Chapter 2

The early civilizations had an idea of the composition of matter.



Egyptian: “kēme” (earth)
Al-kimia: “The art of transformation”
History? --> No single, straight history.


ORIGINS: Possibly the advent of fire and burning
Leading to pottery, metal-works (metallurgy), ancient structure-making

The early civilizations had an idea of the composition of matter.
Matter was seen as continuous, as the four
Greek elements were.
No distinct divisions between fire, water, earth, air
Mixtures of the four gave the properties of being hot or cold, moist, or dry

The early civilizations had an idea of the composition of matter.
 LEUCIPPUS: “There must be tiny particles of water that could not be subdivided.” Observe the
SAND.

DEMOCRITUS: Referred to these particles as atomos ;
Each atom was distinct in size and shape (eg. Water as round balls, Fire as sharp)

The early civilizations had an idea of the composition of matter.


Metallurgy and Alchemy was also common practice in ancient civilizations.
 Fire led to purification and material creation (eg. Glass and metals)
 ALLOYS -> Bronze Age (ca.3000-1200BCE)
 Alloys and lasting metals (eg. gold) were precious

Metallurgy and Alchemy was also common practice in ancient civilizations.



Need: more stable materials like gold
Hypothesis: Perhaps there is a way of converting other materials to gold
(philosopher’s stone)
MORE THAN GOLD: An early philosphical and spiritual discipline linking all things


“Solve et coagula”- “Separate and join together”
Al-kimia: the art of transformation
 Aim to improve life: Elixir of life
 Problems: Non-systematic, vague language and concepts, pseudoscience angle, fraudulence
 14th Century weakening

Finally in 1661, Robert Boyle re-worked the hypothesis of atoms.




“The Sceptical Chymist or Chymico-
Physical Doubts & Paradoxes” by
Robert Boyle in 1661
Thesis: “Matter consisted of atoms and clusters of atoms in motion and that every phenomenon was the result of collisions of particles in motion”
Appeal to EXPERIMENT!
Chemistry should stop being subservient to medicine and alchemy and establish itself as a separate science

According to Boyle, substances are made of elements which are in turn composed of “simple bodies” = atom.
Element - 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 a structure that consists of two or more atoms that are chemically bound together and thus behaves as an independent unit.

According to Boyle, substances are made of elements which are in turn composed of “simple bodies” = atom.
Compound - a substance composed of two or more elements which are chemically combined.
Mixture - a group of two or more elements and/or compounds that are physically intermingled.

According to Boyle, substances are made of elements which are in turn composed of “simple bodies” = atom.

In the 18 th century, several experiments led to the different LAWS we know of today.






Antoine Laurent Lavoisier
1789- “When a chemical reaction is carried out in a closed system, the total mass of the system is not changed.”
Red mercuric oxide  Mercury +
OXYGEN
1 st to use systematic names; 1 st chem bk. ; “father”
Experiments with burning coal
(combustion), and breathing guinea pigs (respiration).
LAW: Matter is neither created nor destroyed in a chemical change. The total mass of the reaction products is always equal to the total mass of the reactants
We cannot create from nothing.
Chemistry is about transformation.

In the 18 th century, several experiments led to the different LAWS we know of today.
Law of Mass Conservation:
The total mass of substances does not change during a chemical reaction.
reactant 1 + reactant 2 total mass calcium oxide + carbon dioxide
=
CaO + CO
2
56.08g + 44.00g
product total mass calcium carbonate
CaCO
3
100.08g

In the 18 th century, several experiments led to the different LAWS we know of today.
Law of Mass Conservation:

In the 18 th century, several experiments led to the different LAWS we know of today.
 Joseph Louis Proust: Copper carbonate always had the same composition
 LAW OF DEFINITE PROPORTIONS or CONSTANT
COMPOSITION: A compound always contains the same elements in certain definite proportions and in no other combinations.
 J.J.Berzelius: Prepared an extensive list of atomic weights; Lead sulfide experiments
 Henry Cavendish: 1783;
 Hydrogen gas + Oxygen gas  Water
 1800: Volta designed a powerful battery W.Nicholson
and A.Carlisle would use to separate water into its elements.

