Structure of an Atom and Mass Spectrum

Lessons

Senior 4 · Chemistry

Structure of an Atom and Mass Spectrum

4 lessons

Lesson 4 of 4

Reading Mass Spectra and Uses of the Mass Spectrometer

25 min

A strip of magnesium ribbon, three peaks

The magnesium ribbon you burn in the school laboratory looks like one pure substance. Put a scrap of it into a mass spectrometer and the recorder draws not one line but three, at m/z 24, 25 and 26. Each line is one isotope, and its height tells you what share of the atoms it makes up. Reading graphs like this one, and calculating from them, is the skill this unit's key competence asks for: calculate the relative atomic mass of an element from a simple mass spectrum.

Definition

Mass spectrum: the output of a mass spectrometer, a graph of ion abundance (y-axis) against mass-to-charge ratio m/z (x-axis). Each vertical line is a peak.

Relative abundance: the y-axis value of a peak. It is given either as a percentage of all the ions, or as a relative intensity on which the tallest peak (the base peak) is set to 100.

Reading the spectrum of an element

For an element, ionised mainly to 1+ ions:

the number of peaks is the number of isotopes;
the m/z of each peak is the relative isotopic mass of that isotope;
the height of each peak is proportional to the abundance of that isotope.

Figure 4.1: The mass spectrum of magnesium

Worked example 1: $A_{r}$ from percentage abundances

Use Figure 4.1 to calculate the relative atomic mass of magnesium to 4 significant figures.

Solution. The percentages add up to 100 (78.99 + 10.00 + 11.01), so divide by 100:

A_r(\mathrm{Mg}) = \frac{(24 \times 78.99) + (25 \times 10.00) + (26 \times 11.01)}{100} = \frac{1895.76 + 250.00 + 286.26}{100} = 24.32

Answer: 24.32. (The periodic table gives 24.31: the true isotopic masses are slightly below the mass numbers.)

Worked example 2: $A_{r}$ from relative intensities

The mass spectrum of lead shows four peaks, with heights measured relative to the tallest peak:

m/z	204	206	207	208
relative intensity	2.7	46.0	42.2	100

Calculate $A_r(\mathrm{Pb})$ .

Solution. The heights do not add up to 100, so divide by their total, $2.7 + 46.0 + 42.2 + 100 = 190.9$ :

A_r = \frac{(204 \times 2.7) + (206 \times 46.0) + (207 \times 42.2) + (208 \times 100)}{190.9} = \frac{39\,562.2}{190.9} = 207.2

Answer: 207.2 (4 s.f.).

Warning

Divide by the total, not by 100, unless the abundances are percentages that add up to 100. Dividing the lead data by 100 would give 395.6, an impossible answer: $A_{r}$ must lie between the lightest and heaviest isotope.

Doubly charged ions. Small peaks at half the usual m/z (for magnesium at 12, 12.5 and 13) come from 2+ ions such as ${}^{25}\mathrm{Mg^{2+}}$ ( $\frac{25}{2} = 12.5$ ). They are not extra isotopes; leave them out when you calculate $A_{r}$ .

Spectra of molecules

When a molecule M is ionised, the ion $\mathrm{M^{+}}$ is called the molecular ion. Its m/z equals the relative molecular mass, $M_{r}$ . Some molecular ions break into smaller pieces, fragment ions, which give peaks at lower m/z. For a simple compound, the peak with the highest m/z (ignoring tiny isotope peaks just above it) is usually the molecular ion. Carbon dioxide, for example, gives $\mathrm{CO_2^{+}}$ at 44, and fragments $\mathrm{CO^{+}}$ at 28, $\mathrm{O^{+}}$ at 16 and $\mathrm{C^{+}}$ at 12: so $M_r(\mathrm{CO_2}) = 44$ .

Chlorine shows how isotopes and molecules combine. A $\mathrm{Cl_2}$ molecule contains two atoms, each of which is chlorine-35 (probability $\frac{3}{4}$ ) or chlorine-37 (probability $\frac{1}{4}$ ). Three molecular ions are possible:

\begin{aligned} {}^{35}\mathrm{Cl}{-}{}^{35}\mathrm{Cl}\ (70)&: \ \tfrac{3}{4} \times \tfrac{3}{4} = \tfrac{9}{16} \\ {}^{35}\mathrm{Cl}{-}{}^{37}\mathrm{Cl}\ (72)&: \ 2 \times \tfrac{3}{4} \times \tfrac{1}{4} = \tfrac{6}{16} \\ {}^{37}\mathrm{Cl}{-}{}^{37}\mathrm{Cl}\ (74)&: \ \tfrac{1}{4} \times \tfrac{1}{4} = \tfrac{1}{16} \end{aligned}

The factor 2 in the middle line appears because the mixed molecule can form in two ways (35 then 37, or 37 then 35). So the molecular ion peaks at 70, 72 and 74 stand in the ratio 9 : 6 : 1, and fragment $\mathrm{Cl^{+}}$ ions give peaks at 35 and 37 in the ratio 3 : 1 (Figure 4.2).

Figure 4.2: The mass spectrum of chlorine gas

Worked example 3: hydrogen chloride

Predict the m/z values and the height ratio of the molecular ion peaks of hydrogen chloride, HCl. Take hydrogen as ${}^{1}\mathrm{H}$ only and chlorine as 75% chlorine-35 and 25% chlorine-37.

Solution. Each molecule has one hydrogen atom (mass 1) and one chlorine atom. $\mathrm{H^{35}Cl^{+}}$ : $1 + 35 = 36$ . $\mathrm{H^{37}Cl^{+}}$ : $1 + 37 = 38$ . With only one chlorine atom per molecule, the ratio is the chlorine isotope ratio, 75 : 25.

