2.5: Isotopes and Average Atomic Mass

Learning Objectives

• Write and interpret symbols that depict the atomic number, mass number, and charge of an atom or ion
• Define the atomic mass unit and average atomic mass
• Calculate average atomic mass and isotopic abundance

Isotopes are atoms of the same element that have the same number of protons but different number of neutrons, therefore they have different masses. The symbol for a specific isotope of any element is written by placing the mass number as a superscript to the left of the element symbol (Figure \(\PageIndex{4}\)). The atomic number is sometimes written as a subscript preceding the symbol, but since this number defines the element’s identity, as does its symbol, it is often omitted. For example, magnesium exists as a mixture of three isotopes, each with an atomic number of 12 and with mass numbers of 24, 25, and 26, respectively. These isotopes can be identified as ²⁴Mg, ²⁵Mg, and ²⁶Mg. These isotope symbols are read as “element, mass number” and can be symbolized consistent with this reading. For instance, ²⁴Mg is read as “magnesium 24,” and can be written as “magnesium-24” or “Mg-24.” ²⁵Mg is read as “magnesium 25,” and can be written as “magnesium-25” or “Mg-25.” All magnesium atoms have 12 protons in their nucleus. They differ only because a ²⁴Mg atom has 12 neutrons in its nucleus, a ²⁵Mg atom has 13 neutrons, and a ²⁶Mg has 14 neutrons.

Figure \(\PageIndex{4}\): The symbol for an atom indicates the element via its usual two-letter symbol, the mass number as a left superscript, the atomic number as a left subscript (sometimes omitted), and the charge as a right superscript.

Information about the naturally occurring isotopes of elements with atomic numbers 1 through 10 is given in Table \(\PageIndex{2}\). Note that in addition to standard names and symbols, the isotopes of hydrogen are often referred to using common names and accompanying symbols. Hydrogen-2, symbolized ²H, is also called deuterium and sometimes symbolized D. Hydrogen-3, symbolized ³H, is also called tritium and sometimes symbolized T.

Table \(\PageIndex{2}\): Nuclear Compositions of Atoms of the Very Light Elements

Element	Symbol	Atomic Number	Number of Protons	Number of Neutrons	Mass (amu)	% Natural Abundance
hydrogen	\(\ce{^1_1H}\) (protium)	1	1	0	1.0078	99.989
	\(\ce{^2_1H}\) (deuterium)	1	1	1	2.0141	0.0115
	\(\ce{^3_1H}\) (tritium)	1	1	2	3.01605	— (trace)
helium	\(\ce{^3_2He}\)	2	2	1	3.01603	0.00013
	\(\ce{^4_2He}\)	2	2	2	4.0026	100
lithium	\(\ce{^6_3Li}\)	3	3	3	6.0151	7.59
	\(\ce{^7_3Li}\)	3	3	4	7.0160	92.41
beryllium	\(\ce{^9_4Be}\)	4	4	5	9.0122	100
boron	\(\ce{^{10}_5B}\)	5	5	5	10.0129	19.9
	\(\ce{^{11}_5B}\)	5	5	6	11.0093	80.1
carbon	\(\ce{^{12}_6C}\)	6	6	6	12.0000	98.89
	\(\ce{^{13}_6C}\)	6	6	7	13.0034	1.11
	\(\ce{^{14}_6C}\)	6	6	8	14.0032	— (trace)
nitrogen	\(\ce{^{14}_7N}\)	7	7	7	14.0031	99.63
	\(\ce{^{15}_7N}\)	7	7	8	15.0001	0.37
oxygen	\(\ce{^{16}_8O}\)	8	8	8	15.9949	99.757
	\(\ce{^{17}_8O}\)	8	8	9	16.9991	0.038
	\(\ce{^{18}_8O}\)	8	8	10	17.9992	0.205
fluorine	\(\ce{^{19}_9F}\)	9	9	10	18.9984	100
neon	\(\ce{^{20}_{10}Ne}\)	10	10	10	19.9924	90.48
	\(\ce{^{21}_{10}Ne}\)	10	10	11	20.9938	0.27
	\(\ce{^{22}_{10}Ne}\)	10	10	12	21.9914	9.25

Atomic Mass

Because each proton and each neutron contribute approximately one amu to the mass of an atom, and each electron contributes far less, the atomic mass of a single atom is approximately equal to its mass number (a whole number). However, the average masses of atoms of most elements are not whole numbers because most elements exist naturally as mixtures of two or more isotopes.

The mass of an element shown in a periodic table or listed in a table of atomic masses is a weighted, average mass of all the isotopes present in a naturally occurring sample of that element. This is equal to the sum of each individual isotope’s mass multiplied by its fractional abundance.

\[\mathrm{average\: mass}=\sum_{i}(\mathrm{fractional\: abundance\times isotopic\: mass})_i\]

For example, the element boron is composed of two isotopes: About 19.9% of all boron atoms are ¹⁰B with a mass of 10.0129 amu, and the remaining 80.1% are ¹¹B with a mass of 11.0093 amu. The average atomic mass for boron is calculated to be:

\[\begin{align*}
\textrm{boron average mass} &=\mathrm{(0.199\times10.0129\: amu)+(0.801\times11.0093\: amu)}\\
&=\mathrm{1.99\: amu+8.82\: amu}\\
&=\mathrm{10.81\: amu}
\end{align*}\]

It is important to understand that no single boron atom weighs exactly 10.8 amu; 10.8 amu is the average mass of all boron atoms, and individual boron atoms weigh either approximately 10 amu or 11 amu.

