Mass Spectroscopy of Elements质谱与同位素

Big Idea · Scale, Proportion, and Quantity (SPQ)AP Science Practice 5 · Mathematical RoutinesAP link · Unit 1~45 min

Learning goals

  • Describe how a mass spectrometer separates isotopes, and read the mass spectrum of an element: peak position gives the isotope mass, peak height gives its abundance
  • Read the neutron count of an isotope from its peak's m/z, and explain how isotopes changed Dalton's atomic theory
  • Calculate an element's average atomic mass from mass spectrum data, and identify the element from its atomic mass and the pattern of its peaks
  • Given the average atomic mass and the masses of two isotopes, work back to their abundances

Before you start

  • Isotopes (1.3.4)
  • Relative atomic mass (1.3.5)
  • A short history of atomic models (1.3.7)
  • Moles and molar mass (U1.1)

Where does 75.8% come from?

Lesson 1.3.5 said natural chlorine is about 75.8% chlorine-35 and 24.2% chlorine-37, which is why its average atomic mass is 35.45. But isotopes behave almost identically in chemical reactions, so chemistry can’t sort them. How were those percentages measured?

With an instrument that lines atoms up by mass: the mass spectrometer. Its output is a mass spectrum, and AP topic 1.2 asks you to read one and use it to calculate the atomic mass on the periodic table.

The mass spectrometer: sorting atoms by mass

A mass spectrometer质谱仪 works in four stages:

  1. Vaporize and ionize. The sample is turned into a gas, and high-energy electrons knock one electron off each atom, giving ions with a 1+ charge, such as ClX+\ce{Cl+}.
  2. Accelerate. An electric field gives all the ions the same kinetic energy and forms them into a beam.
  3. Separate. A magnetic field bends the beam, and lighter ions bend more. (Another design races the ions down a tube, and lighter ones arrive first.) Ions of different mass end up in different places.
  4. Detect. A detector counts how many ions arrive at each position.

The ion’s mass mm divided by its charge zz is the mass-to-charge ratio质荷比, m/zm/z. For an ion with a 1+ charge, m/zm/z is numerically the ion’s mass in amu. Losing one electron changes the mass by a negligible amount, so m/zm/z is the mass of the isotope.

Reading a mass spectrum

On the mass spectrum质谱图 of an element:

  • The horizontal axis is m/zm/z. Each peak is one isotope, and its position is the isotope’s mass, close to its mass number.
  • The vertical axis is relative abundance. The taller the peak, the more atoms of that isotope. An isotope’s abundance丰度 is the percentage of the element’s atoms that are that isotope.

Spectra use two styles for the vertical axis. One labels each peak as a percentage, and the percentages add up to 100%. The other sets the tallest peak to 100 and scales the rest; those numbers are not percentages, and you have to convert them first (see the tool card below).

Reading neutrons from a peak

For a 1+ ion, a peak’s m/zm/z is very close to the isotope’s mass number AA (protons + neutrons). Look up the atomic number ZZ on the periodic table, and

neutrons=A−Z\text{neutrons} = A - Z

Boron (Z=5Z = 5) has peaks at m/zm/z ≈ 10 and 11, so boron-10 has 5 neutrons and boron-11 has 6.

A reminder from lessons 1.3.3 and 1.3.4 about nuclide symbols, XZAX2Z2AX\ce{^{A}_{Z}X}: there are ZZ protons and A−ZA - Z neutrons; a neutral atom has ZZ electrons, a cation fewer, and an anion more. Atoms with the same number of protons but different numbers of neutrons are isotopes. Change the number of protons and you have a different element.

Isotopes and Dalton’s atomic theory

Dalton (1803) proposed that all atoms of a given element are identical in mass (lesson 1.3.7). Mass spectra rule that out directly: the spectrum of natural magnesium has three peaks, so magnesium atoms come in three different masses. Dalton’s other ideas still hold, such as atoms being neither created nor destroyed in reactions, and compounds forming from atoms in whole-number ratios. Only “identical mass” has to change: atoms of the same element have the same number of protons, but their masses can differ.

