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Reading a mass spectrum

The evidence that isotopes exist, and the weighted mean that turns a spectrum into a relative atomic mass.

Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.

1. What you will learn

By the end of this lesson you will be able to read a supplied mass spectrum — which peak is the heavier isotope, which peak the sample holds most of — and turn the mass numbers and abundances it reports into a relative atomic mass, by multiplying each mass by its percentage, adding, and dividing by a hundred. You will also be able to say why that number is almost never a whole one, and why no single atom of the element ever weighs it.

2. What you already have

You can read a nuclide symbol as three counts, and you know that two atoms of one element with different neutron counts are isotopes of it. What you have not yet been shown is how anybody knows that. This lesson is the evidence, and the arithmetic that comes with it.

3. Words for this lesson

4. The number on the table is not a mass of anything

The periodic table gives chlorine a relative atomic mass of 35.5. A mass number counts protons and neutrons and is always a whole number, so 35.5 cannot be one. Two accounts have been offered for numbers like it.

• Account A: every chlorine atom weighs 35.5, and the half is just a consequence of how the scale was set up.
• Account B: chlorine is a mixture of two kinds of atom, and 35.5 is what the mixture averages.

A mass spectrometer settles it. It sorts the atoms of a sample by mass and counts how many it finds at each mass. Here is what it reports for four elements:

Which account do the first three rows support, and what exactly rules the other one out? Be specific: name the peak that Account A predicts and the instrument does not find.

Now the fourth row, which is the one worth arguing about. Fluorine gives a single peak, and the periodic table gives fluorine a relative atomic mass of 19.0 — a whole number. Does that row support Account A after all, or does Account B explain it too? Say which, and why.

Last: chlorine's two masses are 35 and 37, and its table value is 35.5 rather than 36. Boron's are 10 and 11, and its table value is 10.8 rather than 10.5. What, besides the two masses, has to go into the calculation? The answer is in the third column, and the rest of the lesson is the arithmetic for it.

5. The spectrum is the evidence, and the average is the consequence

Put a sample of chlorine into a mass spectrometer and you do not get one peak. You get two:

• a peak at mass $35$, three quarters of the sample;
• a peak at mass $37$, one quarter of it.

There is no peak at $35.5$, and no atom in the sample weighs $35.5$. That number is an average, and it is the number written on the periodic table because it is the number that is useful: a gram of chlorine holds a mixture, and it is the mixture you weigh out.

The average is weighted, because the two isotopes are not equally common. Each mass number counts in proportion to its abundance:

$$A_r = \frac{(35 \times 75) + (37 \times 25)}{100} = \frac{2625 + 925}{100} = \frac{3550}{100} = 35.5$$

Three steps and no others: multiply each mass number by its percentage, add the products, divide by 100. The division is by 100 because the percentages account for the whole sample; if they did not add to 100, they would not be percentages of it.

The same sum works for three isotopes, or for ten. It just has more terms.

6. What the chart can and cannot tell you

A mass spectrum is a bar chart and it settles exactly two kinds of question.

Which isotope is heavier. The bars are laid out in order of increasing mass, so a bar further right is a heavier isotope. That is a matter of position.

Which isotope there is more of. The height of a bar is a count, so a taller bar is a more abundant isotope. That is a matter of comparison.

It does not tell you the chemistry. Every peak in a chlorine spectrum is chlorine, every one of those atoms has seventeen protons and seventeen electrons, and every one of them forms exactly the same compounds. The spectrometer is weighing atoms, and weight is the one property isotopes differ in.

It also does not tell you which element you have, on its own. Two peaks two units apart, three to one, is the shape of a chlorine spectrum — and it is also the shape of a copper spectrum. What identifies the element is where the peaks are, not how they are arranged.

7. The method, step by step, and how to check it

Read each peak's mass number. Left to right, in order of increasing mass.

Read each peak's abundance. As a percentage of the sample.

Check the percentages add to 100. If they do not, a peak is missing, or the figures are not percentages, and dividing by 100 would be wrong.

Multiply each mass number by its abundance. One product per peak.

Add the products, then divide by 100. The result is the relative atomic mass, and it carries no unit.

Check the work. The answer must lie between the lightest and heaviest mass numbers — an average cannot be bigger than everything it averages. It should sit nearest the tallest peak, because the commonest isotope pulls hardest. And if two isotopes happen to be equally common, the answer should equal the plain average; any other answer means a product was slipped.

8. Why each step is allowed

Multiplying each mass number by its percentage is allowed because a percentage is a count out of a hundred. A sample that is 75 percent chlorine-35 holds 75 atoms of mass 35 in every hundred, and $35 \times 75$ is the mass of those 75 atoms.

Adding the products is allowed because the sample is just those atoms together. The sum is the mass of a hundred typical atoms, taken in the proportions the sample actually has.

Dividing by 100 is allowed because the percentages add to 100. The sum was the mass of a hundred atoms, so dividing by a hundred gives the average mass of one.

Using mass numbers in place of true atomic masses is a simplification this course makes. A real chlorine-35 atom has a relative mass of about 34.97, not exactly 35, because the masses of protons and neutrons are not exactly 1 and some mass is lost when a nucleus forms. To the one decimal place this course works to, mass numbers give the right answer.

9. Why the periodic table's numbers are rarely whole

Almost every relative atomic mass in the table is a weighted mean over a mixture, and that is why so few of them are whole numbers.

Bromine is the one that looks like an exception and is not: its two isotopes happen to be equally common, so the weighted mean and the plain average agree. That is a coincidence about bromine and not a rule, and it is the reason a weighted mean has to be done as a weighted mean even when it looks as though it will not matter.

