# 2.3: Isotopes and Atomic Weight

Learning Objectives

• Explain how isotopes differ from one another.
• Understand how isotopes differ in particles and mass.
• Identify the most abundant isotope when given specific values.
• Calculate the atomic mass of an element from the masses and relative percentages of the isotopes of the element.

## Isotopes

Isotopes are atoms of the same element that contain different numbers of neutrons. For these species, the number of electrons and protons remain constant. This difference in neutron amount affects the atomic mass (A) but not the atomic number (Z). In a chemical laboratory, isotopes of an element appear and react the same. For this reason, it is difficult to distinguish between an atom's isotopes. In contrast, nuclear scientists can identify and separate different types of atomic nuclei. The technology required for this process is more sophisticated that what could be found in a typical chemical laboratory.

The element carbon ($$\ce{C}$$) has an atomic number of 6, which means that all neutral carbon atoms contain 6 protons and 6 electrons. In a typical sample of carbon-containing material, 98.89% of the carbon atoms also contain 6 neutrons, so each has a mass number of 12. An isotope of any element can be uniquely represented as $${}_Z^{A}X$$ where X is the atomic symbol of the element. The isotope of carbon that has 6 neutrons is therefore $$\ce{_6^{12}C}$$ The subscript indicating the atomic number is actually redundant because the atomic symbol already uniquely specifies Z. Consequently, ​it is more often written as $$\ce{^{12}C}$$, which is read as “carbon-12.” Nevertheless, the value of $$Z$$ is commonly included in the notation for nuclear reactions because these reactions involve changes in $$Z$$.

Most elements on the periodic table have at least two stable isotopes. For example, in addition to $$\ce{^{12}C}$$, a typical sample of carbon contains 1.11% $$\ce{_6^{13}C}$$, with 7 neutrons and 6 protons, and a trace of $$\ce{_6^{14}C}$$, with 8 neutrons and 6 protons. The nucleus of $$\ce{_6^{14}C}$$ is not stable, however, but undergoes a slow radioactive decay that is the basis of the carbon-14 dating technique used in archeology. Many elements other than carbon have more than one stable isotope; tin, for example, has 10 isotopes. There are about twenty elements that exist in only one isotopic form (sodium and fluorine are examples of these).

An important series of isotopes is found with hydrogen atoms. Most hydrogen atoms have a nucleus with only a single proton. About 1 in 10,000 hydrogen nuclei, however, also has a neutron; this particular isotope is called deuterium. An extremely rare hydrogen isotope, tritium, has 1 proton and 2 neutrons in its nucleus. Figure $$\PageIndex{1}$$ compares the three isotopes of hydrogen.

There are currently over 3,500 isotopes known for all the elements. When scientists discuss individual isotopes, they need an efficient way to specify the number of neutrons in any particular nucleus. A/Z and symbol-mass formats can be used to display periodic table information. When viewing either of these two notations, isotopic differences can be obtained.

The discovery of isotopes required a minor change in Dalton’s atomic theory. Dalton thought that all atoms of the same element were exactly the same.

Look at the A/Z formats for the three isotopes of hydrogen in Table $$\PageIndex{1}$$. Note how the atomic number (bottom value) remains the same while the atomic masses (top number) are varied. All isotopes of a particular element will vary in neutrons and mass. This variance in mass will be visible in the symbol-mass format of same isotopes as well.

Table $$\PageIndex{1}$$
Common Name A/Z formats symbol-mass format Expanded Name
Hydrogen $$\mathrm{^{1}_{1}H}$$ $$\text{H-1}$$ hydrogen-1
Deuterium $$\mathrm{^{2}_{1}H}$$ $$\text{H-2}$$ hydrogen-2
Tritium $$\mathrm{^{3}_{1}H}$$ $$\text{H-3}$$ hydrogen 3

Both A/Z or symbol-mass formats can be utilized to determine the amount of subatomic particles (protons, neutrons, and electrons) contained inside an isotope. When given either format, these mass values should be used to calculate the number of neutrons in the nucleus.

## Atomic Weight

Since most naturally occurring elements samples are mixtures of isotopes, it is useful to use an average weight of an element. The atomic mass of an element is the weighted mass of all the naturally presented isotopes (on earth). To determine the most abundant isotopic form of an element, compare given isotopes to the weighted average on the periodic table. For example, the three hydrogen isotopes in Figure $$\PageIndex{1}$$ are H-1, H-2, and H-3. The atomic mass or weighted average of hydrogen is around 1.008 amu ( look again to the periodic table). Of the three hydrogen isotopes, H-1 is closest in mass to the weighted average; therefore, it is the most abundant. The other two isotopes of hydrogen are quite rare, but are very exciting in the world of nuclear science.

You can calculate the atomic mass (or average mass) of an element provided you know the relative abundances (the fraction of an element that is a given isotope), the element's naturally occurring isotopes, and the masses of those different isotopes. We can calculate this by the following equation:

$\text{Atomic mass} = \left( \%_1 \right) \left( \text{mass}_1 \right) + \left( \%_2 \right) \left( \text{mass}_2 \right) + \cdots \label{eq1}$

Averages like Equation 1 are known as weigthed averages. An element's atomic mass can be calculated provided the relative abundances of the element's naturally occurring isotopes and the masses of those isotopes are known. If all the abundances are not provided, it is safe to assume that all numbers should add up to 100%.

