1.8-1.10 - Mass Spectra And Relative Mass

Reading A Mass Spectrum

Peak position identifies mass-to-charge ratio; peak height measures relative abundance. Keep these two readings separate.

The problem

A sample of an element gives two peaks in a mass spectrum:

The instrument is not saying that one atom has a mass of 35.5. It is saying that two kinds of atom are present, and their ions are detected at different $m/z$ values.

What the instrument data means

For the singly charged isotope ions used here:

• each peak represents ions of one isotope
• $m/z$ is mass-to-charge ratio
• for singly charged ions, $m/z$ has the same numerical value as the isotope mass used in the calculation
• peak height or peak area represents relative abundance

So a peak at $m/z = 35$ with height 75 means that isotope-35 contributes 75 relative parts to the sample. A peak at $m/z = 37$ with height 25 means isotope-37 contributes 25 relative parts. The relative atomic mass is the weighted mean of those isotope masses, not the middle of the two masses.

Calculating Relative Atomic Mass

Weight each isotope mass by its abundance, add the contributions and divide by the total abundance.

Worked example: magnesium isotope data

A sample of magnesium contains three isotopes.

Calculate the relative atomic mass of magnesium from these data.

The chemical model is a weighted mean: the most abundant isotope should pull the answer closest to its mass. Here $\ce{^{24}Mg}$ is much more common than the other two, so the answer should be just above 24, not near 25.

$A_r = (1895.76 + 250.00 + 286.26) / 100$

$A_r = 2432.02 / 100$

$A_r = 24.3202$

To a sensible number of significant figures:

$A_r(\ce{Mg}) = 24.32$

The sense check matters. The answer is close to 24 because almost 79% of the atoms are $\ce{^{24}Mg}$.

Same calculation from peak heights

A mass spectrum of element X has these isotope peaks:

The heights happen to sum to 100, so the calculation is:

$A_r = (199.0 + 881.1) / 100$

$A_r = 10.801$

If the heights had been 50 and 201 instead, the chemical mixture would be similar, but the denominator would be $251$.

Arithmetic mean of repeat results

Three equally weighted repeat determinations give relative atomic masses 10.80, 10.84 and 10.82. Their arithmetic mean is $(10.80+10.84+10.82)/3=10.82$. Divide by the number of measurements. This is different from weighting isotope masses by unequal isotope abundances; a simple mean of isotope masses works only when their abundances are equal.

Finding Abundance From Relative Atomic Mass

Sometimes the relative atomic mass is known and the isotope abundances are unknown. The model is still the weighted mean, but one abundance becomes the unknown.

Worked example: two isotopes

Element Y has two isotopes, $\ce{^{10}Y}$ and $\ce{^{11}Y}$. Its relative atomic mass is 10.81. Calculate the percentage abundance of each isotope.

Let the percentage abundance of $\ce{^{10}Y}$ be $x$.

Then the percentage abundance of $\ce{^{11}Y}$ is $100 - x$.

Use the weighted mean:

$10.81 = [10x + 11(100 - x)] / 100$

Multiply by 100:

$1081 = 10x + 1100 - 11x$

$1081 = 1100 - x$

$x = 19$

So:

$\ce{^{10}Y}$ = 19%

$\ce{^{11}Y}$ = 81%

This is chemically sensible because the relative atomic mass, 10.81, is much closer to 11 than to 10, so the isotope with mass 11 must be more abundant.

Common-error contrast

Tempting but wrong: "10.81 is 81% of the way from 10 to 11, so $\ce{^{10}Y}$ is 81%."

Correct: being 81% of the way towards mass 11 means the heavier isotope is 81% abundant. The average moves towards the isotope that is more common.

Predicting Diatomic Molecule Spectra

Diatomic molecules create a new layer of reasoning because each molecule contains two atoms. The mass peak for the molecule depends on the isotope combination inside that molecule.

