The Mass Spectrum and Relative Atomic Mass
The Mass Spectrum and Relative Atomic Mass Revision
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.
![](https://mmerevise.co.uk/app/uploads/2022/09/Relative-atomic-mass-example-e1663858570269.png)
\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}
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.
![](https://mmerevise.co.uk/app/uploads/2022/09/Mass-spectrum-1-1024x691.png)
![](https://mmerevise.co.uk/app/uploads/2022/09/Mass-spectrum-1-1024x691.png)
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
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)
![](https://mmerevise.co.uk/app/uploads/2022/10/Mass-Specs-e1666264898258.png)
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.
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.
![](https://mmerevise.co.uk/app/uploads/2022/09/Mass-spectrum-4-e1663857240864.png)
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.
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.
![](https://mmerevise.co.uk/app/uploads/2022/09/Mass-spectrum-2-e1663857569461.png)
[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}^+
The Mass Spectrum and Relative Atomic Mass Example Questions
Question 1: The mass spectrum shown is for chlorine \left(\text{Cl}_2\right). Identify the species that causes a peak at \text{m/z } 72.
[1 mark]
Question 2: Lithium has two isotopes of 6\text{-Li} and 7\text{-Li}. The R.A.M of lithium is 6.9. Calculate the percentage abundance of each of these isotopes.
[4 marks]
\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
Question 3: Calculate the relative atomic mass for the sample and identify the element.
[3 marks]
\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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