For example, Boron has two naturally occurring isotopes. In a sample of boron, $$20\%$$ of the atoms are $$\text{B-10}$$, which is an isotope of boron with 5 neutrons and mass of $$10 \: \text{amu}$$. The other $$80\%$$ of the atoms are $$\text{B-11}$$, which is an isotope of boron with 6 neutrons and a mass of $$11 \: \text{amu}$$. How do we calculate the atomic mass of boron?

Boron has two isotopes so we will use the Equation \ref{eq1} and substitute the relative abundances and atomic masses of Boron into Equation \ref{eq1}:

\begin{align} \text{Atomic mass} &= \left( 0.20 \right) \left( 10 \right) + \left( 0.80 \right) \left( 11 \right) \nonumber \\ &= 10.8 \: \text{amu}\nonumber \end{align}\nonumber

The mass of an average boron atom, and thus boron's atomic mass, is $$10.8 \: \text{amu}$$.

Example $$\PageIndex{1}$$: Atomic Weight of Neon

Neon has three naturally occurring isotopes. In a sample of neon, $$90.92\%$$ of the atoms are $$\ce{Ne}$$-20, which is an isotope of neon with 10 neutrons and a mass of $$19.99 \: \text{amu}$$. Another $$0.3\%$$ of the atoms are $$\ce{Ne}$$-21, which is an isotope of neon with 11 neutrons and a mass of $$20.99 \: \text{amu}$$. The final $$8.85\%$$ of the atoms are $$\ce{Ne}$$-22, which is an isotope of neon with 12 neutrons and a mass of $$21.99 \: \text{amu}$$. What is the atomic mass of neon?

Solution

Neon has three isotopes. We will use the equation:

$\text{Atomic mass} = \left( \%_1 \right) \left( \text{mass}_1 \right) + \left( \%_2 \right) \left( \text{mass}_2 \right) + \cdots \nonumber$

Substitute these into the equation, and we get:

\begin{align*} \text{Atomic mass} &= \left( 0.9092 \right) \left( 19.99 \right) + \left( 0.003 \right) \left( 20.99 \right) + \left( 0.0885 \right) \left( 21.99 \right) \\ &= 20.17 \: \text{amu} \end{align*}

The mass of an average neon atom is $$20.17 \: \text{amu}$$

Exercise $$\PageIndex{1}$$

Magnesium has the three isotopes listed in the following table:

Isotope Exact Mass (amu) Percent Abundance (%)
24Mg 23.98504 78.70
25Mg 24.98584 10.13
26Mg 25.98259 11.17

Use these data to calculate the atomic mass of magnesium.

24.31 amu

Applications of Isotopes

During the Manhattan project, the majority of federal funding was dedicated the separation of uranium isotopes. The two most common isotopes of uranium are U-238 and U-235. About 99.3% of uranium is of the U-238 variety, this form is not fissionable and will not work in a nuclear weapon or reaction. The remaining 0.7% is U-235 which is fissionable, but first had to be separated from U-238. This separation process is called enrichment. During World War II, a nuclear facility was built in Oak Ridge, Tennessee to accomplish this project. At the time, the enrichment process only produced enough U-235 for one nuclear weapon. This fuel was placed inside the smaller of the two atomic bombs (Little Boy) dropped over Japan.

Uranium is a natural element that can be found in several different countries. Countries that do not have natural uranium supplies would need to obtain it from one of the countries below. Most nuclear reactors that provide energy rely on U-235 as a source of fuel. Fortunately, reactors only need 2-5% U-235 for the production of megawatts or even gigawatts of power. If the purification process exceeds this level, than it is likely a country is focusing on making nuclear weapons. For example, Manhattan Project scientists enriched U-235 up to 90% in order to produce the Little Boy weapon.

Abbreviations like HEU (highly enriched uranium) and LEU (low-enriched uranium) are used frequently by nuclear scientists and groups. HEU is defined as being over 20% pure U-235 and would not be used in most commercial nuclear reactors. This type of material is used to fuel larger submarines and aircraft carriers. If the purification of U-235 reaches 90%, then the HEU is further classified as being weapons grade material. This type of U-235 could be used to make a nuclear weapon (fission or even fusion based). As for LEU, its U-235 level would be below this 20% mark. LEU would be used for commercial nuclear reactors and smaller, nuclear powered submarines. LEU is not pure enough to be used in a conventional nuclear weapon, but could be used in a dirty bomb. This type of weapon uses conventional explosives like dynamite to spread nuclear material. Unlike a nuclear weapon, dirty bombs are not powerful enough to affect large groups of buildings or people. Unfortunately, the spread of nuclear material would cause massive chaos for a community and would result in casualties.

## Summary

• The isotopes of an element have different masses and are identified by their mass numbers.
• An element's atomic mass is the weighted average of the masses of the isotopes of an element
• An element's atomic mass can be calculated provided the relative abundances of the element's naturally occurring isotopes and the masses of those isotopes are known. If all the abundances are not provided, it is safe to assume that all numbers should add up to 100%.

## Concept Review Exercises

1. Why is the atomic number so important to the identity of an atom?
2. What is the relationship between the number of protons and the number of electrons in an atom?
3. How do isotopes of an element differ from each other?
4. What is the mass number of an element?