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How Atoms Differ
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a. Properties of Subatomic Particles
Symbol Location Relative Charge Relative mass Actual mass (g) Electron Proton Neutron outside the nucleus e- 1 1840 9.10 x g -1 1.673 x g p+ in the nucleus +1 1 1 1.675 x g n0 in the nucleus
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Elements on the Periodic Table
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X b. Atomic Number c. Mass Number Mass number A Z atomic number
the number of protons in an atom Identifies element c. Mass Number represents the total number of protons and neutrons in the nucleus A Z atomic number X Mass number
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d. Isotopes Atoms that have the same number of protons but have a different masses Ex: 3 isotopes of carbon: 12 6 C 13 6 C 14 6 C
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e. Average Atomic Mass … =
the weighted average of the isotopes of that element. Formula: Atomic mass of an element = ( % abundance of Isotope #1 x mass Isotope #1 ) + Isotope #2 …
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Average Atomic Mass The mass of an atom is so small it is difficult to work with, so chemists have developed an atomic standard to compare all the masses The standard is the atomic mass unit (amu) If the mass of an element is not close to a whole number, it is because the atom has several isotopes The atomic mass is the weighted average of the isotopes of that element
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Example 1 Silver has two naturally occurring isotopes. Ag-107 has an abundance of 51.82% and mass of amu. Ag-109 has a relative abundance of 48.18% and a mass of amu. Calculate the average atomic mass of silver.
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Example 1 Silver has two naturally occurring isotopes. Ag-107 has an abundance of 51.82% and mass of amu. Ag-109 has a relative abundance of 48.18% and a mass of amu. Calculate the atomic mass of silver. .5182(106.9 amu) (108.9 amu) (remember to round at the end with more than one operation) **Round to the hundredths for amu values = amu
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Example 2 Rubidium is a soft, silvery-white metal that has two common isotopes, Rb and Rb. If the abundance of 85Rb is 72.2% with amu and the abundance of 87Rb is 27.8% with amu, what is the average atomic mass of rubidium? 85 37 87 37
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Example 2 Rubidium is a soft, silvery-white metal that has two common isotopes, Rb and Rb. If the abundance of 85Rb is 72.2% with amu and the abundance of 87Rb is 27.8% with amu, what is the average atomic mass of rubidium? 85 37 87 37 .722( amu) ( amu) (remember to round at the end with more than one operation) **Round to the hundredths for amu values = amu ≈ amu
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Honors Example 3 Boron has two naturally occurring isotopes. If the abundance of 11B is 80.10% with an amu of , find the abundance of 10B.
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Honors Example 3 Boron has two naturally occurring isotopes. If the abundance of 11B is 80.10% with an amu of , find the amu of 10B. Hint: find the abundance first. 0.801( amu) (X amu) = amu amu (X amu) = amu (remember to round at the end with more than one operation) 0.199 X amu = amu = amu for 10B ≈ amu for 10B **Round to the hundredths for amu values
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