The chlorine pattern

Chlorine has two main isotopes, approximately in a $3 : 1$ abundance ratio:

$\ce{^{35}Cl}$ : $\ce{^{37}Cl}$ = 3 : 1

So the probabilities are:

P($\ce{^{35}Cl}$) = 3/4

P($\ce{^{37}Cl}$) = 1/4

For chlorine molecules, there are three possible masses:

The middle peak is doubled because there are two ways to make a mixed molecule: the lighter isotope can be first or second. The instrument does not care about order, so both mixed combinations contribute to the same $m/z = 72$ peak.

Therefore the predicted molecular-ion peak pattern for $\ce{Cl2}$ is:

m/z 70 : 72 : 74 = 9 : 6 : 1

Worked comparison: equal isotope abundances

Element E has two isotopes, $\ce{^{79}E}$ and $\ce{^{81}E}$, in a $1 : 1$ ratio. The three molecular-ion peaks for $\ce{E2}$ follow from the pair probabilities below.

Masses:

$79 + 79 = 158$

$79 + 81 = 160$

$81 + 81 = 162$

Probabilities:

P(79-79) = 1/2 × 1/2 = 1/4

P(79-81 or 81-79) = 2 × 1/2 × 1/2 = 1/2

P(81-81) = 1/2 × 1/2 = 1/4

So the peak-height ratio is:

m/z 158 : 160 : 162 = 1 : 2 : 1

Common error: atomic and molecular peaks

The $\ce{Cl2}$ peak at $m/z = 72$ is not evidence for a chlorine isotope with mass 36.

A strong explanation says that $m/z = 72$ comes from one $\ce{^{35}Cl}$ atom plus one $\ce{^{37}Cl}$ atom in the same molecule. The peak belongs to a molecule ion, not to a single chlorine atom.

[DIAGRAM: asset_name: Chlorine Molecule Mass Spectrum; asset_slug: edexcel_a_level_chemistry_l003_chlorine_molecule_mass_spectrum; recommended_method: matplotlib; description: Predicted mass spectrum for chlorine molecules, showing m/z 70, 72 and 74 peaks with relative peak heights 9, 6 and 1.]

Using The Molecular Ion Peak

Mass spectrometry can also give the relative molecular mass of a molecule. In this part of the specification, the allowed reasoning is deliberately narrow: use the $m/z$ value for the molecular ion, $\ce{M+}$, to find $M_r$.

What $\ce{M+}$ means

It forms like this:

$\ce{M(g) -> M+(g) + e-}$

For a singly charged molecular ion:

m/z of $\ce{M+}$ = relative molecular mass

So if the molecular ion peak is at $m/z = 46$, the molecule has:

$M_r = 46$

This does not by itself prove the full structure of the molecule. Many different molecules can have the same relative molecular mass. Fragment patterns and high-resolution mass spectrometry are later analytical tools; here the focus is the molecular ion peak giving relative molecular mass.

Worked example: do not confuse $\ce{M+}$ with the base peak

A mass spectrum has major peaks at:

The peak at $m/z = 29$ is the base peak because it is the tallest. The molecular ion peak is stated or identified as $\ce{M+}$ at $m/z = 46$.

Therefore:

$M_r = 46$

The base peak is useful in later organic mass spectrometry because it tells chemists about a stable fragment. For this lesson, the key statement is simply that the $\ce{M+}$ peak gives the relative molecular mass.

Recap

The reasoning chain for mass-spectrometry calculations is compact but very assessable:

• To calculate $A_r$, use a weighted mean: isotope mass multiplied by relative abundance, divided by total abundance.
• To find abundance from $A_r$, set up the same weighted mean with one abundance as the unknown.
• To predict a diatomic molecule spectrum, add isotope masses and use probability to get relative peak heights.
• To find $M_r$ from mass spectrometry for singly charged ions, use the $m/z$ value of the molecular ion, $\ce{M+}$.

Command words call for a visible route. For calculate, show the weighted mean and denominator. For predict, show both the possible masses and the relative peak-height logic. For determine, identify the molecular ion peak before giving $M_r$.