Mass spectrometry data and relative atomic mass
This section is AS content, set on Papers 1 and 3, and it is a calculation section: you need to get from mass spectrometry data to a relative atomic mass, and back again.
On Paper 3 a definition followed by an Ar calculation to a stated accuracy is a typical opening question.
The chemistry behind the numbers is the weighted mean from the previous section; the mass spectrometer is how the weights are measured.
What a mass spectrometer measures
A mass spectrometer sorts particles by mass.
The sample is vaporised and its atoms are ionised, usually by knocking out one electron to form 1+ ions.
The ions are accelerated, separated according to their mass-to-charge ratio, m/z, and detected.
The detector signal for each m/z value is proportional to how many ions arrive, so the height of each line tells you the relative abundance of that ion.
The diagram below shows these stages in a time-of-flight mass spectrometer, a common modern type.
All the positive ions are given the same kinetic energy by the accelerating plates, so lighter ions travel faster and reach the detector first.
For an ion with a charge of 1+, m/z is simply the mass of the ion, and the mass of the electron lost is negligible.
So for an element, each line on the spectrum is one isotope: its m/z value gives the relative isotopic mass, and its height gives the abundance.
The spectrum below is for a sample of copper.
Abundances may be given as percentages (adding up to 100), as relative peak heights (often with the tallest set to 100), or as a ratio.
The method is the same in every case: you divide by the total, whatever it is.
Calculating Ar from abundances
You need to be able to analyse and interpret data from mass spectrometry to calculate relative atomic mass from relative abundance of isotopes and vice versa.
Relative atomic mass is the weighted mean:
You need to know this formula; no equation sheet is given in the exam.
The step list:
1. For each isotope, multiply its relative isotopic mass (the m/z value) by its abundance.
2. Add these products.
3. Divide by the total abundance (100 for percentages; the sum of the heights otherwise).
4. Round only at the end, to the accuracy the question asks for. Ar has no unit.
Three checks catch most errors.
The answer must lie between the lightest and heaviest isotopic masses.
It must lie nearer the most abundant isotope.
And if a percentage is missing, the percentages must add up to 100 before you start.
Use the relative isotopic masses you are given, never the mass numbers.
Copper's isotopes have masses 62.9296 and 64.9278, not 63 and 65, and using the whole numbers moves the answer by a few hundredths, which is exactly the size of the mark when the question asks for two decimal places.
Working backwards: abundances from Ar
"Vice versa" means using a known Ar to find an abundance.
For two isotopes, call the percentage of one of them x, so the other is 100 − x.
Substitute into the weighted mean, set it equal to Ar, and solve the linear equation.
Again use the accurate isotopic masses if they are given.
- Relative abundance: the proportion of the atoms of an element that are a particular isotope, shown by the height of its line in a mass spectrum.
- m/z: the mass-to-charge ratio of an ion; for a 1+ ion it equals the relative mass of the ion.
- Relative atomic mass (learn this wording): the weighted mean mass of an atom of an element relative to one-twelfth of the mass of an atom of carbon-12.
The table shows data from the mass spectrum of a sample of copper.
Isotope
Relative isotopic mass
Relative abundance / %
⁶³Cu
62.9296
69.15
⁶⁵Cu
64.9278
30.85
Calculate the relative atomic mass of copper in this sample.
Give your answer to two decimal places.
(2)
1. Ar = (62.9296 × 69.15 + 64.9278 × 30.85) ÷ 100 (1)
2. = (4351.58 + 2003.02) ÷ 100 = 63.546 = 63.55 (1)
Answer: 63.55.
Correct answer with no working scores 2.
The second mark is lost for 63.5 (one decimal place) or 63.546 (three).
A weaker answer uses 63 and 65 and gets 63.62: the method mark is kept but the answer mark is lost.
In the mass spectrum of a sample of lithium, the line at m/z 6.0151 has a height of 8.2 and the line at m/z 7.0160 has a height of 100.
Calculate the relative atomic mass of lithium.
