4.20: Calculating Average Atomic Mass - Chemistry LibreTexts

Percent Natural Abundance

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

Table \(\PageIndex{1}\): Atomic Masses and Percents of Abundance of Some Natural Isotopes

Element	Isotope (Symbol)	Percent Natural Abundance	Atomic Mass \(\left( \text{amu} \right)\)	Average Atomic Mass \(\left( \text{amu} \right)\)
Hydrogen	\(\ce{_1^1H}\)	99.985	1.0078	1.0080
	\(\ce{_1^2H}\)	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
	\(\ce{_{17}^{37}Cl}\)	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.

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

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 abundance of each isotope, in order to calculate 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 \(\PageIndex{1}\)

Use the atomic masses of each of the two isotopes of chlorine along with their respective percent 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.

\[\begin{array}{ll} \text{chlorine-35} & 0.7577 \times 34.969 = 26.50 \: \text{amu} \\ \text{chlorine-37} & 0.2423 \times 36.966 = 8.957 \: \text{amu} \\ \text{average atomic mass} & 26.50 + 8.957 = 35.46 \: \text{amu} \end{array}\nonumber \]

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

\[\left( 0.7577 \times 34.969 \right) + \left(0.2423 \times 36.966 \right) = 35.46 \: \text{amu}\nonumber \]

Step 3: Think about your result.

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