Percent Abundance
Chapter 2
Audio Summary
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Elements can exist naturally as various isotopes where the number of neutrons varies by each element.
Below is a table defining the notation used on this page.
The standard atomic weight (Ar°(E) where E is an element; unitless; more appropriately called relative atomic mass of the element) of an element represents an weighted mass average of a representative sample of that element. This value is unitless because it is determined from a ratio of the average atomic mass of the atom/element to exactly 1/12th the mass of one carbon-12 atom.
\[A_{\mathrm{r}}^{\circ} = \dfrac{\textrm{avg. mass of atom}}{\frac{1}{12}~\textrm{mass of one carbon-12 atom}}\] The standard atomic weight can have units if it is multiplied by the appropriate conversion factor.
The atomic mass constant (mu) is the unified atomic mass unit (u, or Da), commonly incorrectly referred to as the atomic mass unit (amu), and is given as
\[m_u = 1~\mathrm{u} = 1.660~539~066~60\times 10^{-27}~\mathrm{kg}\]
Therefore, to get units of amu, simply multiply the standard atomic weight by the atomic mass constant, mu.
\[A_{\mathrm{r}}^{\circ}(\mathrm{unitless}) \times m_u = A_{\mathrm{r}}^{\circ}(\mathrm{amu})\]
To get units of kg, multiply the standard atomic weight by the defined kg value.
\[A_{\mathrm{r}}^{\circ}(\mathrm{unitless}) \times \left ( 1.660~539~066~60\times 10^{-27}~\mathrm{kg} \right ) = A_{\mathrm{r}}^{\circ}(\mathrm{kg})\]
We can determine the standard atomic weight (Ar°(E)) for any element by considering the relative atomic mass (Ar(X) where X is an isotope; unitless) for each naturally occurring isotope of that element as well as the percent abundance [x(X) %] or fractional abundance [x(X)] of each isotope (X) by using the following general equation
\[\begin{alignat*}{2} \mathrm{standard~atomic~weight} ~ &= ~ &&\left ( \dfrac{\mathrm{\%~abundance~isotope~1}}{100~\%} \right ) \left ( \mathrm{relative~atomic~mass~of~isotope~1} \right )~~+ \\[1.5ex] &\phantom{=} &&\left ( \dfrac{\mathrm{\%~abundance~isotope~2}}{100~\%} \right ) \left ( \mathrm{relative~atomic~mass~of~isotope~2} \right )~~+ ~\cdots\\[1.5ex] \end{alignat*}\]
or more succinctly
\[\begin{alignat*}{2} A_{\mathrm{r}}^{\circ}(E) ~ &= ~ &&\left ( \dfrac{x \left ( X_1 \! \right ) \%}{100~\%} \right ) ~ A_{\mathrm{r}}(X_1) ~~+ \\[1.5ex] &\phantom{=} &&\left ( \dfrac{x \! \left ( X_2 \right ) \%}{100~\%} \right ) ~ A_{\mathrm{r}}(X_2) ~~+ ~\cdots\\[1.5ex] \end{alignat*}\]
or even more succinctly, for n isotopes,
\[\begin{align*} A_{\mathrm{r}}^{\circ}(E) &= ~ \sum_{i=1}^n \left ( \dfrac{x \left ( X_i \! \right ) \%}{100~\%} \right ) ~ A_{\mathrm{r}}(X_i) \end{align*}\]
Note that the percent abundance of every naturally occurring isotope for an element should sum to 100 %, and, simultaneously, the fractional abundances should sum to 1.
See NIST for a list of naturally occurring isotopes for elements.
Example: Standard Atomic Weight of Boron
Boron has a reported standard atomic weight of 10.81. However, boron naturally exists as two isotopes.