In the 18 th century, several experiments led to the different LAWS we know of today.
Law of Definite Composition:
No matter the source, a particular compound is composed of the same elements in the same parts (fractions) by mass.
Analysis by Mass
(grams/20.0g)
8.0 g calcium
2.4 g carbon
9.6 g oxygen
20.0 g
Mass Fraction
(parts/1.00 part)
0.40 calcium
0.12 carbon
0.48 oxygen
1.00 part by mass
Percent by Mass
(parts/100 parts)
40% calcium
12% carbon
48% oxygen
100% by mass

In the 18 th century, several experiments led to the different LAWS we know of today.
 Elements could combine in in more than one set of proportions.
 If elements A and B react to form two different compounds, the masses of B combined with a fixed mass of A, can be expressed as a ratio of small whole numbers

In the 18 th century, several experiments led to the different LAWS we know of today.
Law of Multiple Proportions: Example: Carbon Oxides A & B
Carbon Oxide I : 57.1% oxygen and 42.9% carbon
Carbon Oxide II : 72.7% oxygen and 27.3% carbon
Assume that you have 100g of each compound.
In 100 g of each compound: g O = 57.1 g for oxide I & 72.7 g for oxide II g C = 42.9 g for oxide I & 27.3 g for oxide II g O g C g O g C
=
=
57.1
42.9
72.7
27.3
= 1.33
= 2.66
2.66 g O/g C in II 2
=
1.33 g O/g C in I 1
2

Finally in 1808, John Dalton presented the atomic theory of matter.
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.

Dalton’s Atomic Theory explained the different mass laws that were established before.
Mass conservation
Atoms cannot be created or destroyed postulate 1 or converted into other types of atoms.
postulate 2
Since every atom has a fixed mass, postulate 3 during a chemical reaction atoms are combined differently and therefore there is no mass change overall.

Dalton’s Atomic Theory explained the different mass laws that were established before.
Definite composition
Atoms are combined in compounds in specific ratios and each atom has a specific mass.
postulate 3 postulate 4
So each element has a fixed fraction of the total mass in a compound.

Dalton’s Atomic Theory explained the different mass laws that were established before.
Multiple proportions
Atoms of an element have the same mass postulate 3 and atoms are indivisible.
postulate 1
So when different numbers of atoms of elements combine, they must do so in ratios of small, whole numbers.

In the 19 th century Cathode rays were studied to learn more about electricity.
Phosphor coated plate detects position of the
CATHODE RAY
Anode
+
Cathode
-
Evacuated tube
Power Supply

In the 19 th century Cathode rays were studied to learn more about electricity.
N
S
Anode
+
Cathode
-

In the 19 th century Cathode rays were studied to learn more about electricity.
+
Anode
+
Cathode
-
-

In the 19 th century Cathode rays were studied to learn more about electricity.
Anode
+
Cathode
-

In the 19 th century Cathode rays were studied to learn more about electricity.
1. Ray bends in magnetic field.
2. Ray bends towards positive plate in electric field.
consists of charged particles
3. Ray is identical for any cathode.
consists of negative particles particles found in all matter

In 1897, J. J. Thompson measured the mass-to-charge ratio of this cathode ray particle…
MASS OF PARTICLE <<<<<< MASS OF HYDROGEN



In 1909, Robert Millikan measured the charge and mass of this small particle (later called electrons )

In 1909, Robert Millikan measured the charge and mass of this small particle (later called electrons ) mass of electron = mass charge
X determined by J.J.
Thomson and others charge
= (-5.686x10
-12 kg/C) X (-1.602x10
-19 C)
= 9.109x10
-31 kg = 9.109x10
-28 g

