Transcript Atomic Structure PWPT

Atomic Structure
I. Atoms
The atom is the basic unit of
matter.
 Smallest
particle of an
element that retains the
characteristics of that
element.
Three Basic Subatomic Particles:
proton
Charge:
+1
Mass
Number: 1 amu
neutron
electron
0
-1
1 amu
0 amu
Location: nucleus nucleus
electron
cloud
What is an amu?
atomic mass unit:
 used
to express very small
masses
 based
on the mass of 1/12 of
a carbon-12 atom
1
amu = 1.66054 x 10-24 grams
The atom is divided into two
separate areas:
 Nucleus
 Electron Cloud
1. Nucleus
 Found in center of an atom
 Small and dense
 Contains protons and
neutrons, so is positively
charged

 Contains
most of the mass of
an atom
 The number of protons is
ALWAYS the same .
 The number of neutrons can
vary.
2. Electron Cloud
 Located outside the nucleus
 NOT dense
 Contains electrons, so
negatively charged
 Mass of electrons is
negligible
 Electrons determine reactivity
Periodic Table
 Each
box on the periodic table
provides three important
pieces of information:
1. Symbol
2. Atomic Number
3. Atomic Mass (Mass #)
 Atomic
number:
The number of protons in the
nucleus
The number of protons identifies
the element
 Examples:
atomic # is 6, so 6 p+
 Sulfur: atomic # is 16, so 16 p+
 Carbon:
Because atoms are electrically neutral,
each atom contains the same number
of protons as electrons.
Examples:
2 p+, so 2 eMagnesium: 12 p+, so 12 eIron: 26 p+, so 26 eHelium:

Atomic mass (mass #):
The number of protons plus the
number of neutrons in an atom

Examples:
 Carbon-12 = 6 p+ + 6 n0
 Mercury-201 = 80 p+ + 121 n0
Mass
Number
Atomic
Number
X
Example:
How many neutrons in C?
Mass Number
(# p+ + # no)
Atomic Number
(# p+ )
12
6
C
12 -6 = 6 neutrons
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II. Ions

An ion is a charged particle

# of p+ does not equal the # of e-

The ONLY way to form an ion is to
change # of ELECTRONS

Positive ions (cations): # e- < # p+

Negative ions (anions): # e- > # p+
Example 1:
Lithium
 Neutral atom: 3 p+ and 3 e Loses one e- to form Li+1 ion
 Ion: 3 p+ and 2 e-
Example 2:
Oxygen
 Neutral atom: 8 p+ and 8 e Gains two e- to form O-2 ion
 Ion: 8 p+ and 10 e-
Remember:
Electrons control behavior in
ordinary chemical reactions!
III. Isotopes
 Atoms
of the same element
that differ from each other
by the number of neutrons
they contain
Most
elements
have at least two
naturally occurring
isotopes.
Isotopes of the same element:
1.
2.
3.
Have the same
atomic number
Have different mass
numbers
Have similar behavior
 When
you are working with a
specific isotope, the mass
number on the periodic table
will not be exactly correct
 You need a way to write the
symbol so that it shows the
atomic number and the correct
atomic mass of that particular
isotope.
1. To write an Isotope (Nuclear) Symbol
Mass
Number
Atomic
Number
X
Examples:
How many neutrons in C-12?
Mass Number
(# p+ + # no)
Atomic Number
(# p+ )
12
6
C
12-6 = 6 neutrons
How many neutrons in C-13?
Mass Number
(# p+ + # no)
Atomic Number
(# p+ )
13
6
C
13-6 = 7 neutrons
2. To name an isotope using a
Hyphen Notation Symbol
X – atomic mass of specific isotope

C-12 is carbon with a mass of 12 amu
(6 protons + 6 neutrons)

C-13 is carbon with a mass of 13 amu
(6 protons + 7 neutrons)
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IV. Relative Abundance and
Average Atomic Mass
 The
naturally occurring
isotopes of each element
are present in specific
amounts known as
relative abundance
The
higher the
% Abundance,
the more common
the isotope.
Average Atomic Mass Mass
# on periodic table
 Weighted
average of the mass
and abundance of the naturally
occurring isotopes of an
element
To Calculate Average Atomic
Mass:
1.
2.
Multiply mass by % abundance for each
isotope
Add all answers from step #1 to determine
average atomic mass
Hint: when calculated,
average atomic mass
should equal the Mass
Number from the Periodic
Table.
Example 1:
Calculate the average atomic
mass for the element Boron.
Boron
– 10
19.8%
Boron
– 11
80.2%
 Step
1: Change % to decimals.
19.8% = .198
80.2% = .802
 Step
2: Multiply mass of each
isotope by its abundance.
10 x .198 = x
11 x .802 = y
 Step
3: Add. x + y =
Answer:
10 x .198 = 1.98
11 x .802 = 8.82 +
10.802
To make life easier, we will always
round atomic masses to the nearest
tenth.
10.8 amu
Example 2:
 Two
isotopes of copper occur
in nature.
 69.17% of copper atoms have
a mass of 62.94 amu
 30.83% have a mass of
64.93amu.
 What is the average atomic
mass?
Answer
62.94 x .6917 = 43.54
64.93 x .3083 = 20.02 +
63.56 amu
63.6 amu
Example 3:

Uranium has 3 isotopes with the following
masses and % abundance. Calculate its
average atomic mass.
234 U
92
0.0058%
235 U
92
0.71%
238 U
92
99.23%
 (234
x .000058) + (235 x .0071) +
(238 x .9923) = 237.84
237.8 amu
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