Average atomic mass

An element’s average atomic mass平均原子质量 is the weighted average of its isotope masses, weighted by abundance written as a decimal:

mˉ=m1f1+m2f2+⋯\bar{m} = m_1 f_1 + m_2 f_2 + \cdots

where f1+f2+⋯=1f_1 + f_2 + \cdots = 1. This is the atomic mass printed on the periodic table, the same number as the molar mass in g/mol from U1.1. For chlorine (isotope masses 34.97 amu and 36.97 amu):

mˉ=34.97×0.7576+36.97×0.2424=35.45 amu\bar{m} = 34.97 \times 0.7576 + 36.97 \times 0.2424 = 35.45\ \text{amu}

The result always lies between the lightest and heaviest isotopes, and closer to the more abundant one.

🔧 Math tool card · Weighted AveragesSkip it if you already know this

This is the weighted average from lesson 1.3.5: multiply each value by its share, then add. Two things to watch when reading a spectrum:

Turn percents into decimals. Write 75.76% as 0.7576 before multiplying. If you multiply by 75.76 instead, divide the total by 100.

Normalize a spectrum whose tallest peak is 100. For chlorine the peak heights are 100 and 32.0, which are not percentages. Add the heights, then divide each by the sum:

f(X35X2235Cl)=100100+32.0=0.758f(\ce{^{35}Cl}) = \frac{100}{100 + 32.0} = 0.758f(X37X2237Cl)=32.0100+32.0=0.242f(\ce{^{37}Cl}) = \frac{32.0}{100 + 32.0} = 0.242

An element's mass spectrum sets the tallest peak to 100, and the only other peak has a height of 25.0. What is the abundance of the isotope with the smaller peak?

Identifying an element: atomic mass and peak pattern

Calculating the average atomic mass from a spectrum and checking the periodic table is the main way to identify an element. The pattern of peaks is a clue too:

  • Iodine has only one stable isotope, so its spectrum is a single peak at m/zm/z = 127. Tellurium’s atomic mass (127.60) is close to iodine’s (126.90), but its spectrum has 8 peaks spread from 120 to 130.
  • Bromine shows two nearly equal peaks at 79 and 81, and its average, 79.90, sits almost exactly between them.
  • Zirconium has 5 peaks. The tallest is at 90, yet the average, 91.22, is well above 90, because the peaks at 92 and 94 aren’t small.

In “Read a spectrum”, click some peaks, switch between the two axis styles, and compare iodine with zirconium. Then try a few mystery spectra in “Identify the element”; tellurium is one of them:

Try it: read mass spectra, calculate the average atomic mass, then identify the element
Vertical axis:

Click a peak to see which isotope it is.

Data: NIST Atomic Weights and Isotopic Compositions (SRD 144).

Example 1: Average atomic mass of magnesium

The mass spectrum of magnesium has three peaks: m/zm/z = 23.985 (78.99%), 24.986 (10.00%), and 25.983 (11.01%). Calculate the average atomic mass of magnesium.

Step-by-step solution

  1. Know what you’re computing. Average atomic mass is a weighted average of the isotope masses, weighted by abundance. Convert the percents to decimals: 0.7899, 0.1000, 0.1101.

  2. Check that the abundances add to 1.

    0.7899+0.1000+0.1101=1.00000.7899 + 0.1000 + 0.1101 = 1.0000

  3. Multiply each mass by its abundance. All masses are in amu:

    23.985×0.7899=18.94624.986×0.1000=2.49925.983×0.1101=2.861\begin{aligned} 23.985 \times 0.7899 &= 18.946 \\ 24.986 \times 0.1000 &= 2.499 \\ 25.983 \times 0.1101 &= 2.861 \end{aligned}

  4. Add them.

    mˉ=18.946+2.499+2.861=24.31 amu\begin{aligned} \bar{m} &= 18.946 + 2.499 + 2.861 \\ &= \boxed{24.31\ \text{amu}} \end{aligned}

  5. Check. 24.31 lies between 24 and 26, very close to the most abundant isotope, magnesium-24, and it matches the 24.31 on the periodic table.