If matter is electrically neutral, atoms must have a positively charged part to counter act the electrons…

Before the turn of the century, Becquerel discovered radioactivity, (a term which was coined by Marie Sklodowska) the spontaneous emission of radiation from unstable elements.
• MARIE SKLODOWSKA married PIERRE CURIE, a
French Physicist and discover radioactive polonium and radium
• 1903 Nobel in Physics
(Becquerel, Curie, Curie)
• Marie Curie: 2 nd Nobel prize in 1911

Ernest Rutherford named the radioactive emissions as alpha (+), beta (-) and gamma

In 1910, Rutherford (and his colleagues Hans Geiger and Ernest
Marsden) discovered that the positive part of the atom is not dispersed but rather concentrated at the center.
1. atoms positive charge is concentrated in the nucleus
2. proton (p) has opposite (+) charge of electron (-)
3. mass of p is 1840 x mass of e (1.67 x 10 -24 g)

In 1910, Rutherford (and his colleagues Hans Geiger and Ernest
Marsden) discovered that the positive part of the atom is not dispersed but rather concentrated at the center.
 RUTHERFORD: The smallest positive-ray particle is the unit of positive charge in the nucleus. This is the PROTON , with a charge equal in magnitude to a an electron, and nearly the same mass as a hydrogen atom

In 1932, James Chadwick measured the mass of He vs. mass of H.
JAMES
CHADWICK and the
NEUTRON
H atoms - 1 p; He atoms - 2 p mass He/mass H should = 2 measured mass He/mass H = 4 neutron (n) is neutral (charge = 0) n mass ~ p mass = 1.67 x 10 -24 g

Revisiting the model: atomic radius ~ 100 pm = 1 x 10 -10 m nuclear radius ~ 5 x 10 -3 pm = 5 x 10 -15 m
If the atom were the blue eagle gym, the nucleus would be a microscopic speck of dust in the center, but weighing millions of tons.

Table 2.2
Properties of the Three Key Subatomic Particles
Name (Symbol)
Charge
Relative Absolute (C)*
Mass
Relative (amu) † Absolute (g)
Location in the Atom
Proton (p + ) 1+ +1.60218 x 10 -19 1.00727
1.67262 x 10 -24
Nucleus
Neutron (n 0 )
Electron (e )
0 0
1-1.60218 x 10 -19
1.00866
1.67493 x 10 -24
Nucleus
0.00054858
9.10939 x 10 -28
Outside
Nucleus
* The coulomb (C) is the SI unit of charge.
† The atomic mass unit (amu) equals 1.66054 x 10 -24 g.

Atomic Symbols, Isotopes, Numbers
A
Z
The Symbol of the Atom or Isotope
X = atomic symbol of the element
A = mass number; A = Z + N
Z = atomic number
(the number of protons in the nucleus)
N = number of neutrons in the nucleus
Isotopes = atoms of an element with the same number of protons, but a different number of neutrons
Figure 2.8

Sample Problem 2.4
Determining the Number of Subatomic Particles in the Isotopes of an Element
PROBLEM: Silicon (Si) is essential to the computer industry as a major component of semiconductor chips. It has three naturally occurring isotopes: 28 Si, 29 Si, and 30 Si. Determine the number of protons, neutrons, and electrons in each silicon isotope.
PLAN: We have to use the atomic number and atomic masses.
SOLUTION: The atomic number of silicon is 14. Therefore,
28 Si has 14p
+
, 14e
and 14n 0 (28 - 14)
29 Si has 14p
+
, 14e
and 15n 0 (29 - 14)
30 Si has 14p
+
, 14e
and 16n 0 (30 - 14)

Sample Problem 2.5
Calculating the Atomic Mass of an Element
PROBLEM: Silver (Ag: Z = 47) has 46 known isotopes, but only two occur naturally, 107 Ag and 109 Ag. Given the following mass spectrometric data, calculate the atomic mass of Ag:
Mass (amu) Abundance (%) Isotope
107 Ag
109 Ag
106.90509
108.90476
51.84
48.16
PLAN: We have to find the weighted average of the isotopic masses, so we multiply each isotopic mass by its fractional abundance and then sum those isotopic portions.

