How to Calculate Relative Atomic Mass from Three Isotopes
Learn how to calculate relative atomic mass from three isotopes using percentage abundances and mass-spectrum intensities. Includes worked examples explaining isotope masses, ionic charge, and common calculation errors.
A mass spectrum shows three peaks at \(m/z=24\), \(25\), and \(26\). The peak at 24 is much taller than the other two. What is the relative atomic mass of the element?
A tempting answer is to average the three numbers:
\[ \frac{24+25+26}{3}=25 \]
But that assumes all three isotopes are equally abundant. They aren’t. The tallest peak represents the isotope that contributes the most atoms to the sample, so it must have the greatest influence on the average.
To calculate relative atomic mass from three isotopes, you need three pieces of information: the mass of each isotope, its relative abundance, and, if the data come from a mass spectrum, the charge of the ions being detected.
01The calculation: three isotopes
The relative atomic mass, \(A_r\), is the weighted average of the masses of an element’s isotopes.
For three isotopes, the calculation is:
\[ A_r=m_1f_1+m_2f_2+m_3f_3 \]
Where:
- \(m_1,m_2,m_3\) are the masses of the three isotopes.
- \(f_1,f_2,f_3\) are their fractional abundances.
- \(A_r\) is the relative atomic mass.
A fractional abundance is simply a percentage written as a decimal.
For example:
\[ 79\%=0.79 \]
If the question gives percentages directly, use:
\[ A_r= \frac{m_1p_1+m_2p_2+m_3p_3}{100} \]
Here, \(p_1,p_2,p_3\) are percentage abundances.
If the question gives relative peak intensities instead, divide each intensity by the total intensity.
The important idea is that every isotope contributes according to how common it is.
Predict before calculating
Suppose an element has three isotopes:
| Isotope mass | Abundance |
|---|---|
| 24 | 79% |
| 25 | 10% |
| 26 | 11% |
Should the relative atomic mass be closer to 24, 25, or 26?
It should be closest to 24 because 79% of the atoms have that mass.
In fact:
\[ \begin{aligned} A_r&=(24)(0.79)+(25)(0.10)+(26)(0.11)\\ &=18.96+2.50+2.86\\ &=24.32 \end{aligned} \]
Notice that the answer is not the midpoint of the three isotope masses. It is pulled towards the most abundant isotope.
This is the central calculation. The next step is understanding where the isotope masses and abundances come from.
02Read the isotope notation correctly
Before interpreting a mass spectrum, make sure you can distinguish mass number, atomic number, and ionic charge.
Consider the following magnesium ion:
\[ {}^{24}_{12}\mathrm{Mg}^{2+} \]
Each part tells you something different.
| Symbol | Meaning |
|---|---|
| 24 | Mass number: protons + neutrons |
| 12 | Atomic number: number of protons |
| \(2+\) | Ion has lost two electrons |
The number of neutrons is:
\[ N=A-Z=24-12=12 \]
The magnesium ion has 10 electrons because it has lost two of its original 12 electrons.
Now consider three magnesium isotopes:
\[ {}^{24}_{12}\mathrm{Mg} \]
\[ {}^{25}_{12}\mathrm{Mg} \]
\[ {}^{26}_{12}\mathrm{Mg} \]
All three have 12 protons. Their neutron numbers are 12, 13, and 14, respectively.
They are different isotopes of the same element because their neutron numbers differ.
An important distinction: changing the charge does not change the isotope. Removing two electrons produces \(\mathrm{Mg}^{2+}\), but the nucleus still contains 12 protons and 12 neutrons for magnesium-24.
For a refresher on the complete notation, see Atomic Number and Mass Number Notation for HSC Chemistry.
03How a mass spectrometer identifies isotopes
Why do three isotopes produce three separate peaks?
Chemists need a way to distinguish atoms that belong to the same element but have different masses. We cannot simply look at a collection of atoms and count their neutrons.
A mass spectrometer solves this problem by turning atoms into ions and separating them according to their mass-to-charge ratios.
Follow the path through the apparatus
A simplified magnetic-sector mass spectrometer works in the following order.
Step 1: The sample enters the instrument.
The sample is introduced into the instrument and converted into gaseous particles. The instrument operates under vacuum to minimise collisions with other particles.