Give your answer to three significant figures.
(2)
1. Total abundance = 8.2 + 100 = 108.2
2. Ar = (6.0151 × 8.2 + 7.0160 × 100) ÷ 108.2 (1)
3. = (49.32 + 701.60) ÷ 108.2 = 6.940 = 6.94 (1)
Answer: 6.94.
A weaker answer divides by 100 because "abundances are percentages" and gets 7.51, which is larger than the heaviest isotope: impossible, so the check would have caught it.
Boron has two isotopes, ¹⁰B (relative isotopic mass 10.0129) and ¹¹B (11.0093).
A sample of boron has a relative atomic mass of 10.81.
Calculate the percentage abundance of each isotope.
(3)
1. Let the percentage of ¹⁰B be x, so ¹¹B is 100 − x.
2. 10.81 = [10.0129x + 11.0093(100 − x)] ÷ 100 (1)
3. 1081 = 1100.93 − 0.9964x, so x = 19.93 ÷ 0.9964 = 20.0 (1)
4. ¹⁰B 20.0%, ¹¹B 80.0% (1)
Answer: ¹⁰B 20.0%, ¹¹B 80.0%.
With mass numbers 10 and 11 the same method gives 19.0%: show the masses you were given in the first line so the examiner can see you used them.
The table shows data for the three isotopes in a sample of silicon.
Isotope
Relative isotopic mass
Relative abundance / %
²⁸Si
27.9769
92.23
²⁹Si
28.9765
³⁰Si
29.9738
3.09
Calculate the relative atomic mass of silicon in the sample.
Give your answer to two decimal places.
(3)
1. Abundance of ²⁹Si = 100 − 92.23 − 3.09 = 4.68% (1)
2. Ar = (27.9769 × 92.23 + 28.9765 × 4.68 + 29.9738 × 3.09) ÷ 100 (1)
3. = (2580.31 + 135.61 + 92.62) ÷ 100 = 28.085 = 28.09 (1)
Answer: 28.09.
TE from an incorrect missing abundance for the last two marks.
Check: 28.09 lies between 27.98 and 29.97, close to the main isotope.
This question is about gallium, a metal used in light-emitting diodes.
Natural gallium contains two isotopes.
Isotope
Relative isotopic mass
Relative abundance / %
⁶⁹Ga
68.9256
60.11
⁷¹Ga
70.9247
(a) State what is meant by the term relative atomic mass.
(2)
- the weighted mean mass of an atom (of an element) (1)
- relative to one-twelfth of the mass of an atom of carbon-12 (1)
(b) Calculate the relative atomic mass of gallium.
Give your answer to three significant figures.
(3)
- abundance of ⁷¹Ga = 100 − 60.11 = 39.89% (1)
- Ar = (68.9256 × 60.11 + 70.9247 × 39.89) ÷ 100 = 69.723 (1)
- = 69.7 to three significant figures (1)
(c) Deduce the number of neutrons in an atom of ⁷¹Ga.
(1)
- 71 − 31 = 40 (1)
Answer: 6 marks in all.
A weaker answer to (a), "the average mass of the isotopes compared with carbon-12", scores 0: "average of the isotopes" is not "weighted mean mass of an atom", and "one-twelfth" is missing.
In (b), 69.72 loses the accuracy mark because four significant figures were given where three were asked.
Before you calculate, add up the abundances and write the total down: divide by that total, not automatically by 100.
After you calculate, check the answer lies between the isotopic masses and nearer the most abundant one, then round to exactly the accuracy the question states.
- Using mass numbers when relative isotopic masses are given. It is tempting because whole numbers are quicker, but the answer then misses a two-decimal-place mark.
- Dividing peak heights by 100. Heights are relative, not percentages, so divide by their sum.
- Forgetting to find the missing abundance first. Always make the percentages total 100.
- Rounding to the wrong accuracy, or giving a unit. "Three significant figures" means 69.7, not 69.72; and Ar has no unit, so writing g mol⁻¹ is penalised.