Sample Problem 2.5
continued
Calculating the Atomic Mass of an Element
SOLUTION: mass portion from 107 Ag = 106.90509 amu x 0.5184 = 55.42 amu mass portion from 109 Ag = 108.90476 amu x 0.4816 = 52.45 amu atomic mass of Ag = 55.42 amu + 52.45 amu = 107.87 amu

The Modern Reassessment of the Atomic Theory
1. All matter is composed of atoms.
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.
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. 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 instruments became sophisticated, more and more elements were discovered. There came a need to organize elements.

Electrons of atoms interact to form chemical bonds 
COMPOUNDS. Electrons of atoms can interact in two ways:
1.
An atom can transfer an electron to another atom forming ions .
Oppositely charged ions are attracted to each other forming
Ionic Compounds
Na +
Cl -
Cations
• positive charges
• Usually form from metals that lose electrons
Anions
• Negative charges
• Usually form from nonmetals that gain electrons

Electrons of atoms interact to form chemical bonds 
COMPOUNDS. Electrons of atoms can interact in two ways:
The resulting ions aggregate
(due to electrostatic forces of attraction) and form a regular pattern of stable solids.

Electrons of atoms interact to form chemical bonds 
COMPOUNDS. Electrons of atoms can interact in two ways:
How many electrons are lost or gained?

Electrons of atoms interact to form chemical bonds 
COMPOUNDS. Electrons of atoms can interact in two ways:
What affects attraction between ions?


2

Sample Problem 2.6
Predicting the Ion an Element Forms
PROBLEM: What monatomic ions do the following elements form?
(a) Iodine (Z = 53) (b) Calcium (Z = 20) (c) Aluminum (Z = 13)
PLAN: Use Z to find the element. Find its relationship to the nearest noble gas. Elements occurring before the noble gas gain electrons and elements following lose electrons.
SOLUTION: I
-
Iodine is a nonmetal in Group 7A(17). It gains one electron to have the same number of electrons as
54
Xe.
Ca 2+ Calcium is a metal in Group 2A(2). It loses two electrons to have the same number of electrons as
18
Ar.
Al 3+ Aluminum is a metal in Group 3A(13). It loses three electrons to have the same number of electrons as
10
Ne.

Electrons of atoms interact to form chemical bonds 
COMPOUNDS. Electrons of atoms can interact in two ways:
1.
An atom can transfer an electron to another atom forming ions . Oppositely charged ions are attracted to each other forming Ionic Compounds
2.
An atom can also share its electron with another atom to form Covalent
Compounds

Electrons of atoms interact to form chemical bonds 
COMPOUNDS. Electrons of atoms can interact in two ways:
Some elements occur as diatomic (or polyatomic) molecules

Electrons of atoms interact to form chemical bonds 
COMPOUNDS. Electrons of atoms can interact in two ways:
Compounds arise when two different atoms share electrons
Polyatomic ions are made up of covalently bonded atoms

A chemical formula shows the type and number of each atom present in the smallest unit of the substance.
1.
The empirical formula shows the relative number of atoms of each element in the compound.
2.
The molecular formula shows the actual number of atoms of each element in the compound
3.
A structural formula shows the number of atoms and the bonds between them; that is, the relative placement and connectinos of atoms in the molecule.

