A.
B.
C.
P rotons – positive charge
N eutrons – no charge
E lectrons – negative charge
1. in constant motion around nucleus of atom; travel in orbitals, or shells

 Atomic number - # of protons in the nucleus of an atom


Atomic mass - # of protons and neutrons in the nucleus
21
Example for Scandium atom
Sc atomic # = # of p +
45.0
= ___
# p + = # e = ___ atomic mass – atomic # = n 0 = ___

• Atomic number - # of protons in the nucleus of an atom
• Atomic mass - # of protons and neutrons in the nucleus
21
• Example for Sc atom
Sc
45.0
atomic # = # of p + = 21
# p + = # e = 21 atomic mass – atomic # = n 0 = 24

 Identity of each element is determined by the number of protons in the nucleus.
 In a neutral atom, the number of electrons equals the number of protons.
 Practice problems pg. 99

A. Atoms of the same element having different atomic masses.
B. Atoms of the same element having different numbers of neutrons.
 Hydrogen has three isotopes:
Protium – 0 neutrons
Deuterium – 1 neutron
Tritium – 2 neutrons

C. Naming
 It is often important to identify one isotope from another. This is done in two ways.
1. Hyphen notation . This gives the element name (which you can use to determine the atomic number) followed by the atomic mass which you can use to determine the number of neutrons.
 Hydrogen – 2

2. Nuclide or nuclear symbol notation
Atomic mass
Atomic number
4
2
He

+
-
0
1. Symbol notation 107 Ag
2. Hyphen notation
Potassium – 39
Potassium – 40
Potassium – 41
Solve practice problems pg. 99,101 #’s11-
14. For #14 add on hyphen notation and symbol notation

3.An average sample of hydrogen in nature contains 99.985% protium,
0.015% deuterium and a miniscule amount of tritium.
 Protium has an atomic mass of
1.007825 amu.
 Deuterium has an atomic mass of
2.014102 amu.

 Multiply the masses of each isotope by its percentage. These are weighted values.
 Add all of these values together.
(Do NOT divide by the number of isotopes. This is unnecessary.)

1.007825 * .99985 = 1.0076738 amu protium
.00015 * 2.014102
= 0.0003021 amu deuterium
1.0076738 + 0.0003021 = 1.00798 amu
 1.00798 amu is a weighted average because most of the hydrogen is protium.




1.00798 amu is a weighted average because most of the hydrogen is protium.
Multiply the masses of each isotope by its percentage. These are weighted values.
Add all of these values together. (Do NOT divide by the number of isotopes. This is unnecessary.)

 The normal way to determine the average mass of something would be to add the two masses and simply divide by two.

 The normal way to determine the average mass of something would be to add the two masses and simply divide by two.
1.007825 + 2.014102 = 3.021927
3.071927 / 2 = 1.5109635 amu

4. What is the atomic mass of silicon if 92.21% of its atoms have mass 27.977u, 4.70% have mass 28.976 u and 3.09% have mass
29.974?
 28.1 u
 .9221 x 27.997 = 25.82 u
 .0470 x 28.976 = 1.36 u
 .0309 x 29.974 = 0.926 u

5. Neon has two isotopes. Neon-20 has a mass of 19.992 u and neon-22 has a mass of
21.991 u. In any sample of 100 neon atoms,
90 will be neon-20 and 10 will be neon-22.
Calculate the average atomic mass of neon.
 20.192 u
 (.90 x 19.992) + (.10 x 21.991)
 17.9928 + 2.1991