Check your understanding
1. Which is the relative atomic mass of a sample of rubidium containing 72.17% ⁸⁵Rb (relative isotopic mass 84.9118) and 27.83% ⁸⁷Rb (86.9092)? (1)
- A 85.47
- B 85.56
- C 85.91
- D 86.35
Total [1]
2. The table shows the relative heights of the lines in the mass spectrum of a sample of magnesium.
m/z
23.9850
24.9858
25.9826
Relative height
100
12.66
13.94
Calculate the relative atomic mass of magnesium in this sample.
Give your answer to two decimal places.
(2)
Total [2]
3. Antimony has two isotopes, ¹²¹Sb (relative isotopic mass 120.9038) and ¹²³Sb (122.9042). The relative atomic mass of antimony is 121.76.
Calculate the percentage abundance of ¹²¹Sb.
Give your answer to three significant figures.
(3)
Total [3]
4. Lead from different ore deposits contains slightly different proportions of its isotopes, so archaeologists measure the isotopes in lead objects to work out where the metal was mined. The table shows data for lead from one object.
Isotope
Relative isotopic mass
Relative abundance / %
²⁰⁴Pb
203.9730
1.35
²⁰⁶Pb
205.9745
25.50
²⁰⁷Pb
206.9759
20.80
²⁰⁸Pb
207.9767
(a) Calculate the percentage abundance of ²⁰⁸Pb.
(1)
(b) Calculate the relative atomic mass of lead in this object.
Give your answer to two decimal places.
(2)
(c) Every lead atom has 82 protons.
Explain why lead from two different ores can have different relative atomic masses.
(2)
Total [5]
5. A student writes: "The relative atomic mass of boron is 10.8, so a boron atom has a mass of 10.8 on the carbon-12 scale."
Explain whether or not this statement is valid.
(2)
Total [2]
6. Indium has two isotopes, ¹¹³In and ¹¹⁵In. Its relative atomic mass is 114.8.
Deduce, without a calculation, which isotope is more abundant.
Justify your answer.
(2)
Total [2]
Answers
1. The only correct answer is A (85.47). B is not correct because it uses the mass numbers 85 and 87 in place of the relative isotopic masses. C is not correct because it is a simple mean of the two isotopic masses, ignoring abundance. D is not correct because it attaches each abundance to the wrong isotope.
2. Example of calculation:
- Ar = (23.9850 × 100 + 24.9858 × 12.66 + 25.9826 × 13.94) ÷ 126.60 (1)
- = 24.305 = 24.31 (1). Correct answer with no working scores 2. Dividing by 100 gives 30.77: award 1 for the products only if the total is wrong.
3. Example of calculation:
- 121.76 = [120.9038x + 122.9042(100 − x)] ÷ 100 (1)
- 12176 = 12290.42 − 2.0004x, so x = 114.42 ÷ 2.0004 (1)
- = 57.2% (1). TE on a slip in rearranging.
4. (a) 100 − 1.35 − 25.50 − 20.80 = 52.35% (1)
- (b) Ar = (203.9730 × 1.35 + 205.9745 × 25.50 + 206.9759 × 20.80 + 207.9767 × 52.35) ÷ 100 (1); = 207.204 = 207.20 (1). TE from (a).
- (c) An explanation that makes reference to the following points:
- Ar is a weighted mean that depends on the relative abundances of the isotopes (1)
- lead from different ores has different proportions of ²⁰⁴Pb, ²⁰⁶Pb, ²⁰⁷Pb and ²⁰⁸Pb, so the weighted mean differs, even though every atom has 82 protons (1)
5. An explanation that makes reference to the following points:
- not valid: 10.8 is a weighted mean of the masses of the ¹⁰B and ¹¹B atoms (1)
- each boron atom has a relative mass close to 10 or close to 11; no single atom has a mass of 10.8 (1)
6. - ¹¹⁵In is more abundant (1)
- because the weighted mean, 114.8, lies much closer to 115 than to 113 (1). M2 dependent on M1.