NAMING IONIC COMPOUNDS
1.
All ionic compounds are named by giving the cation first and then the anion
2.
For Cations, Ion charge is usually equal group number (there are many exceptions). For Anions, ion charge = group number – 8.
3.
Cations usually end with –ium , while anions are named by taking the root name and changing the end with -ide

NAMING IONIC COMPOUNDS
PROBLEM: Name the ionic compound formed from the following pairs of elements:
(a) Magnesium and nitrogen (b) Iodine and cadmium
(c) Strontium and fluorine (d) Sulfur and cesium
PLAN:
SOLUTION:
Use the periodic table to decide which element is the metal and which is the nonmetal. The metal (cation) is named first and we use the -ide suffix on the nonmetal name root.
(a) Magnesium nitride
(b) Cadmium iodide
(c) Strontium fluoride
(d) Cesium sulfide

NAMING IONIC COMPOUNDS
Ionic compounds are formed by an array of alternation cations and anions. There is no sodium chloride “molecule” to speak of. Instead we use formula unit -> similar to an empirical formula
(a) Magnesium nitride
(b) Cadmium iodide
(c) Strontium fluoride
(d) Cesium sulfide
Mg 2+ and N
Ca 2+ and I -
Na + and F -
Cs + and S 2-
EXCEPTIONS: PEROXIDES and MERCURY (I) ions
3-

NAMING IONIC COMPOUNDS
Some metals form more than one ion (usually transition metals)

NAMING IONIC COMPOUNDS
1.
Some ions contain more than one atom
(polyatomic)
2.
There are some naming conventions for oxoanions (an ion with a nonmetal bonded to one or more oxygens).

Sample Problem 2.9
Determining Names and Formulas of Ionic
Compounds of Elements That Form More Than One
Ion
PROBLEM: Give the systematic names for the formulas or the formulas for the names of the following compounds:
PLAN:
SOLUTION:
(a) tin(II) fluoride (b) CrI
3
(d) CoS
(c) ferric oxide
Compounds are neutral. We find the smallest number of each ion which will produce a neutral formula. Use subscripts to the right of the element symbol.
(a) Tin(II) is Sn 2+ ; fluoride is F ; so the formula is SnF
2
.
(b) The anion I is iodide ( I
); 3I chromium(III) iodide.
means that chromium is Cr 3+ . CrI
3 is
(c) Ferric is a common name for Fe 3+ ; oxide is O 2, therefore the formula is Fe
2
O
3
.
(d) Co is cobalt; the anion S is sulfide (2-); the compound is cobalt(II) sulfide.

NAMING IONIC COMPOUNDS
Some ionic compounds have water associated to them and are called hydrates
MgSO
4
* 7H
2
O
CuSO
4
* 5H
2
O

Sample Problem 2.10
Determining Names and Formulas of Ionic
Compounds Containing Polyatomic Ions
PROBLEM: Give the systematic names for the formula or the formulas for the names of the following compounds:
PLAN:
(a) Fe(ClO
4
)
2
(b) Sodium sulfite (c) Ba(OH)
2
8H
2
Note that polyatomic ions have an overall charge so when writing a formula with more than one polyatomic unit, place the ion in a set of parentheses.
O
SOLUTION: (a) ClO
4
is perchlorate; iron must have a 2+ charge. This is iron(II) perchlorate.
(b) Sodium is Na + ; the anion sulfite is SO
3
2per sulfite. The formula is Na
2
SO
3
.
. You need 2 sodium ions
(c) Barium is a 2+ ion while the hydroxide is OH . When water is included in the formula, we use the term “hydrate” and a prefix which indicates the number of waters. So the name is barium hydroxide octahydrate.

Sample Problem 2.11
Recognizing Incorrect Names and Formulas of Ionic
Compounds
PROBLEM: Something is wrong with the second part of each statement.
Provide the correct name or formula.
SOLUTION:
(a) Ba(C
2
H
3
O
2
)
2 is called barium diacetate.
(b) Sodium sulfide has the formula (Na)
2
SO
3
.
(c) Iron(II) sulfate has the formula Fe
2
(SO
4
)
3
.
(d) Cesium carbonate has the formula Cs
2
(CO
3
).
(a) Barium is always a 2+ ion and acetate is 1-. The “di-” is unnecessary. The correct name is barium acetate.
(b) An ion of a single element does not need parentheses. Sulfide is
S 2, not SO
3
2. The correct formula is Na
2
S.
(c) Since sulfate has a 2- charge, only 1 Fe 2+ is needed. The formula should be FeSO
4
.
(d) Since only one carbonate is needed in this formula, the parentheses are unnecessary. The correct formula is Cs
2
CO
3
.

