Chapter 2
The Components of Matter
1. Development of atomic structure
2. Atomic number and atomic mass
3. Isotopes
4. Atomic weight
5. Atoms and moles
6. The periodic table
2.7 & 2.8 is not required.
General features of the atom today.
•The atom is an electrically neutral, spherical entity.
•The atom is composed of nucleus and electrons.
•The atomic nucleus consists of protons and neutrons.
More than one electron
surround the necleus.
Electrons occupy various
orbitals.
Nucleus (red-dot) is positively charged.
Electron cloud is negatively charged.
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The Modern Reassessment of the Atomic Theory
• All matter is composed of atoms. The atom is the smallest body that
retains the unique identity of the element.
• 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.
• 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.
• Compounds are formed by the chemical combination of two or more
elements in specific ratios.
Properties of the Three Key Subatomic Particles
Charge
Name(Symbol)
Mass
Relative Absolute(C)* Relative(amu)†
Absolute(g)
Location
in the Atom
1+ +1.60218×10-19
1.00727
1.67262 × 10-24
Nucleus
Neutron (n0)
0
0
1.00866
1.67493 × 10-24
Nucleus
Electron (e-)
1-
-1.60218×10-19
0.00054858 9.10939 × 10-28
Outside
Nucleus
Proton (p+)
* The coulomb (C) is the SI unit of charge.
†
The atomic mass unit (amu) equals 1.66054 × 10-24 g.
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Atomic Symbols
A
X
Z
The Symbol of the
Atom or Isotope
12
C
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The Symbol of
carbon Atom
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
A
X
Z
The Symbol of the Atom or Isotope
Isotope = atoms of an element with the same
number of protons, but a different number
of neutrons
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Summary of Mass Terminology
Definition
Unit
Isotopic mass
Mass of an isotope of an element
amu
Atomic mass
Average of the masses of the naturally
occurring isotopes of an element weighted
according to their abundance
amu
Molecular
(or formula) mass
(also called
molecular weight)
Sum of the atomic masses of the atoms (or
ions) in a molecule (or formula unit)
amu
Molar mass (M)
(also called grammolecular weight)
Mass of 1 mole of chemical entities (atoms,
ions, molecules, formula units)
g/mol
Term
(also called
atomic weight)
Sample Problem
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 isoltopes:
28Si, 29Si, and 30Si. 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
Isotope
Proton
Electron
28Si
14p+
14e-
14n0 (28-14)
29Si
14p+
14e-
15n0 (29-14)
14p+
14e-
16n0 (30-14)
30Si
Neutron
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Sample Problem
Calculating the Atomic Mass of an Element
Silver (Ag: Z = 47) has 46 known isotopes, but only two occur naturally,
107Ag and 109Ag. Given the following mass spectrometric data, calculate the
atomic mass of Ag:
PROBLEM:
Isotope
PLAN:
Mass (amu)
Abundance (%)
107Ag
106.90509
51.84
109Ag
108.90476
48.16
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.
SOLUTION:
mass(g) of each
isotope
portion of atomic mass
from each isotope
multiply by fractional
abundance of each isotope
atomic mass
add isotopic portions
mass portion from 107Ag =
106.90509amu × 0.5184 = 55.42amu
mass portion from 109Ag = 108.90476amu × 0.4816 = 52.45amu
atomic mass of Ag = 55.42amu + 52.45amu = 107.87amu
Mole (mol) - the amount of a substance that contains the
same number of entities as there are atoms in exactly 12 g of
carbon-12.
This amount is 6.022×1023. The number is called Avogadro’s
number and is abbreviated as N.
One mole (1 mol) contains 6.022 × 1023 entities (to four
significant figures)
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Summary of Mass Terminology
Definition
Unit
Isotopic mass
Mass of an isotope of an element
amu
Atomic mass
Average of the masses of the naturally
occurring isotopes of an element weighted
according to their abundance
amu
Molecular
(or formula) mass
(also called
molecular weight)
Sum of the atomic masses of the atoms (or
ions) in a molecule (or formula unit)
amu
Molar mass (M)
(also called grammolecular weight)
Mass of 1 mole of chemical entities (atoms,
ions, molecules, formula units)
g/mol
Term
(also called
atomic weight)
Sample Problem
Calculating the Mass and the Number of Atoms in a Given
Number of Moles of an Element
PROBLEM: (a) Silver (Ag) is used in jewelry and tableware but no longer in U.S. coins.
