 # Isotopes  Atoms with the same number of protons but different numbers of neutrons  Ex) Carbon 12 vs. Carbon 14  These atoms have a different mass 

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Isotopes  Atoms with the same number of protons but different numbers of neutrons  Ex) Carbon 12 vs. Carbon 14  These atoms have a different mass  Chemically alike because still have the same number of protons  Atoms with the same number of protons but different numbers of neutrons  Ex) Carbon 12 vs. Carbon 14  These atoms have a different mass  Chemically alike because still have the same number of protons

Isotopes of Hydrogen  Hydrogen -1 simply called hydrogen  Hydrogen - 2 called deuterium  Hydrogen - 3 called tritium  Hydrogen -1 simply called hydrogen  Hydrogen - 2 called deuterium  Hydrogen - 3 called tritium

Development of AMUs  Atomic Mass Units (AMUs)  Protons have a mass of 1 amu  Neutrona have a mass of 1 amu  Electrons have a mass of 0 amu  Atomic Mass Units (AMUs)  Protons have a mass of 1 amu  Neutrona have a mass of 1 amu  Electrons have a mass of 0 amu

Atomic Mass  The weighted average mass of the isotopes in a naturally occurring sample of the element  Don’t confuse with “mass number”  To calculate atomic mass you need 3 pieces of information  1. The number of stable isotopes  2.The mass of each isotope  3.The natural percent abundance of each isotope  The weighted average mass of the isotopes in a naturally occurring sample of the element  Don’t confuse with “mass number”  To calculate atomic mass you need 3 pieces of information  1. The number of stable isotopes  2.The mass of each isotope  3.The natural percent abundance of each isotope

Atomic Mass  Example Problem - Calculate the atomic mass for element X. One isotope has a mass of 10 amus ( 10 X) and is 20% abundant. The other has a mass number of 11 amus ( 11 X) and an abundance of 80%.  To solve: Multiply the mass number times the abundance than add them together.  Example Problem - Calculate the atomic mass for element X. One isotope has a mass of 10 amus ( 10 X) and is 20% abundant. The other has a mass number of 11 amus ( 11 X) and an abundance of 80%.  To solve: Multiply the mass number times the abundance than add them together.

Atomic Mass  10 x 0.20 = 2.0  11 x 0.80 = 8.8  Add 2.0 + 8.8 = 10.8  The atomic mass of element X is 10.8 amus  10 x 0.20 = 2.0  11 x 0.80 = 8.8  Add 2.0 + 8.8 = 10.8  The atomic mass of element X is 10.8 amus

Atomic Mass  Your turn. Solve:  What is the atomic mass of Element Z? The isotopes are 16 Z, 17 Z, 18 Z; with percent abundances of 99.759, 0.037, 0.204.  Your turn. Solve:  What is the atomic mass of Element Z? The isotopes are 16 Z, 17 Z, 18 Z; with percent abundances of 99.759, 0.037, 0.204.

Atomic Mass  Answer  16 x 0.99759 = 15.961  17 x 0.00037 = 0.0063  18 x 0.00204 = 0.0367  15.961 + 0.0063 + 0.0367 = 16.004  Tha atomic mass of element Z is 16.004 amus  Answer  16 x 0.99759 = 15.961  17 x 0.00037 = 0.0063  18 x 0.00204 = 0.0367  15.961 + 0.0063 + 0.0367 = 16.004  Tha atomic mass of element Z is 16.004 amus

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