size. Normally separating a precipitate from a solution, or particles from an air stream.Crystallization : Separation is based upon differences in solubility of components in a mixture.

Distillation : separation is based upon differences in volatility.

Extraction : Separation is based upon differences in solubility in different solvents (major material).

Chromatography : Separation is based upon differences in solubility in a solvent versus a stationary phase.

Fig. 2.D

Filtration

Crystallization

Fig. 2.E

Fig. 2.F

Fig. 2.G

Procedure for Column Chromatography

Fig. 2.H

Fig. 2.I A&B