# 4.3: Isotopes and Atomic Mass

Difficulty Level: At Grade Created by: CK-12

## Lesson Objectives

• Define atomic number.
• Define mass number.
• Understand how isotopes differ from one another and be able to designate them by various methods.
• Be able to calculate the average atomic mass of an element.

## Lesson Vocabulary

• atomic mass
• atomic mass unit
• atomic number
• isotope
• mass number
• nuclide

## Introduction

Atoms are the fundamental building blocks of all matter and are composed of protons, neutrons, and electrons. Because atoms are electrically neutral, the number of positively charged protons must be equal to the number of negatively charged electrons. One of Dalton’s points in his atomic theory was that all atoms of a given element are identical. In this section, we will see how this is not strictly true, thanks to variability in the number of neutrons that an atom may contain.

## Atomic Number

The atomic number (Z) of an element is the number of protons in the nucleus of each atom of that element. An atom can be classified as a particular element based solely on its atomic number. For example, any atom with an atomic number of 8 (its nucleus contains 8 protons) is an oxygen atom, and any atom with a different number of protons would be a different element. The periodic table (Figure below) displays all of the known elements and is arranged in order of increasing atomic number. In this table, an element’s atomic number is indicated above the elemental symbol. Hydrogen, at the upper left of the table, has an atomic number of 1. Every hydrogen atom has one proton in its nucleus. Next on the table is helium, whose atoms have two protons in the nucleus. Lithium atoms have three protons, beryllium atoms have four, and so on.

Since atoms are neutral, the number of electrons in an atom is equal to the number of protons. Hydrogen atoms all have one electron occupying the space outside of the nucleus.

The periodic table of the elements.

## Mass Number

Rutherford’s experiment showed that the vast majority of the mass of an atom is concentrated in its nucleus, which is composed of protons and neutrons. The mass number is defined as the total number of protons and neutrons in an atom. Consider Table below, which shows data from the first six elements of the periodic table.

Atoms of the First Six Elements
Name Symbol Atomic Number Protons Neutrons Electrons Mass Number
Hydrogen H 1 1 0 1 1
Helium He 2 2 2 2 4
Lithium Li 3 3 4 3 7
Beryllium Be 4 4 5 4 9
Boron B 5 5 6 5 11
Carbon C 6 6 6 6 12

Consider the element helium. Its atomic number is 2, so it has two protons in its nucleus. Its nucleus also contains two neutrons. Since 2 + 2 = 4, we know that the mass number of the helium atom is 4. Finally, the helium atom also contains two electrons, since the number of electrons must equal the number of protons. This example may lead you to believe that atoms have the same number of protons and neutrons, but a further examination of Table above will show that this is not the case. Lithium, for example has three protons and four neutrons, giving it a mass number of 7.

Knowing the mass number and the atomic number of an atom allows you to determine the number of neutrons present in that atom by subtraction.

Number of neutrons = mass number - atomic number

Atoms of the element chromium (Cr) have an atomic number of 24 and a mass number of 52. How many neutrons are in the nucleus of a chromium atom? To determine this, you would subtract as shown:

52 - 24 = 28 neutrons in a chromium atom

The composition of any atom can be illustrated with a shorthand notation using the atomic number and the mass number. Both are written before the chemical symbol, with the mass number written as a superscript and the atomic number written as a subscript. The chromium atom discussed above would be written as:

5224Cr\begin{align*}{}_{24}^{52}\text{Cr}\end{align*}

Another way to refer to a specific atom is to write the mass number of the atom after the name, separated by a hyphen. The above atom would be written as chromium-52.

## Isotopes

As stated earlier, not all atoms of a given element are identical. Specifically, the number of neutrons in the nucleus can vary for many elements. As an example, naturally occurring carbon exists in three forms, which are illustrated in Figure below.

Nuclei of the three isotopes of carbon: Almost 99% of naturally occurring carbon is carbon-12, whose nucleus consists of six protons and six neutrons. Carbon-13 and carbon-14, with seven or eight neutrons, respectively, have a much lower natural abundance.

