4.7 Atomic Mass4.7 Atomic Mass

Even the largest atoms have very small Even the largest atoms have very small masses (Fluorine – 3.155 x 10masses (Fluorine – 3.155 x 10-23-23) ) Masses of single atoms can be determined Masses of single atoms can be determined

using using mass spectrometersmass spectrometers, but these , but these masses would be inconvenient to work with.masses would be inconvenient to work with.

It is more useful to compare the relative It is more useful to compare the relative masses of atoms using an isotope of masses of atoms using an isotope of carbon, carbon-12, as a basis.carbon, carbon-12, as a basis.

4.7 Atomic Mass4.7 Atomic Mass

Carbon-12 was assigned a mass of Carbon-12 was assigned a mass of exactly 12.00000 amuexactly 12.00000 amu

atomic mass unit (amu)atomic mass unit (amu) – one- – one-twelfth the mass of a carbon-12 twelfth the mass of a carbon-12 atomatom

Since carbon-12 has a mass number Since carbon-12 has a mass number of 12 (6 protons and 6 neutrons), of 12 (6 protons and 6 neutrons), the mass of a single proton or the mass of a single proton or neutron is approximately 1 amu.neutron is approximately 1 amu.

4.7 Atomic Mass4.7 Atomic Mass

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

4.7 Atomic Mass4.7 Atomic Mass

In nature, most elements occur as a In nature, most elements occur as a mixture of two or more isotopesmixture of two or more isotopes Each isotope has a fixed mass and a Each isotope has a fixed mass and a

natural percent abundancenatural percent abundance

atomic massatomic mass – a – a weighted average weighted average mass mass of the atoms in a naturally of the atoms in a naturally occurring sample of the elementoccurring sample of the element This reflects both the mass and the relative This reflects both the mass and the relative

abundance of the isotopes as they occur in abundance of the isotopes as they occur in naturenature

4.7 Atomic Mass4.7 Atomic Mass Example: Hydrogen-Atomic mass: 1.0079 Example: Hydrogen-Atomic mass: 1.0079

amuamu Three isotopes: Hydrogen-1, Hydrogen-2, Three isotopes: Hydrogen-1, Hydrogen-2,

Hydrogen-3Hydrogen-3 Hydrogen-1 has a mass of 1.0078 and a Hydrogen-1 has a mass of 1.0078 and a

natural abundance of 99.98%natural abundance of 99.98% Hydrogen-2 and hydrogen-3 occur in trace Hydrogen-2 and hydrogen-3 occur in trace

amountsamounts The slight difference in the mass of The slight difference in the mass of

hydrogen-1 and the average atomic mass hydrogen-1 and the average atomic mass takes into account the other two isotopes.takes into account the other two isotopes.

4.8 Calculating the 4.8 Calculating the Atomic MassAtomic Mass

Because the atomic mass must Because the atomic mass must reflect both the masses and the reflect both the masses and the relative natural abundances of the relative natural abundances of the isotopes, you must know:isotopes, you must know: The number of stable isotopes of that The number of stable isotopes of that

elementelement The mass of each isotopeThe mass of each isotope The natural percent abundance of each The natural percent abundance of each

isotopeisotope

4.8 Calculating Atomic 4.8 Calculating Atomic MassMass

Example: Example:

Element X has two natural isotopes. Element X has two natural isotopes. The isotope with mass The isotope with mass 10.012 amu10.012 amu has a relative abundance of has a relative abundance of 19.91 %.19.91 %. The isotope with mass The isotope with mass 11.009 amu11.009 amu has a relative abundance of has a relative abundance of 80.09%.80.09%. Calculate the average atomic Calculate the average atomic mass and determine what mass and determine what element it is.element it is.

4.8 Calculating Atomic 4.8 Calculating Atomic MassMass

Solution:Solution: Find the mass that each isotope Find the mass that each isotope

contributes to the weighted average contributes to the weighted average by by multiplying the mass by its relative multiplying the mass by its relative abundanceabundance. Then . Then add the productsadd the products..

X 10.012 amu x 0.1991 = 1.993 amuX 10.012 amu x 0.1991 = 1.993 amu

X 11.009 amu x 0.8009 = X 11.009 amu x 0.8009 = 8.817 amu8.817 amu

Total: 10.810 amuTotal: 10.810 amu

Element X is boron.Element X is boron.

1010

1111

4.8 Concept Practice4.8 Concept Practice

20. The element copper contains the 20. The element copper contains the naturally occurring isotopes C and naturally occurring isotopes C and C. The relative abundances and C. The relative abundances and atomic masses are 69.2% atomic masses are 69.2%

(mass = 62.93 amu) and 30.8% (mass = 62.93 amu) and 30.8% (mass = 64.93 amu), respectively. (mass = 64.93 amu), respectively. Calculate the average atomic mass Calculate the average atomic mass of copper.of copper.

6329

6529

4.8 Concept Practice4.8 Concept Practice

20. Solution:20. Solution:

62.93 amu x 0.692 = 43.5 amu62.93 amu x 0.692 = 43.5 amu

64.93 amu x 0.308 = 64.93 amu x 0.308 = 20.0 amu20.0 amu

Total: 63.5 amuTotal: 63.5 amu