Unit 3:
Atomic
Structure
A. Subatomic Particles
ATOM
ATOM
NUCLEUS
NUCLEUS
PROTONS
PROTONS
POSITIVE
POSITIVE
CHARGE
CHARGE
NEUTRONS
NEUTRONS
NEUTRAL
NEUTRAL
CHARGE
CHARGE
ELECTRON
ELECTRONS
CLOUD
NEGATIVE CHARGE
ELECTRONS
NEGATIVE
CHARGE
Atomic Number
equal in a
equals
the
of...
Most
of#the
atom’s
massatom
neutral
(Mass Number)
A. Subatomic Particles
ATOMIC NUMBER
• # of protons
•Identifies the element
• # of electrons in
neutral atoms
AVERAGE ATOMIC
MASS
• mass of protons +
neutrons
19
K
Potassium
39.0983
B. Mass Number
MASS NUMBER =
protons + neutrons
•round off the average
atomic mass
• always a whole
number
• NOT on the periodic
table!!!!!
19
K
Potassium
39.0983
WHAT IS THE MASS NUMBER FOR POTASSIUM?
39
B. Mass Number
Q: Carbon has 6
protons & 6
neutrons.What
is it’s mass
number?
A: 12
(Carbon-12)
© Addison-Wesley Publishing Company, Inc.
C. Isotopes
 Atoms of the same element with
different mass numbers (This means
they have different # of neutrons.
 Nuclear symbol:
Mass #
Atomic #
56
26
Hyphen notation: Iron-56
Fe
HOW DO THESE ELEMENTS DIFFER?
SAME ELEMENT, BUT
DIFFERENT # OF
NEUTRONS  ISOTOPES
C. Isotopes
 To find # of neutrons:
mass number - atomic number
Mass #
Atomic #
# of Neutrons = 6
12
6
C
Example
 Chlorine-37
• atomic #:
17
• mass #:
37
• # of protons:
17
• # of electrons:
17
• # of neutrons:
20
37
17
Cl
D. Ions
 Atoms with a (+) or (-) charge
due to a gain or loss of
electron(s)
Positive Ion:
loss of eEx: Ca 2+
Negative Ion:
gain of eEx: Cl -
Lost 2 e-
Gained 1 e-
Examples
 Br
–
 Al
Loses 3 e3+ _________
 Mg
Gains 1 e__________
2+ ________
Loses 2 e-
1 e Na + Loses
_________
O
I
2- __________
Gains 2 e-
Gains 1 e- ____________
E. Relative Atomic Mass
 Carbon-12 atom has a mass of
= 1.992 × 10-23 g (very small number)
 So we use atomic mass unit (amu)
 1 amu = 1/12 the mass of a
12C
atom
 1 p+ = 1.007276 amu
1 n = 1.008665 amu
1 e- = 0.0005486 amu
F. Average Atomic Mass
 A weighted average of all isotopes
 Given on the Periodic Table
Avg.
Atomic
Mass
(mass)(% )  (mass )(% )

100
 round to 2 decimal places
F. Average Atomic Mass
 EX: Calculate the avg. atomic mass of
oxygen if its abundance in nature is
99.76% 16O, 0.04% 17O, and 0.20% 18O.
Avg.
(16)(99.76 )  (17)(0.04)  (18)(0.20)
 16.00
Atomic 
100
amu
Mass
F. Average Atomic Mass
 EX: Find chlorine’s average atomic mass
if approximately 8 of every 10 atoms are
chlorine-35 and 2 are chlorine-37.
Avg.
Atomic
Mass
(35)(8)  (37)(2)

 35.40 amu
10