Step 2: The particles are ionised.
An electron beam removes electrons from some of the atoms, producing positive ions.
For example:
\[ \mathrm{Mg}\rightarrow\mathrm{Mg}^{+}+e^- \]
The ions can then be accelerated using an electric field.
Step 3: The ions are separated.
The positive ions enter a magnetic field.
Their paths bend according to their mass-to-charge ratios. For ions with the same charge, lighter ions follow more strongly curved paths than heavier ions under the same instrument conditions.
This allows different isotopes to be separated.
Step 4: A detector records the ions.
The separated ions reach a detector, which records their signals.
The instrument produces a mass spectrum.
There are two important pieces of information on this graph:
- Horizontal axis: mass-to-charge ratio, \(m/z\).
- Vertical axis: relative intensity, which indicates the relative abundance of the detected ions in a simplified HSC mass spectrum.
The position of a peak tells you about the ion’s mass-to-charge ratio. Its height indicates how abundant that ion is relative to the others.
The instrument does not directly calculate relative atomic mass. It gives you the information needed to calculate it.
04The important trap: mass-to-charge ratio is not always isotope mass
This is where students can lose marks.
A mass spectrum measures mass-to-charge ratio, not mass alone.
The relationship is:
\[ \frac{m}{z}= \frac{\text{ion mass}}{\text{charge number}} \]
Here, \(z\) is the magnitude of the ionic charge, not the atomic number \(Z\).
For a singly charged positive ion:
\[ z=1 \]
Therefore, the mass-to-charge ratio is approximately equal to the isotope mass in atomic mass units.
For a doubly charged ion:
\[ z=2 \]
The mass-to-charge ratio is approximately half the isotope mass.
For example:
| Ion | Approximate mass | Charge | \(m/z\) |
|---|---|---|---|
| \({}^{24}\mathrm{Mg}^{+}\) | 24 | +1 | 24 |
| \({}^{24}\mathrm{Mg}^{2+}\) | 24 | +2 | 12 |
| \({}^{25}\mathrm{Mg}^{+}\) | 25 | +1 | 25 |
| \({}^{25}\mathrm{Mg}^{2+}\) | 25 | +2 | 12.5 |
| \({}^{26}\mathrm{Mg}^{+}\) | 26 | +1 | 26 |
| \({}^{26}\mathrm{Mg}^{2+}\) | 26 | +2 | 13 |
Notice something important: magnesium-24 has not suddenly become magnesium-12 when it forms a \(2+\) ion.
Its mass number remains 24. Only its mass-to-charge ratio changes.
At HSC calculation precision, the small mass difference caused by removing electrons is ignored.
Decision rule: Before using a peak position as an isotope mass, check the charge state. If the ions are all singly charged, you can generally use the peak positions directly. If they are doubly charged, multiply the \(m/z\) values by two.
05Worked example: calculating relative atomic mass from three magnesium isotopes
Question
An illustrative mass spectrum of magnesium contains three singly charged isotope ions. The following data are obtained:
| Isotope ion | Isotopic mass (u) | Relative intensity |
|---|---|---|
| \({}^{24}\mathrm{Mg}^{+}\) | 23.985 | 790 |
| \({}^{25}\mathrm{Mg}^{+}\) | 24.986 | 100 |
| \({}^{26}\mathrm{Mg}^{+}\) | 25.983 | 110 |
Calculate the relative atomic mass of magnesium.
Step 1Identify the isotope masses.
All three ions have a charge of \(+1\).
Therefore, their mass-to-charge ratios correspond approximately to their isotope masses.
We can use the supplied isotopic masses directly.
Step 2Calculate the total relative intensity.
\[ I_{\text{total}}=790+100+110 \]
\[ I_{\text{total}}=1000 \]
Step 3Convert the intensities into fractional abundances.
For magnesium-24:
\[ f_{24}=\frac{790}{1000}=0.790 \]
For magnesium-25:
\[ f_{25}=\frac{100}{1000}=0.100 \]
For magnesium-26:
\[ f_{26}=\frac{110}{1000}=0.110 \]
Check:
\[ 0.790+0.100+0.110=1.000 \]
The fractional abundances are correct.