Each carbon atom has the same number of protons (6), which is equal to its atomic number. Each carbon atom also contains six electrons, allowing the atom to remain electrically neutral. However the number of neutrons varies from six to eight. Isotopes are atoms that have the same atomic number but different mass numbers due to a change in the number of neutrons. The three isotopes of carbon can be referred to as carbon-12 (12 6C\begin{align*}{}_{\ 6}^{12}\text{C}\end{align*}), carbon-13 (13 6C\begin{align*}{}_{\ 6}^{13}\text{C}\end{align*}), and carbon-14 (14 6C\begin{align*}{}_{\ 6}^{14}\text{C}\end{align*}). Naturally occurring samples of most elements are mixtures of isotopes. Carbon has only three natural isotopes, but some heavier elements have many more. Tin has ten stable isotopes, which is the most of any element. The term nuclide refers to the nucleus of a given isotope of an element. The nucleus of a given carbon atom will be one of the three possible nuclides discussed above.

While the presence of isotopes affects the mass of an atom, it does not affect its chemical reactivity. Chemical behavior is governed by the number of electrons and the number of protons. Carbon-13 behaves chemically in exactly the same way as the more plentiful carbon-12.

Sample Problem 4.2: Composition of an Atom

How many protons, neutrons, and electrons are present in each of the nuclides below?

1. Iron (Fe): atomic number = 26, mass number = 56
2. Iodine-127 (atomic number = 53)
3. 3115P\begin{align*}{}_{15}^{31}\text{P}\end{align*}

Step 1: List the known and unknown quantities and plan the problem.

Known

• Atomic number and mass number for each

Unknown

• Number of protons, electrons, and neutrons

Each shows a different way to specify an isotope of an atom. Use the definitions of atomic number and mass number to calculate the numbers of protons, neutrons, and electrons.

Step 2: Calculate

Number of protons = atomic number

1. 26
2. 53
3. 15

Number of electrons = number of protons

1. 26
2. 53
3. 15

Number of neutrons = mass number - atomic number

1. 56 - 26 = 30
2. 127 - 53 = 74
3. 31 - 15 = 16

For each atom, the results are consistent with the definitions of atomic number and mass number.

Do the practice problems below. If necessary, refer to the periodic table in Figure above for the atomic number or symbol of the given element.

Practice Problems
1. How many protons, neutrons, and electrons are there in the atom 19 9F\begin{align*}{}_{\ 9}^{19}\text{F}\end{align*}?
2. How many protons, neutrons, and electrons are there in an atom of lead-207?
3. A certain atom has an atomic number of 36 and a mass number of 84. Write out the designation for this isotope in both nuclide symbol form and in hyphenated form.
4. An atom has a mass number of 59 and contains 32 neutrons in its nucleus. What element is it?

## Atomic Mass

The masses of individual atoms are very, very small. However, using a modern device called a mass spectrometer, it is possible to measure such minuscule masses. An atom of oxygen-16, for example, has a mass of 2.66 × 10−23 g. While comparisons of masses measured in grams would have some usefulness, it is far more practical to have a system that will allow us to more easily compare relative atomic masses. Scientists decided on using the carbon-12 nuclide as the reference standard by which all other masses would be compared. By definition, one atom of carbon-12 is assigned a mass of exactly 12 atomic mass units (amu). An atomic mass unit is defined as a mass equal to one twelfth the mass of an atom of carbon-12. The mass of any isotope of any element is expressed in relation to the carbon-12 standard. For example, one atom of helium-4 has a mass of 4.0026 amu. An atom of sulfur-32 has a mass of 31.972 amu.

The carbon-12 atom has six protons and six neutrons in its nucleus for a mass number of 12. Since the nucleus accounts for nearly all of the mass of the atom, a single proton or single neutron has a mass of approximately 1 amu. However, as seen by the helium and sulfur examples, the masses of individual atoms are not quite whole numbers. This is because an atom’s mass is affected very slightly by the interactions of the various particles within the nucleus and also includes the small mass added by each electron.

As stated in the section on isotopes, most elements occur naturally as a mixture of two or more isotopes. Table below shows the naturally occurring isotopes of several elements along with the percent natural abundance of each.

