# 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**,** A _{r }**, 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}

**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{\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)

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 (M_{r}) 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 M_{r} 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 M_{r} 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 M_{r}, **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.

**[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.

## You May Also Like...

### MME Learning Portal

Online exams, practice questions and revision videos for every GCSE level 9-1 topic! No fees, no trial period, just totally free access to the UK’s best GCSE maths revision platform.