Step 4Calculate the weighted average.
\[ A_r=m_1f_1+m_2f_2+m_3f_3 \]
Substitute the values:
\[ \begin{aligned} A_r={}&(23.985)(0.790)\\ &+(24.986)(0.100)\\ &+(25.983)(0.110) \end{aligned} \]
Calculate each contribution:
\[ \begin{aligned} A_r={}&18.94815\\ &+2.49860\\ &+2.85813 \end{aligned} \]
Therefore:
\[ \boxed{A_r=24.30488} \]
To three decimal places:
\[ \boxed{A_r=24.305} \]
The relative atomic mass is dimensionless, so it has no unit.
Step 5Check whether the answer makes sense.
The answer is between the smallest and largest isotope masses.
It is also much closer to 23.985 than to either heavier isotope because magnesium-24 makes up 79% of the sample.
That is exactly what we would expect.
06Worked example: when all three ions have a +2 charge
Now change one condition.
The three isotope peaks still have the same relative intensities, but the ions are doubly charged.
Question
A simplified mass spectrum contains three peaks at:
| \(m/z\) | Relative intensity |
|---|---|
| 12.0 | 790 |
| 12.5 | 100 |
| 13.0 | 110 |
All three peaks correspond to doubly charged magnesium ions.
Using the approximate isotope mass numbers, calculate the relative atomic mass.
Step 1Correct for the ionic charge.
All ions have a charge of \(+2\).
Therefore:
\[ \text{Isotope mass}\approx\frac{m}{z}\times2 \]
The isotope masses are:
\[ 12.0\times2=24 \]
\[ 12.5\times2=25 \]
\[ 13.0\times2=26 \]
We have identified the three isotope mass numbers.
Step 2Calculate the fractional abundances.
The total intensity is:
\[ 790+100+110=1000 \]
Therefore:
\[ f_1=0.790,\qquad f_2=0.100,\qquad f_3=0.110 \]
Step 3Calculate relative atomic mass.
\[ \begin{aligned} A_r={}&(24)(0.790)\\ &+(25)(0.100)\\ &+(26)(0.110) \end{aligned} \]
\[ A_r=18.96+2.50+2.86 \]
\[ \boxed{A_r=24.32} \]
The answer is an approximate relative atomic mass because we used whole-number mass numbers rather than precise isotopic masses.
The trap here is using 12, 12.5, and 13 directly in the weighted-average equation. That would give 12.16, which is incorrect because those numbers represent mass-to-charge ratios, not isotope masses.
07A quick decision process for exam questions
Different questions can present the same isotope information in different ways.
Use this table to decide what to do before starting the calculation.
| Information supplied | What you should do |
|---|---|
| Percentage abundance | Divide each percentage by 100 |
| Fractional abundance | Use the fractions directly |
| Relative intensities | Divide each intensity by the total |
| Relative intensity ratio | Normalise the ratio so the fractions total 1 |
| Mass numbers | Use them as approximate isotope masses |
| Precise isotopic masses | Use the supplied masses instead of rounding to mass numbers |
| Singly charged ions | Use the \(m/z\) values directly, approximately |
| Doubly charged ions | Multiply \(m/z\) by 2 to recover the isotope mass |
Before submitting your answer, check three things:
- Do the fractional abundances add to 1?
- Is the calculated relative atomic mass between the smallest and largest isotope masses?
- Is the answer closer to the more abundant isotope?
If one of these checks fails, revisit your working.
For more practice with the general weighted-average method, see Relative Atomic Mass from Isotopic Composition.
08Questions and solutions
Question 1
An element has three naturally occurring isotopes with the following composition:
| Isotopic mass | Percentage abundance |
|---|---|
| 10.0 | 70.0% |
| 11.0 | 20.0% |
| 12.0 | 10.0% |
Calculate the relative atomic mass of the element.
Solution 1
The relative atomic mass is \(\boxed{10.40}\).
We need to calculate a weighted average because the three isotopes have different abundances.
