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# 4.7 Atomic Mass Even the largest atoms have very small masses (Fluorine – 3.155 x 10 -23 ) Even the largest atoms have very small masses (Fluorine – 3.155.

## Presentation on theme: "4.7 Atomic Mass Even the largest atoms have very small masses (Fluorine – 3.155 x 10 -23 ) Even the largest atoms have very small masses (Fluorine – 3.155."— Presentation transcript:

4.7 Atomic Mass Even the largest atoms have very small masses (Fluorine – 3.155 x 10 -23 ) Even the largest atoms have very small masses (Fluorine – 3.155 x 10 -23 ) Masses of single atoms can be determined using mass spectrometers, but these masses would be inconvenient to work with. Masses of single atoms can be determined using mass spectrometers, but these masses would be inconvenient to work with. It is more useful to compare the relative masses of atoms using an isotope of carbon, carbon-12, as a basis. It is more useful to compare the relative masses of atoms using an isotope of carbon, carbon-12, as a basis.

4.7 Atomic Mass Carbon-12 was assigned a mass of exactly 12.00000 amu Carbon-12 was assigned a mass of exactly 12.00000 amu atomic mass unit (amu) – one-twelfth the mass of a carbon-12 atom Since carbon-12 has a mass number of 12 (6 protons and 6 neutrons), the mass of a single proton or neutron is approximately 1 amu. Since carbon-12 has a mass number of 12 (6 protons and 6 neutrons), the mass of a single proton or neutron is approximately 1 amu.

4.7 Atomic Mass Since most of the mass of an atom depends on the number of protons and neutrons in the nucleus you may expect the atomic mass to be a whole number – this is not the case. Since most of the mass of an atom depends on the number of protons and neutrons in the nucleus you may expect the atomic mass to be a whole number – this is not the case.

4.7 Atomic Mass In nature, most elements occur as a mixture of two or more isotopes In nature, most elements occur as a mixture of two or more isotopes Each isotope has a fixed mass and a natural percent abundance Each isotope has a fixed mass and a natural percent abundance atomic mass – a weighted average mass of the atoms in a naturally occurring sample of the element This reflects both the mass and the relative abundance of the isotopes as they occur in nature This reflects both the mass and the relative abundance of the isotopes as they occur in nature

4.7 Atomic Mass Example: Hydrogen-Atomic mass: 1.0079 amu Example: Hydrogen-Atomic mass: 1.0079 amu Three isotopes: Hydrogen-1, Hydrogen-2, Hydrogen-3 Three isotopes: Hydrogen-1, Hydrogen-2, Hydrogen-3 Hydrogen-1 has a mass of 1.0078 and a natural abundance of 99.98% Hydrogen-1 has a mass of 1.0078 and a natural abundance of 99.98% Hydrogen-2 and hydrogen-3 occur in trace amounts Hydrogen-2 and hydrogen-3 occur in trace amounts The slight difference in the mass of hydrogen-1 and the average atomic mass takes into account the other two isotopes. The slight difference in the mass of hydrogen-1 and the average atomic mass takes into account the other two isotopes.

4.8 Calculating the Atomic Mass Because the atomic mass must reflect both the masses and the relative natural abundances of the isotopes, you must know: Because the atomic mass must reflect both the masses and the relative natural abundances of the isotopes, you must know: The number of stable isotopes of that element The number of stable isotopes of that element The mass of each isotope The mass of each isotope The natural percent abundance of each isotope The natural percent abundance of each isotope

4.8 Calculating Atomic Mass Example: Example: Element X has two natural isotopes. The isotope with mass 10.012 amu has a relative abundance of 19.91 %. The isotope with mass 11.009 amu has a relative abundance of 80.09%. Calculate the average atomic mass and determine what element it is.

4.8 Calculating Atomic Mass Solution: Solution: Find the mass that each isotope contributes to the weighted average by multiplying the mass by its relative abundance. Then add the products. Find the mass that each isotope contributes to the weighted average by multiplying the mass by its relative abundance. Then add the products. X 10.012 amu x 0.1991 = 1.993 amu X 11.009 amu x 0.8009 = 8.817 amu Total: 10.810 amu Total: 10.810 amu Element X is boron. 10 11

4.8 Concept Practice 20. The element copper contains the naturally occurring isotopes C and C. The relative abundances and atomic masses are 69.2% (mass = 62.93 amu) and 30.8% (mass = 64.93 amu), respectively. Calculate the average atomic mass of copper. (mass = 62.93 amu) and 30.8% (mass = 64.93 amu), respectively. Calculate the average atomic mass of copper. 63 29 65 29

4.8 Concept Practice 20. Solution: 62.93 amu x 0.692 = 43.5 amu 64.93 amu x 0.308 = 20.0 amu Total: 63.5 amu Total: 63.5 amu

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