# The Mass Spectrum and Relative Atomic Mass

A LevelAS LevelAQA

## The Mass Spectrum and Relative Atomic Mass

A mass spectrometer creates the mass spectrum. It can be used to find the relative atomic mass of an element.

The relative atomic mass, Ar , is the weighted average mass of an atom of an element, relative to one twelfth of the mass of an atom of carbon-12.

## Calculating Relative Atomic Mass

There are two equations that can help to calculate relative atomic mass.

$\text{Relative atomic mass}= \dfrac{\Sigma\left(\text{isotopic mass}\times\%\text{ abundance}\right)}{100}$

$\text{Relative atomic mass}= \dfrac{\Sigma\left(\text{isotopic mass}\times\text{relative abundance}\right)}{\text{total relative abundance}}$

If given percentage abundance, the first equation is used, if given relative abundance the second equation is used.

Example: Calculate the relative atomic mass of this element using its % abundance values.

\begin{aligned}A_{r}&=\dfrac{\Sigma\left(\text{isotopic mass}\times\%\text{ abundance}\right)}{100}\\ &=\dfrac{\left(54\times5.8\right)+\left(56\times91.8\right)+\left(57\times2.1\right)+\left(58\times0.3\right)}{100}\\&= \dfrac{313.2+5140.8+119.7+17.4}{100}\\& = \dfrac{5591.1}{100} \\&= 55.91\end{aligned}

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## Using Mass Spectra to Calculate Relative Atomic Mass

You could also be asked to calculate relative atomic mass based on a mass spectrum. Mass spectra are created from data collected by the mass spectrometer. It shows the relative abundance against m/z, the mass charge ratio.

Example: Calculate the relative atomic mass of this element based on its mass spectrum.

$A_{r}=\dfrac{\Sigma\left(\text{isotopic mass}\times\text{relative abundance}\right)}{\text{total relative abundance}}$

$A_{r}=\dfrac{\left(23\times10\right)+\left(100\times11\right)}{100+23}= \dfrac{230+1100}{123}= \dfrac{2330}{123} = 10.81$
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## Chlorine and Bromine

Chlorine and Bromine are both diatomic molecules with two isotopes.

• Chlorine: $\text{Cl}^{35} \left(75\%\right)$ and $\text{Cl}^{37} \left(25\%\right)$
• Bromine: $\text{Br}^{79} \left(50\%\right)$ and $\text{Br}^{81} \left(50\%\right)$

The height of the peak for bromine at $\text{m/z} 160$ will be due to the fact that there is double the probability of a $\text{Br}^{81}\text{Br}^{79}$ molecule.

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## Fragmentation

If a molecule goes through the mass spectrometer it can be ionised by one of two processes, electron impact or electrospray ionisation.

If electron impact ionisation is used, a molecule may break up into fragments which will cause multiple peaks.

Relative molecular mass (Mr) is the average mass of a molecule of an element or a compound compared to one twelfth the mass of an atom of carbon.

To find the Mr of the molecule, you will need to find the peak with the biggest m/z value.

The figure below shows the mass spectra for pentane, by looking at the peak with the biggest m/z value, we can work out that the ion has an Mr of $72$.

If electrospray ionisation is used, there will be no fragmentation and so there will be only one peak caused by the intact molecule. Because the molecule will have a $\text{H}^+$ ion added to it, to determine the Mr, $1$ will need to be subtracted to account for the $\text{H}^+$ ion.

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## Example: Calculating Relative Atomic Mass Using Mass Spectra

Calculate the relative atomic mass of this element based on its mass spectrum and give the species responsible for the peak at $\text{m/z }24$.

[3 marks]

\begin{aligned}A_{r}&= \dfrac{\Sigma\left(\text{isotopic mass}\times\%\text{ abundance}\right)}{100}\\&=\dfrac{\left(24\times79\right)+\left(25\times10\right)+\left(26\times11\right)}{100}\\&= \dfrac{1896+250+286}{100}\\ &= \dfrac{2432}{100} = 24.32\end{aligned}

Species responsible for peak =  $^{24}\text{Mg}^+$

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## The Mass Spectrum and Relative Atomic Mass Example Questions

$^{35}\text{Cl}^{37}\text{Cl}^+$

$\text{Relative atomic mass} = \dfrac{\Sigma\left(\text{isotopic mass}\times\%\text{ abundance}\right)}{100}$

\begin{aligned}6.90&=\frac{(6\times y)+(7\times[100-y])}{100}\\\text{}\\ 690&=6y+(7\times[100-y])\\ 690&=6y+700-7y\\690&=-y+700\\-10&=-y\\10&=y\end{aligned}

$\%\text{ abundance } 6\text{-Li} = 10$

$\%\text{ abundance } 7\text{-Li} = 90$

$\text{Relative atomic mass} = \dfrac{\Sigma\left(\text{isotopic mass}\times\%\text{ abundance}\right)}{100}$

$\text{Relative atomic mass} = \dfrac{\left(78\times0.4\right)+\left(80\times2.3\right)+\left(82\times11.6\right)+\left(83\times11.5\right)+\left(84\times57\right)+\left(85\times17.3\right)}{100}$

$\text{Relative atomic mass} = \dfrac{31.2+184+951.2+954.5+4788+1470.5}{100}$

$\text{Relative atomic mass} = \dfrac{8379.4}{100} = 83.8$

The element is krypton.

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