NAMING ACIDS
1) Binary acid 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.
is
The oxoanion prefixes “ hypo” and “ per” are retained. Thus, BrO
4
perbromate , and HBrO
4 is perbromic acid; IO
2
is iodite , and HIO
2 is iodous acid.

Sample Problem 2.12
Determining Names and Formulas of Anions and
Acids
PROBLEM: Name the following anions and give the names and formulas of the acids derived from them:
(a) Br (b) IO
3
(c) CN (d) SO
4
2(e) NO
2
-
SOLUTION:
(a) The anion is bromide; the acid is hydrobromic acid, HBr.
(b) The anion is iodate; the acid is iodic acid, HIO
3
.
(c) The anion is cyanide; the acid is hydrocyanic acid, HCN.
(d) The anion is sulfate; the acid is sulfuric acid, H
2
SO
4
.
(e) The anion is nitrite; the acid is nitrous acid, HNO
2
.

NAMING BINARY COVELENT (nonmetal – nonmetal)
COMPOUNDS
1) The element with lower group number is the first word in the name
(EXCEPTION: When compounds have both oxygen and halogen, halogen is named first)
2) IF both elements are of the same group, higher period is named first
3) Use of prefixes to indicate number of atoms

Sample Problem 2.13
Determining Names and Formulas of Binary
Covalent Compounds
PROBLEM: (a) What is the formula of carbon disulfide?
(b) What is the name of PCl
5
?
(c) Give the name and formula of the compound whose molecules each consist of two N atoms and four O atoms.
SOLUTION: (a) Carbon is C, sulfide is sulfur S and di- means 2. The formula is
CS
2
.
(b) P is phosphorous, Cl is chloride, the prefix for 5 is penta-. The name of PCl
5 is phosphorous pentachloride.
(c) N is nitrogen and is in a lower group number than O (oxygen).
Therefore the formula is N
2
O
4
. Its name is dinitrogen tetraoxide.

Sample Problem 2.14
PROBLEM:
Recognizing Incorrect Names and Formulas of
Binary Covalent Compounds
Explain what is wrong with the name of formula in the second part of each statement and correct it:
(a) SF
4 is monosulfur pentafluoride.
(b) Dichlorine heptaoxide is Cl
2
O
6
.
(c) N
2
O
3 is dinitrotrioxide.
SOLUTION: (a) The prefix mono- is not needed for one atom of the first element; the prefix for four is tetra-. So the name is sulfur tetrafluoride.
(b) Hepta- means 7; the formula should be Cl
2
O
7
.
(c) The first element is given its elemental name so N
2
O
3 trioxide.
is dinitrogen

MOLECULAR MASS
1) The molecular mass is the sum of all the atomic masses in a given compound
2) For ionic compounds, the molecular mass is calculated as the
“formula” mass – use of the lowest integer relationships between each component.
CALCULATE THE MOLECULAR WEIGHT OF THE FOLLOWING:
(a) Tetraphosphorous trisulfide (b) Ammonium nitrate
MOLECULAR MODELS

Sample Problem 2.16
PROBLEM:
Using Molecular Depictions to Determine Formula,
Name, and Mass
Each circle contains a representation of a binary compound.
Determine its formula, name, and molecular (formula) mass.

MIXTURES are combinations of two or more elements or compounds that can be separated by physical means
Physically mixed, therefore can be separated by physical means; in this case by a magnet.
Allowed to react chemically, therefore cannot be separated by physical means.

MIXTURES are combinations of two or more elements or compounds that can be separated by physical means
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.