Answer: peaks at m/z 36 and 38 in the ratio 3 : 1. A pair of peaks two units apart in a 3 : 1 ratio is the fingerprint of one chlorine atom in a molecule.

What mass spectrometers are used for

Table 4.1: Uses of mass spectrometry

Use	What the spectrometer measures	Example
Relative atomic masses	the masses and abundances of an element's isotopes	the $A_{r}$ values printed in the periodic table
Identifying compounds	the molecular ion ( $M_{r}$ ) and the pattern of fragment peaks, compared with a library of known spectra	forensic analysis of powders; testing food and water for pesticides
Drug testing in sport	the compounds in a urine sample, first separated by gas chromatography and then identified by their spectra	anti-doping laboratories
Radiocarbon dating	the ratio of carbon-14 to carbon-12 atoms, counted directly (accelerator mass spectrometry)	dating wood, bone or cloth from the last 50 000 years
Space science	the gases and compounds in the air or soil of another world	the Sample Analysis at Mars (SAM) instrument on NASA's Curiosity rover
Medicine	small molecules in a few drops of blood	screening newborn babies for inherited metabolic disorders

Tip

Set out every $A_{r}$ calculation the same way: (1) write $\sum(\text{m/z} \times \text{height})$ as one fraction; (2) divide by $\sum \text{heights}$ ; (3) check the answer lies between the lightest and heaviest isotope, nearer the tallest peak; (4) give the answer to the number of significant figures the data allow.

Allery Chemistry: Calculating Relative Atomic Mass from Mass Spectra (6 min). Two exam-style spectra are worked through. Pause after each spectrum appears and try the calculation before you watch the solution.

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Practice

A mass spectrum of rubidium shows peaks at m/z 85 (72.2%) and 87 (27.8%). Calculate $A_r(\mathrm{Rb})$ .
The spectrum of an element X has peaks at m/z 28 (relative intensity 100), 29 (5.1) and 30 (3.4). Calculate $A_r(\mathrm{X})$ and identify X from the periodic table.
The spectrum of neon has a large peak at 20, a very small one at 21, a peak at 22 and two tiny peaks at 10 and 11. Explain the peaks at 10 and 11.
Bromine is about 50% bromine-79 and 50% bromine-81. Predict the m/z values and height ratio of the molecular ion peaks of $\mathrm{Br_2}$ .
Explain why the mass spectrum of hydrogen chloride has peaks at 36 and 38 but not at 37.
The mass spectrum of a gas has its highest peak at m/z 44 and fragment peaks at 29, 15 and 14. It contains only carbon, hydrogen and oxygen. Suggest two possible molecular formulae with $M_{r} = 44$ , and say which fits the fragment at 15 (a $\mathrm{CH_3^{+}}$ ion).
Evaluate this student's claim: "A mass spectrometer measures the mass of each ion."

Show answersHide answers

$\frac{(85 \times 72.2) + (87 \times 27.8)}{100} = \frac{6137.0 + 2418.6}{100} = 85.56$ .
$\frac{(28 \times 100) + (29 \times 5.1) + (30 \times 3.4)}{108.5} = \frac{3049.9}{108.5} = 28.11$ ; X is silicon ( $A_{r} = 28.09$ ).
They are ${}^{20}\mathrm{Ne^{2+}}$ ( $\frac{20}{2} = 10$ ) and ${}^{22}\mathrm{Ne^{2+}}$ ( $\frac{22}{2} = 11$ ): doubly charged ions, not isotopes.
158 (79 + 79), 160 (79 + 81, two ways) and 162 (81 + 81) in the ratio $\frac{1}{4} : \frac{2}{4} : \frac{1}{4}$ , that is 1 : 2 : 1.
Hydrogen is almost all mass 1 and chlorine is 35 or 37, so the molecules are $\mathrm{H^{35}Cl}$ (36) and $\mathrm{H^{37}Cl}$ (38). No common combination adds up to 37.
$\mathrm{CO_2}$ and $\mathrm{C_2H_4O}$ (for example ethanal, $\mathrm{CH_3CHO}$ ) both have $M_{r} = 44$ . Carbon dioxide has no hydrogen, so it cannot give $\mathrm{CH_3^{+}}$ (15); the peak at 15 and the peak at 29 ( $\mathrm{CHO^{+}}$ ) fit $\mathrm{C_2H_4O}$ . (Propane, $\mathrm{C_3H_8}$ , also has $M_{r} = 44$ but contains no oxygen.)
Not quite: it measures the mass-to-charge ratio. For a 1+ ion, m/z equals the mass, but a 2+ ion appears at half its mass, so an ion of mass 24 and charge 2+ gives a peak at 12.

Quick check

An element gives peaks at m/z 69 (60%) and 71 (40%). What is its relative atomic mass?

Key points

For an element: number of peaks = number of isotopes; m/z = isotopic mass; height ∝ abundance.
$A_r = \frac{\sum(\text{m/z} \times \text{height})}{\sum \text{heights}}$ ; divide by 100 only when the heights are percentages.
Peaks at half the usual m/z are 2+ ions, not isotopes.
The molecular ion $\mathrm{M^{+}}$ gives $M_{r}$ ; $\mathrm{Cl_2}$ gives 70 : 72 : 74 in the ratio 9 : 6 : 1.
Mass spectrometry finds $A_{r}$ , identifies compounds, tests for drugs, dates ancient carbon and analyses other planets.

1A strip of magnesium ribbon, three peaks

2Reading the spectrum of an element

3Spectra of molecules

4What mass spectrometers are used for

A strip of magnesium ribbon, three peaks

Reading the spectrum of an element

Spectra of molecules

What mass spectrometers are used for