How many grams of Ag are in 0.0342 mol of Ag?
(b) Iron (Fe), the main component of steel, is the most important metal in
industrial society. How many Fe atoms are in 95.8 g of Fe?
PLAN:
(a) To convert mol of Ag to g we have to use
the #g Ag/mol Ag, the molar mass M.
SOLUTION 0.0342mol Ag x 107.9 g Ag = 3.69g Ag
:
mol Ag
PLAN: (b) To convert g of Fe to atoms we first have to
find the #mols of Fe and then convert mols to
atoms.
mol Fe
SOLUTION 95.8g Fe x
= 1.72mol Fe
55.85g Fe
:
1.72mol Fe x
6.022x1023atoms Fe
mol Fe
= 1.04x1024 atoms
Fe
amount(mol) of Ag
multiply by M of Ag
(107.9g/mol)
mass(g) of Ag
mass(g) of Fe
divide by M of Fe
(55.85g/mol)
amount(mol) of Fe
multiply by 6.022x1023
atoms/mol
atoms of Fe
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Sample Problem
Calculating the Moles and Number of Formula Units in
a Given Mass of a Compound
Ammonium carbonate is white solid that decomposes with warming.
PROBLEM: Among its many uses, it is a component of baking powder, first
extinguishers, and smelling salts.
How many formula unit are in 41.6 g of ammonium carbonate?
PLAN: After writing the formula for the
compound, we find its M by adding the masses
of the elements. Convert the given mass, 41.6 g
to mols using M and then the mols to formula
units with Avogadro’s number.
SOLUTION:
The formula is (NH4)2CO3.
mass(g) of (NH4)2CO3
divide by M
amount(mol) of (NH4)2CO3
multiply by 6.022x1023
formula units/mol
number of (NH4)2CO3 formula units
M = (2 x 14.01 g/mol N)+(8 x 1.008 g/mol H)
+(12.01 g/mol C)+(3 x 16.00 g/mol O) = 96.09 g/mol
41.6 g (NH4)2CO3 x
mol (NH4)2CO3
x
6.022x1023 formula units (NH4)2CO3
96.09 g (NH4)2CO3
=
mol (NH4)2CO3
2.61x1023 formula units (NH4)2CO3
Interconverting Moles, Mass, and Number of Chemical Entities
no. of grams
g
Mass (g) = no. of moles x
1 mol
No. of moles = mass (g) x
1 mol
M
no. of grams
No. of entities = no. of moles x
6.022x1023 entities
1 mol
1 mol
No. of moles = no. of entities x
6.022x1023 entities
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The modern periodic table.
In the periodic table, the horizontal rows are periods, the columns are
groups/families.
All elements in a group have similar chemical and physical properties.
• Representative elements: Group A
elements
• Transition elements: Groups 3-12
• Inner transition elements:
lanthanide and actinide
• Metallic elements: left of the
stairstep line
• Nonmetallic elements: right of the
stairstep line
• Metalloids: adjacent to the
stairstep line (not Al)
•
•
•
•
•
•
•
•
Alkali metals: 1A
Alkaline earth metals: 2A
Chalcogens: 6A
Halogens: 7A
Noble gases: 8A
Rare earth metals: lanthanide
Coinage metals: Cu, Ag, Au
Noble metals: Ru, Os, Rh, Ir,
Pd, Pt, Ag, Au, Hg
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Metals, metalloids, and nonmetals.
Cadmium
Copper
Lead
Chromium
Bismuth
Arsenic
Silicon
Antimony
Chlorine
Bromine
Sulfur
Iodine
Boron
Tellurium
Carbon
(graphite)
Take home message
Tools of the Laboratory
The Mass Spectrometer and Its Data
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Take home message
Laboratory preparation of molar solutions.
A
•Weigh the solid needed.
•Transfer the solid to a
volumetric flask that contains
about half the final volume of
solvent.
C Add solvent until the solution
reaches its final volume.
B Dissolve the solid
thoroughly by swirling.
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