Mass Spectrometry

While Dalton’s model stated that all atoms of an element are identical, this is not exactly correct. Atoms of the same element can have varying numbers of neutrons (isotopes), even if they have the same number of protons (and electrons, in a neutral atom). Mass spectrometry is a technique to separate species by mass. As such, it can detect isotopes and their ratios in an element.

Example 1.

Mass spectrometry data shows that copper is composed of two stable isotopes in the following percent abundances: \(70\%\) \(\require{mhchem}\ce{^{63}Cu}\) and \(30\%\) \(\require{mhchem}\ce{^{65}Cu}\). Based on these values, what is the average atomic mass of copper?

A. \(63.6\text{ amu}\)

B. \(6360\text{ amu}\)

C. \(1.36\text{ amu}\)



A. \(63.6\text{ amu}\)

The average atomic mass of an element (the one shown on the periodic table) is a weighed average of the isotopes.

\(\begin{align}\text{atomic mass} &= [\text{fraction of isotope A}\times\text{mass of isotope A}] + [\text{fraction of isotope B}\times\text{mass of isotope B}] \\ \text{atomic mass} &= [0.70\times 63\text{ amu}] + [0.30\times 65\text{ amu}] = 63.6\text{ amu} \end{align}\)

Example 2.

Chlorine is composed of two stable isotopes: \(\require{mhchem}\ce{^{35}Cl}\) and \(\require{mhchem}\ce{^{37}Cl}\). If the average atomic mass of \(\require{mhchem}\ce{Cl}\) is \(35.45\text{ amu}\), what is the natural abundance of \(\require{mhchem}\ce{^{37}Cl}\)?

A. \(77\%\)

B. \(23\%\)

C. \(51\%\)



B. \(23\%\) \(\require{mhchem}\ce{^{37}Cl}\)

Since there are two stable isotopes, the sum of their fractions must add to the total (\(1\)). The equation can be rearranged to solve for the fraction of one isotope at a time.

 \(\begin{align}\text{fraction } ^{35}\text{Cl } + \text{fraction } ^{37}\text{Cl } &= 1\\ 1 - \text{fraction } ^{37}\text{Cl }& = \text{fraction } ^{35}\text{Cl } \end{align}\)

\(\begin{align}\text{atomic mass} &= [\text{fraction of }^{35} \text{Cl}\times\text{mass of }^{35}\text{Cl}] + [\text{fraction of }^{37} \text{Cl}\times\text{mass of }^{37}\text{Cl}] \\ 35.45\text { amu} &= [(1-\text{fraction of }^{37} \text{Cl})\times 35\text{ amu}] + [\text{fraction of }^{37} \text{Cl}\times 37\text{ amu}] \end{align}\)

For simplicity, let fraction of \(\require{mhchem}\ce{^{37}Cl} = x\)

\(\begin{align}35.45\text { amu} &= [(1-x)\times 35\text{ amu}] + [x\times 37\text{ amu}] \\35.45\text { amu} &= 35-35x+37x \\0.45 &= 2x\\0.225 &=x = 22.5\%\end{align}\)

Example 3.

How many neutrons are in the nucleus of an atom of \(\require{mhchem}\ce{^{13}C}\)?

A. \(13\)

B. \(6\)

C. \(7\)



C. \(7\)

The mass number (\(13\)) is the sum of the number of protons and neutrons. All carbon atoms will have six protons, so there will be \(7\) neutrons.