The shape of the lesson
By the end of the lesson, learners can
- One standard atom: Load an element into the mass spectrometer and read off each isotope's m/z value and abundance.
- A r from a mass spectrum: Multiply each isotope's mass by its abundance, add the products, and divide by the total abundance.
- Molecules and formula units: Add up the A r values for every atom in the formula, taking care with brackets and water of crystallisation.
- Standard: 1/12 of the mass of one 12C atom.
- Ar: sum of (mass × abundance) ÷ total abundance.
- Mr: add the Ar values; a number outside a bracket multiplies everything inside.
- No units: relative masses are ratios.
One picture every idea returns to
All relative masses are measured against 1/12 of the mass of one atom of carbon-12 (the unified atomic mass unit, u). Relative isotopic mass is the mass of one atom of an isotope on that scale. Relative atomic mass Ar is the weighted mean mass of the atoms of an element. Relative molecular mass Mr adds the Ar values in a molecule; for giant ionic compounds the same sum is called relative formula mass. None of them has units.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Weighed against carbon-12.. Atoms are far too light to weigh in grams, so chemists compare them with one agreed standard: an atom of carbon-12, fixed at exactly 12. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: One standard atom | Load an element into the mass spectrometer and read off each isotope's m/z value and abundance. Project tab 1 of the lab (Stoichiometry lab, mass spectrum) and let a student drive it while the class predicts. | Whole class, then pairs |
| 13 to 21 min | Teach and check: A r from a mass spectrum | Multiply each isotope's mass by its abundance, add the products, and divide by the total abundance. | Pairs |
| 21 to 30 min | Teach and check: Molecules and formula units | Add up the A r values for every atom in the formula, taking care with brackets and water of crystallisation. | Numbered Heads Together |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 13 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Molecules and formula units, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Numbered Heads Together
Teams of four number off. Put up a question, give the team time to make sure every member can answer, then call a number: that student answers for the team, so nobody can leave the thinking to someone else.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| A molecule such as Cl 2 gives three molecular ion peaks, at m/z 70, 72 and 74, because each atom can be either isotope. | With 75% 35 Cl and 25% 37 Cl the chances are 0.75 × 0.75, 2 × 0.75 × 0.25 and 0.25 × 0.25, a ratio of 9: 6: 1. Do not forget the factor of 2 for the mixed molecule, which can form two ways. |
Relative masses: worksheet
6 questions, 13 marks. Written in the style of the papers; no past paper question is reproduced.
Define relative atomic mass.
A sample of magnesium contains 78.99% 24Mg, 10.00% 25Mg and 11.01% 26Mg. Calculate the relative atomic mass of magnesium to three significant figures.
Gallium has two isotopes, 69Ga and 71Ga. Its relative atomic mass is 69.72. Calculate the percentage abundance of each isotope.
Calculate the relative molecular mass of ethanol, C2H5OH, and the relative formula mass of ammonium sulfate, (NH4)2SO4. (Ar: H 1.0, C 12.0, N 14.0, O 16.0, S 32.1)
Natural chlorine is 75% 35Cl and 25% 37Cl. Predict the m/z values and the relative heights of the molecular ion peaks in the mass spectrum of Cl2. Show your reasoning.
Explain why the term relative formula mass, and not relative molecular mass, is used for sodium chloride.
Answer key and marking notes
- The weighted mean mass of an atom of an element,
- relative to 1/12 of the mass of an atom of carbon-12.
- Sum of mass × abundance: 24 × 78.99 + 25 × 10.00 + 26 × 11.01 = 1895.76 + 250.00 + 286.26 = 2432.02.
- Ar = 2432.02 ÷ 100 = 24.3.
- Let x be the percentage of 69Ga: 69x + 71(100 − x) = 6972.
- 7100 − 2x = 6972, so x = 64.0. 69Ga is 64.0% and 71Ga is 36.0%.
- Ethanol: 2 × 12.0 + 6 × 1.0 + 16.0 = 46.0.
- Ammonium sulfate: 2 × (14.0 + 4.0) + 32.1 + 4 × 16.0 = 36.0 + 32.1 + 64.0 = 132.1.
- Possible molecules: 35Cl35Cl (m/z 70), 35Cl37Cl (m/z 72) and 37Cl37Cl (m/z 74).
- Probabilities: 0.75 × 0.75 = 0.5625; 2 × 0.75 × 0.25 = 0.375 (the mixed molecule forms two ways); 0.25 × 0.25 = 0.0625.
- Relative heights 70 : 72 : 74 = 9 : 6 : 1.
- Sodium chloride is a giant ionic lattice of Na+ and Cl− ions; it contains no discrete molecules.
- The mass is therefore quoted for one formula unit, NaCl, the simplest ratio of ions in the lattice.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
Which statement is the definition of relative isotopic mass?
- Athe mass of one atom of a particular isotope, relative to 1/12 of the mass of one atom of carbon-12
- Bthe weighted mean mass of the atoms of an element, relative to 1/12 of the mass of one atom of carbon-12
- Cthe total number of protons and neutrons in the nucleus of one atom of the isotope
- Dthe mass in grams of one mole of atoms of a particular isotope
Natural neon is 90.5% 20Ne, 0.3% 21Ne and 9.2% 22Ne. What is the relative atomic mass of neon, to three significant figures?
