Download Isotopes and Average Atomic Mass

Unit 3
Atomic Structure and
Periodicity
Dalton’s Atomic Theory
1. All matter is composed of
ATOMS
_____________.
2. Atoms of the same element are
IDENTICAL
_______________.
Atoms of different
DIFFERENT
elements are _______________.
3. Atoms of different elements can
PHYSICALLY
_______________
mix together or can
_______________
combine with one
CHEMICALLY
another in simple whole-number
COMPOUNDS
ratios to form _______________.
4. Chemical change involves a
REARRANGEMENT
____________________
of atoms.
Charges and relative masses of the
three main subatomic particles.
• PROTON
charge = +1;
mass = 1 amu;
• NEUTRON
no charge;
mass = 1 amu;
• ELECTRON
charge = -1;
mass = 1/1846 amu
IN A NEUTRAL ATOM
• The atomic number is the number of
PROTONS and the number of
__________
ELECTRONS
_____________.
(in a charged atom, # p = #e)
• The mass number is the total number
PLUS
of protons __________
neutrons.
• To find the number of neutrons,
SUBTRACT the atomic number from
__________
the mass number.
An atom is identified as platinum – 195.
(a) What does the number represent?
Mass number
(a) Symbolize this atom
using superscripts
and subscripts.
Mass number = p + n
195
Pt
78
Atomic number = # protons
ISOTOPES
Isotopes of the same element
• identical number of protons
• different number of neutrons, therefore…
different masses and mass numbers
List the number of protons, neutrons, and
electrons in each pair of isotopes.
(1)
Li-6, Li-7
Li-6: 3 p+, 3 e-, 3 nO
Li-7: 3 p+, 3 e-, 4 nO
(2) Ca-42, Ca-44
Ca-42: 20 p+, 20 e-, 22 nO
Ca-44: 20 p+, 20 e-, 24 nO
AVERAGE ATOMIC MASS
• The average atomic mass on the
periodic table = the weighted
average of all the isotopes for that
element
•Each isotope exists in nature in
different abundances (the abundance
of Li – 6 is 7.5%; the abundance of Li – 7 is
92.5%)
to calculate the average
atomic mass of an element:
1. Multiply the abundance (in decimal
form, ex. 92.5% = .925) by the
mass of the isotope
2. repeat step #1 for each isotope
3. Add all calculations together (do
not round)
Example…Cesium has three known isotopes:
Cs – 133, Cs – 132, and Cs – 134.
Their abundances in nature are 75%, 20%,
and 5% respectively.
What is the average atomic mass of cesium?
Steps #1, #2 and #3 can be performed together:
(.75)(133)
99.75
+
(.20)(132)
+
26.4
132.85 amu
+
+
(.05)(134)
6.7
Using the data for nitrogen listed in Table
4, p. 82, calculate the average atomic
mass of nitrogen.
N – 14 and N – 15
Abundances, 99.63% and .37%, respectively
Avg. Atomic Mass =
(.9963)(14) + (0.0037)(15)
13.9482
+
.0555
=
=
14.0037
amu