Let us calculate the reported standard atomic weight of boron.
\[\begin{align*} A_{\mathrm{r}}^{\circ}(E) &= \left ( \dfrac{x \! \left ( X_1 \right )\%}{100~\%} \right ) \left [ A_{\mathrm{r}} \! \left ( X_1 \right ) \right ] + \left ( \dfrac{x \! \left ( X_2 \right )\%}{100~\%} \right ) \left [ A_{\mathrm{r}} \! \left ( X_2 \right ) \right ] \longrightarrow \\[3ex] A_{\mathrm{r}}^{\circ}(\mathrm{B}) &= \left ( \dfrac{x \! \left ( \mathrm{^{10}B} \right ) \%}{100~\%} \right ) \left [ A_{\mathrm{r}} \! \left (\mathrm{^{10}B} \right ) \right ] + \left ( \dfrac{x \! \left (\mathrm{^{11}B} \right ) \%}{100~\%} \right ) \left [ A_{\mathrm{r}} \! \left (\mathrm{^{11}B}\right ) \right ] \\[3ex] &= \left ( \dfrac{19.9~\%}{100~\%} \right ) \left ( 10.0129 \right ) + \left ( \dfrac{80.1~\%}{100~\%} \right ) \left ( 11.0093 \right ) \\[3ex] &= 1.9\bar{9}25 + 8.8\bar{1}84\\[3ex] &= 10.8\bar{1}09 \\[3ex] &= 10.81 \end{align*}\]
Suppose we wanted to calculate the percent abundance of the isotope boron-10. Rearrange the equation and solve for the percent abundance of boron-10.
\[\begin{align*} A_{\mathrm{r}}^{\circ} \! \left (\mathrm{B} \right ) &= \left ( \dfrac{x \left ( \mathrm{^{10}B} \right ) \%}{100~\%} \right ) \left [ A_{\mathrm{r}} \! \left ( \mathrm{^{10}B} \right ) \right ] + \left ( \dfrac{x \left ( \mathrm{^{11}B} \right ) \%}{100~\%} \right ) \left [ A_{\mathrm{r}} \! \left (\mathrm{^{11}B} \right ) \right ] \longrightarrow \\[6ex] x \! \left ( \mathrm{^{10}B} \right ) \% &= \dfrac{\Biggr[A_{\mathrm{r}}^{\circ} \! \left (\mathrm{B} \right ) - \Biggl\{ \left ( \dfrac{x \! \left (\mathrm{^{11}B} \right ) \% }{100~%} \right ) \left [ A_{\mathrm{r}} \! \left (\mathrm{^{11}B} \right ) \right ] \Biggl\}\Biggr]} {A_{\mathrm{r}} \! \left (\mathrm{^{10}B}\right )} \times 100~\% \\[3ex] &= \dfrac{\Biggr[10.81 - \Biggl\{ \left ( \dfrac{80.1~\%}{100~\%} \right ) \left ( 11.0093 \right ) \Biggl\}\Biggr]} {10.0129} \times 100~\% \\[3ex] &= \dfrac{\left( 10.81 - 8.8\bar{1}84 \right)} {10.0129} \times 100~\% \\[3ex] &= \dfrac{1.9\bar{9}16} {10.0129} \times 100~\% \\[3ex] &= 19.\bar{8}90~\%\\[3ex] &= 19.9~\% \end{align*}\]
Example: Standard Atomic Weight of Neon
Neon has a reported standard atomic weight of 20.18. However, neon naturally exists as three isotopes. Suppose we had the following information.
Let us calculate the percent abundance of neon-20 and neon-22. Let x represent the abundance of neon-20 and let y represent the abundance of neon-22. Here I use the fractional abundances [x(X)] for the isotopes of neon. Let a and b represent the unknown fractional abundances for 20Ne and 22Ne, respectively.
\[\begin{align*} A_{\mathrm{r}}^{\circ} \left ( \mathrm{Ne} \right ) &= x \left (\mathrm{^{20}Ne} \right ) ~ A_{\mathrm{r}} \left ( \mathrm{^{20}Ne} \right ) + x \left (\mathrm{^{21}Ne} \right ) ~ A_{\mathrm{r}} \left ( \mathrm{^{21}Ne} \right ) + x \left (\mathrm{^{22}Ne} \right ) ~ A_{\mathrm{r}} \left ( \mathrm{^{22}Ne} \right ) \\[1.5ex] 20.18 &= x \left (\mathrm{^{20}Ne} \right ) \left ( 19.9924 \right ) + \left ( 0.0027 \right ) \left ( 20.9938 \right ) + x \left (\mathrm{^{22}Ne} \right ) \left ( 21.9914 \right ) \end{align*}\]
Since the fractional abundances should sum to 1 (exactly), we know that
\[\begin{align*} x \! \left ( \mathrm{^{20}Ne} \right ) + x \! \left ( \mathrm{^{21}Ne} \right ) + x \! \left ( \mathrm{^{22}Ne} \right ) &= 1 \\[1.5ex] x \! \left ( \mathrm{^{22}Ne} \right ) &= 1 - 0.0027 - x \! \left ( \mathrm{^{20}Ne} \right ) \\[1.5ex] &= 0.9973 - x \left ( \mathrm{^{20}Ne} \right ) \end{align*}\]
We can now substitute 22Ne in terms of 20Ne to give us a one-variable equation. Solve for the relative isotopic mass of 20Ne.