- A20.2
- B21.0
- C20.0
- D21.8
The mass spectrum shows the two isotopes of element X. The peak heights are relative abundances, not percentages. What is the relative atomic mass of X?
- A69.8
- B70.0
- C70.2
- D14.0
The relative atomic masses of most elements are not whole numbers. What is the main reason?
- Amost elements are a mixture of isotopes, and Ar is a weighted mean of their masses
- Bthe mass of the electrons is added to the mass of the nucleus
- Crelative atomic masses are measured in grams, which are rounded
- Dthe carbon-12 standard changes slightly from sample to sample
What is the relative formula mass of hydrated sodium carbonate, Na2CO3·10H2O? (Ar: H 1.0, C 12.0, O 16.0, Na 23.0)
- A286.0
- B106.0
- C124.0
- D263.0
What is the relative formula mass of calcium nitrate, Ca(NO3)2? (Ar: N 14.0, O 16.0, Ca 40.1)
- A164.1
- B102.1
- C150.1
- D100.1
Chlorine is about 75% 35Cl and 25% 37Cl. The molecular ion region of the mass spectrum of an organic compound, containing only carbon, hydrogen and chlorine, shows peaks at m/z 98, 100 and 102 with relative heights 9 : 6 : 1. What does this show about one molecule of the compound?
- Ait contains two chlorine atoms
- Bit contains one chlorine atom
- Cit contains three chlorine atoms
- Dit contains no chlorine atoms, since the peaks are 2 apart
Element X exists as diatomic molecules, X2, and has two isotopes of relative isotopic mass 63 and 65. The molecular ion peaks of X2 are at m/z 126, 128 and 130 with relative heights 9 : 12 : 4. What is the relative atomic mass of X?
- A63.8
- B127.6
- C64.0
- D63.6
Quiz key
Relative isotopic mass refers to a single isotope and compares the mass of one of its atoms with 1/12 of the mass of a carbon-12 atom, so it has no units. The weighted mean over all the isotopes is the definition of relative atomic mass, a different quantity. The number of protons plus neutrons is the mass number, which is a count, not a mass, even though the values are close. The mass in grams of one mole is the molar mass, which has units of g mol−1.
Sum of mass × abundance = 20 × 90.5 + 21 × 0.3 + 22 × 9.2 = 1810 + 6.3 + 202.4 = 2018.7. Dividing by 100 gives 20.187, which is 20.2. The value 21.0 is the simple mean of 20, 21 and 22 and ignores how common each isotope is. 20.0 is just the most abundant isotope and ignores the heavier ones. 21.8 comes from pairing the abundances with the wrong isotopes (90.5% with mass 22).
Weighted sum = 69 × 12.0 + 71 × 8.0 = 828 + 568 = 1396. The total abundance is 12.0 + 8.0 = 20.0, so Ar = 1396 ÷ 20.0 = 69.8. It lies nearer 69 because that peak is taller. 70.0 is the plain average of 69 and 71. 70.2 swaps the heights, giving the heavier isotope the larger share. 14.0 comes from dividing 1396 by 100, as if the heights were percentages.
An element such as chlorine is a mixture of atoms of different masses (35Cl and 37Cl), and Ar is their abundance-weighted mean, so it falls between the whole-number isotope masses, for example 35.5. Electrons have a tiny mass and cannot explain a value such as 35.5. Relative masses have no units at all, so they are not measured in grams. The carbon-12 standard is fixed by definition as exactly 12 and does not vary.
Na2CO3 = 2 × 23.0 + 12.0 + 3 × 16.0 = 106.0. The ten waters add 10 × 18.0 = 180.0, so the total is 286.0. 106.0 leaves out the water of crystallisation completely. 124.0 adds only one water molecule, ignoring the 10 in front of H2O. 263.0 counts only one sodium atom (23.0 + 12.0 + 48.0 + 180.0).
Each nitrate group, NO3, is 14.0 + 3 × 16.0 = 62.0. There are two of them, so 40.1 + 2 × 62.0 = 164.1. 102.1 ignores the bracket and counts only one nitrate group. 150.1 doubles the oxygen atoms but not the nitrogen (40.1 + 14.0 + 6 × 16.0). 100.1 drops the 3, treating the formula as Ca(NO)2.
With two Cl atoms the combinations are 35 + 35, 35 + 37 (which can occur two ways) and 37 + 37, with probabilities 0.752 : 2 × 0.75 × 0.25 : 0.252 = 9 : 6 : 1, giving three peaks 2 units apart. A compound such as C2H4Cl2 (24 + 4 + 70 = 98) fits. One chlorine atom gives only two peaks, M and M+2, in the ratio 3 : 1. Three chlorine atoms give four peaks (M to M+6, ratio 27 : 27 : 9 : 1). The gap of 2 mass units is exactly the difference between 35Cl and 37Cl, so it points to chlorine rather than ruling it out.
Take 25 molecules: 9 are 63-63, 12 are 63-65 and 4 are 65-65. Number of 63-atoms = 2 × 9 + 12 = 30; number of 65-atoms = 12 + 2 × 4 = 20. So the isotopes are 60% and 40%, and Ar = (63 × 30 + 65 × 20) ÷ 50 = 3190 ÷ 50 = 63.8. 127.6 is the weighted mean mass of the X2 molecule, which must be halved for one atom. 64.0 is the simple average of 63 and 65. 63.6 uses only the two end peaks (9 : 4) as the isotope ratio and ignores the mixed molecules at m/z 128.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.