First, convert the percentages into fractions:
\[ 0.700,\qquad0.200,\qquad0.100 \]
Substitute into the weighted-average equation:
\[ \begin{aligned} A_r={}&(10.0)(0.700)\\ &+(11.0)(0.200)\\ &+(12.0)(0.100) \end{aligned} \]
\[ A_r=7.00+2.20+1.20 \]
\[ \boxed{A_r=10.40} \]
The result is closer to 10.0 because the lightest isotope makes up 70% of the sample.
A simple average of the three masses would give 11.0, incorrectly assuming equal abundance.
Question 2
An element has three singly charged isotopes with the following precise isotopic masses:
| Isotopic mass (u) | Relative intensity |
|---|---|
| 62.93 | 6 |
| 63.93 | 2 |
| 65.93 | 2 |
Calculate the relative atomic mass of the element.
Explain why the relative intensities must be converted into fractional abundances before calculating the answer.
Solution 2
The relative atomic mass is \(\boxed{63.73}\).
The intensities are given as a ratio of 6:2:2, not as percentages.
We must first calculate the total intensity:
\[ I_{\text{total}}=6+2+2=10 \]
Convert each intensity into a fraction:
\[ f_1=\frac{6}{10}=0.6 \]
\[ f_2=\frac{2}{10}=0.2 \]
\[ f_3=\frac{2}{10}=0.2 \]
The fractions total 1, as required.
Now calculate:
\[ \begin{aligned} A_r={}&(62.93)(0.6)\\ &+(63.93)(0.2)\\ &+(65.93)(0.2) \end{aligned} \]
\[ A_r=37.758+12.786+13.186 \]
\[ \boxed{A_r=63.73} \]
The temptation is to treat 6, 2, and 2 as percentage abundances. That would be incorrect because they are relative intensity values.
The correct approach is to divide each intensity by the total intensity, producing fractions of 0.6, 0.2, and 0.2.
Question 3
An element M has exactly three naturally occurring isotopes with mass numbers 40, 42, and 44.
A mass spectrum produces the following peaks:
| \(m/z\) | Relative intensity |
|---|---|
| 20 | 7 |
| 21 | 2 |
| 22 | 1 |
All detected ions have a charge of \(+2\).
Calculate the approximate relative atomic mass of element M.
Solution 3
The approximate relative atomic mass is \(\boxed{40.8}\).
The important detail is that the ions are doubly charged. We cannot use the \(m/z\) values directly as isotope masses.
First, multiply each peak position by two:
\[ 20\times2=40 \]
\[ 21\times2=42 \]
\[ 22\times2=44 \]
These are the three isotope mass numbers.
Next, calculate the total intensity:
\[ 7+2+1=10 \]
The fractional abundances are:
\[ 0.7,\qquad0.2,\qquad0.1 \]
Now calculate the weighted mean:
\[ \begin{aligned} A_r={}&(40)(0.7)\\ &+(42)(0.2)\\ &+(44)(0.1) \end{aligned} \]
\[ A_r=28.0+8.4+4.4 \]
\[ \boxed{A_r=40.8} \]
The key distinction is between mass number and mass-to-charge ratio.
The mass number of the first isotope is 40, even though the instrument records an \(m/z\) value of 20.
Question 4
A magnesium sample contains three isotopes with the following abundances:
| Isotope | Percentage abundance |
|---|---|
| Magnesium-24 | 78.99% |
| Magnesium-25 | 10.00% |
| Magnesium-26 | 11.01% |
A student uses the mass numbers 24, 25, and 26 to calculate a relative atomic mass of 24.32.
A second student is given the following more precise isotopic masses:
| Isotope | Precise isotopic mass (u) |
|---|---|
| Magnesium-24 | 23.9850 |
| Magnesium-25 | 24.9858 |
| Magnesium-26 | 25.9826 |
The second student obtains approximately 24.305.
Explain why the two answers differ and identify which answer should be used when precise isotopic masses are supplied.
Solution 4
Both calculations are mathematically valid, but the result using precise isotopic masses is more accurate for the supplied data.
The difference occurs because mass numbers are whole-number counts of protons and neutrons. They are not exact measurements of isotope mass.
First, calculate using the mass numbers.
