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Isotopes
Isotopes
Na
23
11
sodium-23
Mass number
Atomic number
Isotopes
Isotopes: Atoms with the same number of protons,
but different numbers of neutrons.
Isotopes are atoms of the same element (same
atomic number) with different mass numbers
Isotopes of chlorine
35Cl 37Cl
17 17
chlorine 35 chlorine 37
Isotopes
Example: Carbon isotopes
12 13 14
C C C
6 6 6
98.9% 1.1% >0.1%
Isotopes
Atomic mass
-Listed on the periodic table
-Gives the mass of “average” atom of each element
compared to 12C
-Average atomic mass based on all the isotopes and
their abundance %
-Atomic mass is not a whole number…it’s a weighted
average
Na
22.99
Isotopes
Atomic mass calculations:
Gallium is a metallic element found in small lasers
used in compact disc players. In a sample of
gallium, there is 60.2% of gallium-69 (68.9 u)
atoms and 39.8% of gallium-71 (70.9 u) atoms.
What is the atomic mass of gallium?
Mass1 x Abundance1 + Mass2 x Abundance2 = Average atomic mass
69.7u = Average atomic mass
68.9u x 60.2% + 70.9u x 39.8% = Average atomic mass
Isotopes
Atomic mass calculations analogy:
A chemistry student’s grade is weighted. Tests
are worth 50.%, labs are 25%, and homework is
worth 25%. A student's test average is 85.0%, lab
average is 77.0%, and homework is 91.0%.
What is the student’s average?
Isotopes
Atomic mass calculations:
A sample of boron consists of 10B (mass 10.0 u) and
11B (mass 11.0 u). If the average atomic mass of B
is 10.8 u, what is the % abundance of each boron
isotope?
Isotopes
A sample of boron consists of 10B (mass 10.0 u) and 11B (mass 11.0 u). If the average
atomic mass of B is 10.8 u, what is the % abundance of each boron isotope?
Assign X and Y values:
X = % 10B Y = % 11B
Determine Y in terms of X
X + Y = 1 therefore Y = 1 - X
Solve for X:
X (10.0) + (1 - X )(11.0) = 10.8
10.0X + 11.0 – 11.0X = 10.8
10.0X – 11.0X = 10.8 –11.0
- 1.0X = - 0.2
X = 0.2
X = 20%
Y = 80%
.: the % abundances of 10B and 11B are 20% and 80%, respectively

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13 isotopes

  • 3. Isotopes Isotopes: Atoms with the same number of protons, but different numbers of neutrons. Isotopes are atoms of the same element (same atomic number) with different mass numbers Isotopes of chlorine 35Cl 37Cl 17 17 chlorine 35 chlorine 37
  • 4. Isotopes Example: Carbon isotopes 12 13 14 C C C 6 6 6 98.9% 1.1% >0.1%
  • 5. Isotopes Atomic mass -Listed on the periodic table -Gives the mass of “average” atom of each element compared to 12C -Average atomic mass based on all the isotopes and their abundance % -Atomic mass is not a whole number…it’s a weighted average Na 22.99
  • 6. Isotopes Atomic mass calculations: Gallium is a metallic element found in small lasers used in compact disc players. In a sample of gallium, there is 60.2% of gallium-69 (68.9 u) atoms and 39.8% of gallium-71 (70.9 u) atoms. What is the atomic mass of gallium? Mass1 x Abundance1 + Mass2 x Abundance2 = Average atomic mass 69.7u = Average atomic mass 68.9u x 60.2% + 70.9u x 39.8% = Average atomic mass
  • 7. Isotopes Atomic mass calculations analogy: A chemistry student’s grade is weighted. Tests are worth 50.%, labs are 25%, and homework is worth 25%. A student's test average is 85.0%, lab average is 77.0%, and homework is 91.0%. What is the student’s average?
  • 8. Isotopes Atomic mass calculations: A sample of boron consists of 10B (mass 10.0 u) and 11B (mass 11.0 u). If the average atomic mass of B is 10.8 u, what is the % abundance of each boron isotope?
  • 9. Isotopes A sample of boron consists of 10B (mass 10.0 u) and 11B (mass 11.0 u). If the average atomic mass of B is 10.8 u, what is the % abundance of each boron isotope? Assign X and Y values: X = % 10B Y = % 11B Determine Y in terms of X X + Y = 1 therefore Y = 1 - X Solve for X: X (10.0) + (1 - X )(11.0) = 10.8 10.0X + 11.0 – 11.0X = 10.8 10.0X – 11.0X = 10.8 –11.0 - 1.0X = - 0.2 X = 0.2 X = 20% Y = 80% .: the % abundances of 10B and 11B are 20% and 80%, respectively