Atomic Masses and Percent Abundances of Some Natural Isotopes
Element Isotope (symbol) Percent natural abundance Atomic mass (amu) Average atomic mass (amu)
Hydrogen

11H\begin{align*}{}_1^1\text{H}\end{align*}

21H\begin{align*}{}_1^2\text{H}\end{align*}

31H\begin{align*}{}_1^3\text{H}\end{align*}

99.985

0.015

negligible

1.0078

2.0141

3.0160

1.0079
Carbon

126C\begin{align*}{}_{6}^{12}\text{C}\end{align*}

136C\begin{align*}{}_{6}^{13}\text{C}\end{align*}

146C\begin{align*}{}_{6}^{14}\text{C}\end{align*}

98.89

1.11

trace

12.000

13.003

14.003

12.011
Oxygen

168O\begin{align*}{}_{8}^{16}\text{O}\end{align*}

178O\begin{align*}{}_{8}^{17}\text{O}\end{align*}

188O\begin{align*}{}_{8}^{18}\text{O}\end{align*}

99.759

0.037

0.204

15.995

16.995

17.999

15.999
Chlorine

3517Cl\begin{align*}{}_{17}^{35}\text{Cl}\end{align*}

3717Cl\begin{align*}{}_{17}^{37}\text{Cl}\end{align*}

75.77

24.23

34.969

36.966

35.453
Copper

6329Cu\begin{align*}{}_{29}^{63}\text{Cu}\end{align*}

6529Cu\begin{align*}{}_{29}^{65}\text{Cu}\end{align*}

69.17

30.83

62.930

64.928

63.546

For some elements, one particular isotope is much more abundant than any other isotopes. For example, naturally occurring hydrogen is nearly all hydrogen-1, and naturally occurring oxygen is nearly all oxygen-16. For many other elements, however, more than one isotope may exist in substantial quantities. Chlorine (atomic number 17) is a yellowish-green toxic gas. About three quarters of all chlorine atoms have 18 neutrons, giving those atoms a mass number of 35. About one quarter of all chlorine atoms have 20 neutrons, giving those atoms a mass number of 37. Were you to simply calculate the arithmetic average of the precise atomic masses, you would get approximately 36.

(34.969+36.966)/2=35.968 amu\begin{align*}(34.969 + 36.966) / 2 = 35.968 \ \text{amu}\end{align*}

As you can see, the average atomic mass given in the last column of Table above is significantly lower. Why? The reason is that we need to take into account the natural abundance percentages of each isotope in order to calculate what is called the weighted average. The atomic mass of an element is the weighted average of the atomic masses of the naturally occurring isotopes of that element. The sample problem below demonstrates how to calculate the atomic mass of chlorine.

Sample Problem 4.3: Calculating Atomic Mass

Use the atomic masses of each of the two isotopes of chlorine along with their percent abundances to calculate the average atomic mass of chlorine.

Step 1: List the known and unknown quantities and plan the problem.

Known

• chlorine-35: atomic mass = 34.969 amu and % abundance = 75.77%
• chlorine-37: atomic mass = 36.966 amu and % abundance = 24.23%

Unknown

• Average atomic mass of chlorine

Change each percent abundance into decimal form by dividing by 100. Multiply this value by the atomic mass of that isotope. Add together the results for each isotope to get the average atomic mass.

Step 2: Calculate

chlorine-35chlorine-37average atomic mass0.7577×34.969=26.50 amu0.2423×36.966=8.957 amu26.50+8.957=35.45 amu\begin{align*} & \text{chlorine-35} && 0.7577 \times 34.969 = 26.50 \ \text{amu} \\ & \text{chlorine-37} && 0.2423 \times 36.966 = 8.957 \ \text{amu} \\ & \text{average atomic mass} && 26.50 + 8.957 = 35.45 \ \text{amu} \end{align*}

Note: Applying significant figure rules results in the 35.45 amu result without excessive rounding error. In one step:

(0.7577×34.969)+(0.2423×36.966)=35.45 amu\begin{align*}(0.7577 \times 34.969) + (0.2423 \times 36.966) = 35.45 \ \text{amu}\end{align*}

The calculated average atomic mass is closer to 35 than to 37 because a greater percentage of naturally occurring chlorine atoms have a mass number of 35. It agrees with the value from Table above.