Convert the percentages to fractions:
\[ 0.7899,\qquad0.1000,\qquad0.1101 \]
Then:
\[ \begin{aligned} A_r={}&(24)(0.7899)\\ &+(25)(0.1000)\\ &+(26)(0.1101) \end{aligned} \]
\[ A_r=18.9576+2.5000+2.8626 \]
\[ A_r=24.3202 \]
Therefore:
\[ \boxed{A_r\approx24.32} \]
Now use the precise isotopic masses:
\[ \begin{aligned} A_r={}&(23.9850)(0.7899)\\ &+(24.9858)(0.1000)\\ &+(25.9826)(0.1101) \end{aligned} \]
\[ A_r=18.9457515+2.4985800+2.86068426 \]
\[ \boxed{A_r\approx24.305} \]
The two answers differ because the precise isotopic masses are not exactly equal to their corresponding mass numbers.
If a question supplies precise isotopic masses, use them. If it only supplies mass numbers, use those as approximations.
Do not replace more precise information with whole-number approximations unnecessarily.
Question 5
An element contains exactly three naturally occurring isotopes with mass numbers 24, 25, and 26.
A mass spectrometry investigation determines that the element has a relative atomic mass of 24.40.
A student claims that the isotope composition must be:
- 60% isotope 24
- 40% isotope 25
- 0% isotope 26
The student argues that this is the only combination capable of producing an average of 24.40.
Is the student correct?
Construct a different valid isotope composition that produces the same relative atomic mass. Explain why knowing the relative atomic mass alone is insufficient to determine all three isotope abundances.
Solution 5
The student is incorrect. More than one isotope composition can produce a relative atomic mass of 24.40.
For example, both of the following mixtures produce the same result:
| Isotope mass | Mixture A | Mixture B |
|---|---|---|
| 24 | 65% | 70% |
| 25 | 30% | 20% |
| 26 | 5% | 10% |
| Total | 100% | 100% |
Let’s verify both mixtures.
Mixture A:
\[ \begin{aligned} A_r={}&(24)(0.65)\\ &+(25)(0.30)\\ &+(26)(0.05) \end{aligned} \]
\[ A_r=15.60+7.50+1.30 \]
\[ \boxed{A_r=24.40} \]
Mixture B:
\[ \begin{aligned} A_r={}&(24)(0.70)\\ &+(25)(0.20)\\ &+(26)(0.10) \end{aligned} \]
\[ A_r=16.80+5.00+2.60 \]
\[ \boxed{A_r=24.40} \]
Both mixtures have exactly the same relative atomic mass, even though their isotope abundances are different.
Why is this possible?
There are three unknown isotope abundances:
\[ f_{24},\qquad f_{25},\qquad f_{26} \]
We know that their total must equal 1:
\[ f_{24}+f_{25}+f_{26}=1 \]
We also know their weighted average:
\[ 24f_{24}+25f_{25}+26f_{26}=24.40 \]
These are two independent equations containing three unknowns.
Rearranging the first equation gives:
\[ f_{24}=1-f_{25}-f_{26} \]
Substitute into the weighted-average equation:
\[ 24(1-f_{25}-f_{26}) +25f_{25}+26f_{26}=24.40 \]
Simplifying:
\[ 24+f_{25}+2f_{26}=24.40 \]
Therefore:
\[ f_{25}+2f_{26}=0.40 \]
This equation allows multiple valid combinations of isotope abundances.
For example, if \(f_{26}=0.05\), then:
\[ f_{25}=0.40-2(0.05)=0.30 \]
And:
\[ f_{24}=1-0.30-0.05=0.65 \]
This gives Mixture A.
If \(f_{26}=0.10\), we obtain Mixture B.
The deeper point is that a relative atomic mass gives you the average composition, not the complete isotope distribution. You need additional experimental information, such as the individual peak intensities from a mass spectrum, to determine all three abundances.
This is why the mass spectrum matters. It provides information that the average alone cannot.
09Where to go next
Calculating relative atomic mass from three isotopes involves more than substituting numbers into a formula.
You need to identify the isotope masses, interpret their relative abundances, and check whether the instrument’s charge state changes the peak positions.
Once you can do those three things, the calculation itself is straightforward.
The next useful skill is interpreting a complete mass spectrum: identifying which peaks belong to different isotopes, distinguishing charge states, and using the resulting abundance data to explain the relative atomic mass shown on the periodic table.