1. Atomic Number, Mass Number, Atomic Mass and Isotopes

2. Atomic Number (Z):
is the number of protons in the nucleus of the atom. Z=#p+
The number of protons (atomic number) determine the identity of an element.

3. What about the electrons? How many?
Atoms have no overall electrical charge (neutral) so, an atom must have as many electrons as there are protons in its nucleus.
# p+ =# e-
The atomic number of an element also equals the number of electrons in a neutral atom of that element
Z =#e-

4. Ex: Sodium 11 11 11 What is the atomic number of Sodium?
How many protons does sodium have?
How many electrons does sodium have? 11 11 11

5. A=#p+ + #n0 Mass Number (A):
The sum of the protons and neutrons in the nucleus.
A=#p+ + #n0

6. Standard atomic Notation

7. Other ways to write elements:36
17Cl Cl-36
Mass Number
Mass Number
Atomic Number

8. Nucleons:
protons and neutrons

9. Isotopes
Atoms of the same element with
Same number of protons
Same number of electrons
Different number of neutrons
They have different mass numbers because they have different numbers of neutrons, but they have the same atomic number because they have the same number of protons.

10. Isotopes Remember: Isotopes have: same # of protons
Do isotopes have similar chemical properties? Why?
Yes – electrons are the same
Do they have similar physical properties? Why?
NO – mass is different.
Remember:
Isotopes have:
same # of protons
different # of neutrons

12. Example: Isotopes of Carbon and Hydrogen
protium deuterium tritium
H H H
Isotopes of Hydrogen
Isotopes of Carbon

13. Atomic Mass Unit
is a unit used to compare the masses of atoms and has the symbol u or amu.

14. is approximately equal to the mass of a single proton or neutron.
1 amu or u
is approximately equal to the mass of a single proton or neutron.

15. Carbon-12
Chemists have defined the carbon-12 atom as having a mass of 12 atomic mass units.
For carbon: Mass = 12.01
99% 12C AND 1% 13C found in nature
12.01 = average mass based on abundance NOT just 12

16. Atomic Mass
is the weighted average mass of all the naturally occurring isotopes of that element.

18. How do we deal with the fact that some atoms have several isotopes?
The solution is to measure the mass of each individual isotope in the element and then calculate the weighted average.
The weighted average is known as the atomic mass of the element

19. MARBLE ACTIVITY

20. Types of calculations
Finding atomic mass, given mass and abundance of each
Finding the mass of one of the isotope given atomic mass and abundance.
Finding abundance given atomic mass and mass of each
NOTE – if atomic mass is not given for 2 and 3 above use the mass in the periodic table.

21. Problem 1- finding atomic mass
An element consist of two isotopes. Isotope A has an abundance of percent, and its mass is atomic mass units. Isotope B has an abundance of percent, and its mass is atomic mass units. What is the atomic mass of the element?
Atomic mass =
(14.00u x ) + ( u x )
= 14.25

22. Problem 2 – finding mass of one isotope
Bromine has two naturally occuring isotopes. Bromine-79 has a mass of amu and is 50.69% abundant. Using the atomic mass reported in the periodic table, determine the mass of bromine-81, the other isotope of bromine.
Atomic mass = (MBr79 x Ra) + (MBr81x Ra)
Ra for Br81 = 100 – = 49.31
79.90(mass from periodic table) = ( x ) + (MBr81 x )
79.90 = Br
79.90 – = Br
= Br
80.91 amu = Br

23. Problem – 3. finding relative abundance
Gallium consists of two naturally occuring isotopes with masses of and amu. The average atomic mass of Ga is amu. Calculate the abundance of each isotope.
Note – if atomic mass is not given use the mass in the periodic table.

24. Answer should be:69Ga: 60.3%, 71Ga: 39.7%
Average atomic mass = (mass 69Ga)(x) + (mass 71Ga)(y) Where x and y are the relative abundances of 69Ga and 71Ga respectively. So, = (x) (y)
Cant have 2 unknowns?????
Let the abundance of 1 isotope = x, then the abundance of the other will be 1-x . In this way the problem will have 1 unknown. (Substitute y for 1-x)
So = (68.926) (x) + (70.925)(1-x)                                     =  68.926x x                                 = x                                           x =
= x 100 = 60.3%