Practice Problem
1. The element bromine consists of two naturally occurring isotopes. The isotope with a mass of 78.92 amu has a percent abundance of 50.69%, while the isotope with a mass of 80.92 amu has a percent abundance of 49.31%. Calculate the average atomic mass of bromine.

The atomic masses for each element on the periodic table are average atomic masses. For later calculations involving atomic mass, we will use these values and round each one to four significant figures.

## Lesson Summary

• The atomic number of an element is equal to the number of protons in its nucleus.
• The mass number of an element is equal to the sum of the protons and the neutrons in its nucleus.
• Isotopes are atoms of the same element that have a different mass number because of a variation in the number of neutrons.
• The average atomic mass of an element can be calculated from the atomic masses and percent natural abundances of each naturally occurring isotope.

## Lesson Review Questions

### Recall

1. Why are all atoms electrically neutral?
2. How many protons are in the nucleus of each of the following atoms?
1. neon
2. gold
3. strontium
4. uranium
3. What part of Dalton’s atomic theory is disproved by the existence of isotopes?
4. Which isotope is used as the reference standard for the atomic mass unit?

### Apply Concepts

1. The average atomic mass of all naturally occurring lithium atoms is 6.941 amu. The two isotopes of lithium are lithium-6 and lithium-7. Are these isotopes equally common? If not, which is more plentiful in nature, and how do you know?

### Think Critically

1. A certain atom contains 28 protons, 28 electrons, and 31 neutrons. Provide the following:
1. atomic number
2. mass number
3. name of element
2. How many protons, neutrons, and electrons are in an atom of cesium-133?
3. Complete Table below:
Table for Problem 8
Isotope Nuclide Symbol Atomic Number Mass Number
sodium-23
7533As\begin{align*}{}_{33}^{75}\text{As}\end{align*}
silver-108
1. Which of the following is an isotope of 4018Ar\begin{align*}{}_{18}^{40}\text{Ar}\end{align*}? Explain.
1. 4020Ca\begin{align*}{}_{20}^{40}\text{Ca}\end{align*}
2. 3918Ar\begin{align*}{}_{18}^{39}\text{Ar}\end{align*}
3. 4018Ar\begin{align*}{}_{18}^{40}\text{Ar}\end{align*}
2. Fill in Table below:
Table for Problem 10
Isotope Number of Protons Number of Electrons Number of Neutrons Nuclide Symbol
hydrogen-1
hydrogen-2
beryllium-9
aluminum-27
1. Fill in Table below:
Table for Problem 11
Element Symbol Atomic Number Mass Number # of Protons # of Electrons # of Neutrons Nuclide Symbol
Nitrogen 14
B 11
30 35
77 116
5626Fe\begin{align*}{}_{26}^{56}\text{Fe}\end{align*}
1. The element tungsten (W) is known best as a metal that is used as filaments for light bulbs. Naturally occurring tungsten consists of the five isotopes shown below. Calculate the atomic mass of tungsten.

tungsten-180tungsten-182tungsten-183tungsten-184tungsten-186atomic mass=179.947 amuatomic mass=181.948 amuatomic mass=182.950 amuatomic mass=183.951 amuatomic mass=185.954 amupercent abundance=0.12%percent abundance=26.50%percent abundance=14.31%percent abundance=30.64%percent abundance=28.43%\begin{align*} \text{tungsten-180} && \text{atomic mass} = 179.947 \ \text{amu} &&& \text{percent abundance} = 0.12\% \\ \text{tungsten-182} && \text{atomic mass} = 181.948 \ \text{amu} &&& \text{percent abundance} = 26.50\% \\ \text{tungsten-183} && \text{atomic mass} = 182.950 \ \text{amu} &&& \text{percent abundance} = 14.31\% \\ \text{tungsten-184} && \text{atomic mass} = 183.951 \ \text{amu} &&& \text{percent abundance} = 30.64\% \\ \text{tungsten-186} && \text{atomic mass} = 185.954 \ \text{amu} &&& \text{percent abundance} = 28.43\% \\ \end{align*}

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