4.19: Calculating Atomic Mass

Introduction

You have a pile of rocks to move and need to decide what equipment you want to rent to move them. If the rocks are fairly small, you can get a shovel to pick them up. Larger rocks could be moved by hand, but big boulders will need some sort of mechanical scoop. The amount of each kind of rock will also determine how much time you will need to get the job done. Knowing the relative amounts of large, medium, and small rocks can be very useful in deciding how to approach the job.

Percent Natural Abundance

Most elements occur naturally as a mixture of two or more isotopes. The table below shows the naturally isotopes of several elements, along with the percent natural abundance of each.

Element Isotope (Symbol) Percent Natural Abundance Atomic Mass $$\left( \text{amu} \right)$$ Average Atomic Mass $$\left( \text{amu} \right)$$
Atomic Masses and Percent Abundances of Some Natural Isotopes
Hydrogen $$\ce{_1^1H}$$ 99.985 1.0078 1.0079
$$\ce{_1^2H}$$ 0.015 2.0141
$$\ce{_1^3H}$$ negligible 3.0160
Carbon $$\ce{_6^{12}C}$$ 98.89 12.000 12.011
$$\ce{_6^{13}C}$$ 1.11 13.003
$$\ce{_6^{14}C}$$ trace 14.003
Oxygen $$\ce{_8^{16}O}$$ 99.759 15.995 15.999
$$\ce{_8^{17}O}$$ 0.037 16.995
$$\ce{_8^{18}O}$$ 0.204 17.999
Chlorine $$\ce{_{17}^{35}Cl}$$ 75.77 34.969 35.453
$$\ce{_{17}^{38}Cl}$$ 24.23 36.966
Copper $$\ce{_{29}^{63}Cu}$$ 69.17 62.930 63.546
$$\ce{_{29}^{65}Cu}$$ 30.83 64.928

For some elements, one particular isotope predominates greatly over the other isotopes. 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 more 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 36.

$\frac{\left( 34.969 + 36.966 \right)}{2} = 35.968 \: \text{amu}$

Clearly the actual average atomic mass from the last column of the table is significantly lower. Why? We need to take into account the percent natural abundances 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.

Example 4.19.1

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

Solution:

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

Known

• Chlorine-35: atomic mass $$= 34.969 \: \text{amu}$$ and percent abundance $$= 75.77\%$$
• Chlorine-37: atomic mass $$= 36.966 \: \text{amu}$$ and percent 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 for each isotope to get the average atomic mass.

Step 2: Calculate

$\begin{array}{ll} \text{chlorine-35} & 0.7577 \times 34.969 = 26.50 \: \text{amu} \\ \text{chlorine-37} & 0.2423 \times 36.966 = 8.957 \: \text{amu} \\ \text{average atomic mass} & 26.50 + 8.957 = 35.45 \: \text{amu} \end{array}$

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

$\left( 0.7577 \times 34.969 \right) + \left(0.2423 \times 36.966 \right) = 35.45 \: \text{amu}$