Example 2: From average atomic mass back to abundances

Copper has just two stable isotopes, copper-63 (62.93 amu) and copper-65 (64.93 amu). The average atomic mass of copper is 63.55 amu. Find the abundance of each isotope.

Step-by-step solution

  1. Choose a variable. With only two isotopes, the abundances add up to 1: let x be the abundance of copper-63, so copper-65 is 1 − x.

  2. Write the weighted-average equation. All masses are in amu:

    62.93x+64.93(1−x)=63.5562.93x + 64.93(1 - x) = 63.55

  3. Expand and solve.

    64.93−2.00x=63.552.00x=1.38x=0.690\begin{aligned} 64.93 - 2.00x &= 63.55 \\ 2.00x &= 1.38 \\ x &= 0.690 \end{aligned}

  4. State both abundances.

    Cu-63:69.0%Cu-65:31.0%\begin{aligned} \text{Cu-63} &: \boxed{69.0\%} \\ \text{Cu-65} &: \boxed{31.0\%} \end{aligned}

  5. Substitute back.

    62.93×0.690+64.93×0.310=43.42+20.13=63.55\begin{aligned} &62.93 \times 0.690 + 64.93 \times 0.310 \\ &= 43.42 + 20.13 = 63.55 \end{aligned}

  6. Is it reasonable? 63.55 is closer to 62.93, so copper-63 should be the majority, and it is. The measured values are 69.15% and 30.85%; the small difference comes from rounding the masses and the average to two decimal places.

Practice: which isotope is more abundant?
  1. Boron: ¹⁰B and ¹¹B, average atomic mass 10.81
  2. Chlorine: ³⁵Cl and ³⁷Cl, average atomic mass 35.45
  3. Copper: ⁶³Cu and ⁶⁵Cu, average atomic mass 63.55
  4. Lithium: ⁶Li and ⁷Li, average atomic mass 6.94
  5. Gallium: ⁶⁹Ga and ⁷¹Ga, average atomic mass 69.72
  6. Rubidium: ⁸⁵Rb and ⁸⁷Rb, average atomic mass 85.47
  7. Iridium: ¹⁹¹Ir and ¹⁹³Ir, average atomic mass 192.22
  8. Thallium: ²⁰³Tl and ²⁰⁵Tl, average atomic mass 204.38
Practice: how many neutrons does this atom have?
  1. The peak at m/z = 20 in the spectrum of neon (Ne, Z = 10)
  2. The peak at m/z = 22 in the spectrum of neon
  3. The peak at m/z = 24 in the spectrum of magnesium (Mg, Z = 12)
  4. The peak at m/z = 26 in the spectrum of magnesium
  5. The peak at m/z = 28 in the spectrum of silicon (Si, Z = 14)
  6. The peak at m/z = 37 in the spectrum of chlorine (Cl, Z = 17)
  7. A X37X2237ClX−\ce{^{37}Cl-} ion
  8. A X1123X211223NaX+\ce{^{23}_{11}Na+} ion
  9. A X2040X220240Ca\ce{^{40}_{20}Ca} atom

Common mistakes

Mistake 1: “A taller peak means a heavier isotope.” A peak’s position (horizontal axis) is its mass; its height (vertical axis) is its abundance. Chlorine-37 is heavier than chlorine-35, but its peak is much shorter.

Mistake 2: “The average atomic mass is the m/z of the tallest peak” or “the plain average of the isotope masses”. Neither. For chlorine the tallest peak is at 35 and the plain average is 36; the real value, 35.45, is a weighted average. Zirconium makes it obvious: the tallest peak is at 90, but the average is 91.22.