Example \(\PageIndex{1}\): Calculation of Average Atomic Mass

A meteorite found in central Indiana contains traces of the noble gas neon picked up from the solar wind during the meteorite’s trip through the solar system. Analysis of a sample of the gas showed that it consisted of 91.84% ²⁰Ne (mass 19.9924 amu), 0.47% ²¹Ne (mass 20.9940 amu), and 7.69% ²²Ne (mass 21.9914 amu). What is the average mass of the neon in the solar wind?

Solution

\[\begin{align*}
\mathrm{average\: mass} &=\mathrm{(0.9184\times19.9924\: amu)+(0.0047\times20.9940\: amu)+(0.0769\times21.9914\: amu)}\\
&=\mathrm{(18.36+0.099+1.69)\:amu}\\
&=\mathrm{20.15\: amu}
\end{align*}\]

The average mass of a neon atom in the solar wind is 20.15 amu. (The average mass of a terrestrial neon atom is 20.1796 amu. This result demonstrates that we may find slight differences in the natural abundance of isotopes, depending on their origin.)

Exercise \(\PageIndex{1}\)

A sample of magnesium is found to contain 78.70% of ²⁴Mg atoms (mass 23.98 amu), 10.13% of ²⁵Mg atoms (mass 24.99 amu), and 11.17% of ²⁶Mg atoms (mass 25.98 amu). Calculate the average mass of a Mg atom.

Answer

24.31 amu

We can also do variations of this type of calculation, as shown in the next example.

Example \(\PageIndex{2}\): Calculation of Percent Abundance

Naturally occurring chlorine consists of ³⁵Cl (mass 34.96885 amu) and ³⁷Cl (mass 36.96590 amu), with an average mass of 35.453 amu. What is the percent composition of Cl in terms of these two isotopes?

Solution

The average mass of chlorine is the fraction that is ³⁵Cl times the mass of ³⁵Cl plus the fraction that is ³⁷Cl times the mass of ³⁷Cl.

\[\mathrm{average\: mass=(fraction\: of\: ^{35}Cl\times mass\: of\: ^{35}Cl)+(fraction\: of\: ^{37}Cl\times mass\: of\: ^{37}Cl)}\]

If we let x represent the fraction that is ³⁵Cl, then the fraction that is ³⁷Cl is represented by 1.00 − x.

(The fraction that is ³⁵Cl + the fraction that is ³⁷Cl must add up to 1, so the fraction of ³⁷Cl must equal 1.00 − the fraction of ³⁵Cl.)

Substituting this into the average mass equation, we have:

\[\begin{align*}
\mathrm{35.453\: amu} &=(x\times 34.96885\: \ce{amu})+[(1.00-x)\times 36.96590\: \ce{amu}]\\
35.453 &=34.96885x+36.96590-36.96590x\\
1.99705x &=1.513\\
x&=\dfrac{1.513}{1.99705}=0.7576
\end{align*}\]

So solving yields: x = 0.7576, which means that 1.00 − 0.7576 = 0.2424. Therefore, chlorine consists of 75.76% ³⁵Cl and 24.24% ³⁷Cl.

Exercise \(\PageIndex{2}\)

Naturally occurring copper consists of ⁶³Cu (mass 62.9296 amu) and ⁶⁵Cu (mass 64.9278 amu), with an average mass of 63.546 amu. What is the percent composition of Cu in terms of these two isotopes?

Answer

69.15% Cu-63 and 30.85% Cu-65

Figure \(\PageIndex{5}\): Analysis of zirconium in a mass spectrometer produces a mass spectrum with peaks showing the different isotopes of Zr.

The occurrence and natural abundances of isotopes can be experimentally determined using an instrument called a mass spectrometer. Mass spectrometry (MS) is widely used in chemistry, forensics, medicine, environmental science, and many other fields to analyze and help identify the substances in a sample of material. In a typical mass spectrometer (Figure \(\PageIndex{5}\)), the sample is vaporized and exposed to a high-energy electron beam that causes the sample’s atoms (or molecules) to become electrically charged, typically by losing one or more electrons. These cations then pass through a (variable) electric or magnetic field that deflects each cation’s path to an extent that depends on both its mass and charge (similar to how the path of a large steel ball bearing rolling past a magnet is deflected to a lesser extent that that of a small steel BB). The ions are detected, and a plot of the relative number of ions generated versus their mass-to-charge ratios (a mass spectrum) is made. The height of each vertical feature or peak in a mass spectrum is proportional to the fraction of cations with the specified mass-to-charge ratio. Since its initial use during the development of modern atomic theory, MS has evolved to become a powerful tool for chemical analysis in a wide range of applications.

Video \(\PageIndex{1}\): Watch this video from the Royal Society for Chemistry for a brief description of the rudiments of mass spectrometry.

Summary

Isotopes of an element are atoms with the same atomic number but different mass numbers; isotopes of an element, therefore, differ from each other only in the number of neutrons within the nucleus. When a naturally occurring element is composed of several isotopes, the atomic mass of the element represents the average of the masses of the isotopes involved. A chemical symbol identifies the atoms in a substance using symbols, which are one-, two-, or three-letter abbreviations for the atoms.

Key Equations

• \(\mathrm{average\: mass}=\sum_{i}(\mathrm{fractional\: abundance \times isotopic\: mass})_i\)

Glossary

atomic mass

average mass of atoms of an element, expressed in amu

atomic mass unit (amu)

(also, unified atomic mass unit, u, or Dalton, Da) unit of mass equal to \(\dfrac{1}{12}\) of the mass of a ¹²C atom

atomic number (Z)

number of protons in the nucleus of an atom

Dalton (Da)

alternative unit equivalent to the atomic mass unit

mass number (A)

sum of the numbers of neutrons and protons in the nucleus of an atom

unified atomic mass unit (u)

alternative unit equivalent to the atomic mass unit

Contributors and Attributions

• Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/85abf193-2bd...a7ac8df6@9.110).

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