\[\begin{align*} 20.18 &= x \! \left ( \mathrm{^{20}Ne} \right ) \left ( 19.9924 \right ) + \left ( 0.0027 \right ) \left ( 20.9938 \right ) + \left [ 0.9973 - x \! \left ( \mathrm{^{20}Ne} \right ) \right ] \left ( 21.9914 \right ) \\[1.5ex] 20.18 &= x \! \left ( \mathrm{^{20}Ne} \right ) \left ( 19.9924 \right ) + 0.05\bar{6}68 + \left [ 0.9973 - x \! \left ( \mathrm{^{20}Ne} \right ) \right ] \left ( 21.9914 \right ) \\[1.5ex] 20.1\bar{2}33 &= \left ( 19.9924 \right ) ~ x \! \left ( \mathrm{^{20}Ne} \right ) + \left [ 0.9973 - x \! \left ( \mathrm{^{20}Ne} \right ) \right ] \left ( 21.9914 \right ) \\[1.5ex] 20.1\bar{2}33 &= \left ( 19.9924 \right ) ~ x \! \left ( \mathrm{^{20}Ne} \right ) + 21.9\bar{3}20 - \left ( 21.9914 \right ) ~ x \! \left ( \mathrm{^{20}Ne} \right ) \\[1.5ex] -1.8\bar{0}87 &= \left ( 19.9924 \right ) ~ x \! \left ( \mathrm{^{20}Ne} \right ) - \left ( 21.9914 \right ) ~ x \! \left ( \mathrm{^{20}Ne} \right ) \\[1.5ex] -1.8\bar{0}87 &= \left ( -1.999\bar{0}00 \right ) ~ x \! \left ( \mathrm{^{20}Ne} \right ) \\[1.5ex] x \! \left ( \mathrm{^{20}Ne} \right ) &= 0.90\bar{4}80 \end{align*}\]
Convert the relative isotopic mass of 20Ne to a percent.
\[\begin{align*} x \! \left ( \mathrm{^{20}Ne} \right ) \% &= x \! \left ( \mathrm{^{20}Ne} \right ) \times 100~\% \\[1.5ex] &= 0.90\bar{4}80 \times 100~\% \\[1.5ex] &= 90.\bar{4}80~\% \\[1.5ex] &= 90.5~\% \end{align*}\]
We have now determined the percent abundance for neon-20. We can now find the fractional abundance for neon-22.
\[\begin{align*} x \! \left ( \mathrm{^{22}Ne} \right ) &= 0.9973 - x \! \left ( \mathrm{^{20}Ne} \right ) \\[1.5ex] &= 0.9973 - 0.90\bar{4}80 \\[1.5ex] &= 0.092\bar{5}0 \end{align*}\]
Convert the fractional abundance of neon-22 to a percent.
\[\begin{align*} x \left ( \mathrm{^{22}Ne} \right ) \% &= x \left ( \mathrm{^{22}Ne} \right ) \times 100~\% \\[1.5ex] &= 0.092\bar{5}0 \times 100~\% \\[1.5ex] &= 9.25~\% \end{align*}\]
The percent abundances for the three isotopes of neon, in summary:
Rationalize the Error
Why do the percent abundances not add up to exactly 100%? One source of error was the standard atomic weight of neon used in this procedure (20.18). The value used was only reported to two decimal places (obtained off the periodic table). Perhaps a standard atomic weight with more precision would have resulted with a more accurate set of percent abundances. According to NIST, the standard atomic weight of neon is 20.1797(6).