Mistake 3: “Each peak is a different element, a different oxidation state, or a different sublevel.” In the mass spectrum of one element, every peak is an isotope of that element: same number of protons, different numbers of neutrons. The peaks have nothing to do with oxidation states, and peaks sorted by sublevel belong to a different technique, the photoelectron spectroscopy of U1.6. AP only uses mass spectra made of single atoms with a 1+ charge. A real sample of chlorine gas also gives ClX2X+\ce{Cl2+} peaks at 70, 72, and 74; spectra like that aren’t tested.

Quick check

The mass spectrum of an element has just two peaks: m/z = 107 (51.8%) and m/z = 109 (48.2%). The element is

The mass spectrum of an element has three peaks: m/z = 20 (90.5%), 21 (0.3%), and 22 (9.3%). Which statement is correct?

The mass spectrum of natural magnesium has three peaks. This is inconsistent with which part of Dalton's atomic theory?

The mass spectrum of an element has 8 peaks between m/z = 120 and 130, with the two tallest at 128 and 130. The element is

AP-style free response

AP-style free response

The mass spectrum of naturally occurring boron has two peaks: m/zm/z = 10.01 (19.9%) and m/zm/z = 11.01 (80.1%).

(a) Explain why the average atomic mass of boron, 10.81, is not a whole number, and why it is closer to 11 than to 10.

(b) How do the boron atoms that produce the two peaks compare in numbers of protons, neutrons, and electrons?

(c) Boron extracted from a certain mineral has an average atomic mass of 10.76. Is this boron richer or poorer in boron-10 than natural boron? Support your answer by calculating its abundance of boron-10.

Model answer and scoring

(a) Natural boron is a mixture of two isotopes, and its average atomic mass is the weighted average of their masses, so it falls between 10.01 and 11.01 rather than at a whole number. Boron-11 is far more abundant (80.1%) than boron-10 (19.9%), so the average is pulled toward 11.

(b) Both kinds of atom have 5 protons (both are boron) and 5 electrons (neutral atoms). Boron-10 has 5 neutrons, and boron-11 has 6.

(c) Let xx be the abundance of boron-10:

10.01x+11.01(1−x)=10.7610.01x + 11.01(1 - x) = 10.76

So x=0.25x = 0.25: about 25% boron-10, more than the 19.9% in natural boron. Qualitatively, 10.76 is closer to 10.01 than 10.81 is, so the lighter isotope makes up a larger share.

Scoring: (a) uses the idea of a weighted average tied to the two abundances; (b) states all three: same protons, same electrons, different neutrons; (c) sets up the weighted-average equation, gets about 25%, and connects “richer” to the data.

Summary

  • A mass spectrometer ionizes atoms to 1+ ions and separates them by mass-to-charge ratio, m/zm/z; for 1+ ions, m/zm/z is the isotope mass.
  • In an element’s mass spectrum, each peak is one isotope: position gives mass, height gives abundance.
  • Average atomic mass mˉ=∑mifi\bar{m} = \sum m_i f_i, with abundances as decimals that add up to 1. If the tallest peak is set to 100, divide each height by the sum of the heights first.
  • With only two isotopes, the average atomic mass gives a one-variable equation for the abundances.
  • A peak’s m/zm/z ≈ mass number, so neutrons = m/z−Zm/z - Z. Isotopes disproved Dalton’s idea that all atoms of an element have the same mass.
  • Identify an element from both its average atomic mass and its peak pattern (iodine has one peak, tellurium eight).

AP tip: Expect a mass spectrum with a request to estimate the average atomic mass, identify the element, read the neutron count of a peak, or explain why the average sits near one isotope, and also the reverse: given an element, pick its spectrum. The numbers are usually friendly, so estimating with mass numbers and checking the periodic table is enough on multiple choice; on free response, write out the weighted-average setup. AP only tests spectra of a single element made of 1+ single-atom ions, not molecular fragments or mixtures of elements.

Practice this chapter: Unit 1 · Atomic Structure and Properties · 12 questionsPractice this topic: U1.2 · Mass Spectroscopy of Elements · 6 questions

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