The abundances above are the rounded ones a school table gives. A real instrument reports them to several decimal places, and the course rounds them so that every average in it terminates.

10. In the world: the uranium for American reactors

Natural uranium is a mixture of two main isotopes: about 99.3 percent uranium-238 and 0.7 percent uranium-235. Only uranium-235 keeps the chain reaction going in an ordinary power reactor, so the fuel for American nuclear plants has to be enriched to about 3 to 5 percent uranium-235. The enrichment plant in Eunice, New Mexico does that, and mass spectrometry is how its operators know what they have made.

The weighted mean shows why the job is so hard. Natural uranium's relative atomic mass works out as $(238 \times 99.3 + 235 \times 0.7) \div 100 = 237.98$. Fuel enriched to 4 percent works out as $(238 \times 96 + 235 \times 4) \div 100 = 237.88$. The whole difference between ore and reactor fuel is a tenth of a unit in the average mass, because the two isotopes differ by only three neutrons in more than two hundred particles.

That is why enrichment uses thousands of centrifuges in long chains, each nudging the mixture a tiny step, and why a spectrometer reading individual peaks — rather than a balance reading the average — is the only way to check the product. The peaks tell the operators the abundance directly; the average hides it.

11. In the world: checking where honey came from

Food scientists at the Department of Agriculture use isotope ratios measured by mass spectrometry to detect honey diluted with cheap corn syrup. Corn and flowering plants take up carbon-13 in slightly different proportions, so the weighted mean of the carbon in the honey shifts when syrup is added.

12. Where this goes wrong

A mass number counts particles in a nucleus and is always a whole number; a relative atomic mass is an average over the isotopes an element actually comes as, and almost never is. Chlorine has no atom of mass 35.5 in it. It has atoms of 35 and atoms of 37, in a fixed proportion, and 35.5 is what that mixture weighs on average.

The plain average is used instead of the weighted one. Averaging $35$ and $37$ gives $36$, and chlorine's relative atomic mass is $35.5$. The plain average is what you would get if the two isotopes were equally common, and the whole point of the spectrum is that they are not.

The tallest peak is taken as the answer. The most abundant isotope of chlorine has mass $35$, and chlorine's relative atomic mass is not $35$. The tallest peak pulls the average hardest; it does not decide it.

The division is done by the number of peaks. Dividing by 2 because there are two isotopes gives the plain average again. The division is by 100, because the weights are percentages.

A taller peak is read as a heavier isotope. Height is how common an isotope is; position along the chart is how heavy it is.

13. Copper, from its spectrum

1. Read the two peaks.

   $63 \text{ at } 75\%; \ 65 \text{ at } 25\%$

   Mass numbers and abundances from the spectrum.

2. Check the percentages.

   $75 + 25 = 100$

   So dividing by 100 will be right.

3. Weight the copper-63 peak.

   $63 \times 75 = 4725$

   Each mass counts in proportion to its abundance.

4. Weight the copper-65 peak.

   $65 \times 25 = 1625$

   The same for the second peak.

5. Add and divide by 100.

   $(4725 + 1625) \div 100 = 63.5$

   The mass of a hundred atoms, back to one.

14. Three peaks instead of two

1. Read the three peaks of magnesium.

   $24 \text{ at } 80\%; \ 25 \text{ at } 10\%; \ 26 \text{ at } 10\%$

   One more term than before; nothing else changes.

2. Check the percentages.

   $80 + 10 + 10 = 100$

   Check the weights before you use them.

3. Weight the magnesium-24 peak.

   $24 \times 80 = 1920$

   One product per peak.

4. Weight the other two.

   $25 \times 10 = 250; \ 26 \times 10 = 260$

   The smaller peaks still count.

5. Add the products.

   $1920 + 250 + 260 = 2430$

   The mass of a hundred atoms.

6. Divide by 100 and check.

   $2430 \div 100 = 24.3$

   Near 24, the tallest peak, as it should be.

15. Boron, and the check against the plain average

1. Read the two peaks.

   $10 \text{ at } 20\%; \ 11 \text{ at } 80\%$

   From the spectrum.

2. Check the percentages.

   $20 + 80 = 100$

   The whole sample is accounted for.

3. Weight the boron-10 peak.

   $10 \times 20 = 200$

   Mass number times percentage.

4. Weight the boron-11 peak.

   $11 \times 80 = 880$

   The commoner isotope contributes far more.

5. Add and divide by 100.

   $(200 + 880) \div 100 = 10.8$

   The relative atomic mass.

6. Compare with the plain average.

   $(10 + 11) \div 2 = 10.5$

   Too low: it ignores how common boron-11 is.

7. Check the range.

   $10 < 10.8 < 11$

   An average lies between what it averages.

16. Your turn: a spectrum of gallium shows mass $69$ at $60\%$ and mass $71$ at $40\%$. What is its relative atomic mass?

1. Weight each mass number.

   $69 \times 60 = 4140; \ 71 \times 40 = 2840$

   Weight each value by how common it is.

2. Add the two products.

   $4140 + 2840 = 6980$

   The mass of a hundred atoms.

3. Your turn: work this step out. Its working is at the end of the packet.

   Divide by 100.

17. Guided practice

A mass spectrometer is run on a sample of chlorine. Each bar is one isotope; the bars stand in order of increasing mass from left to right, and the height of a bar is how much of that isotope the sample holds. Mark the bar for the isotope there is most of.

This task has no paper form; do it on a device.

18. Guided practice

Complete the worked solution: a spectrum shows mass number $72$ at $70\%$ and mass number $74$ at $30\%$. Find each isotope's contribution before dividing, and the relative atomic mass.

1. Weight the lighter isotope.

   $(\text{lighter mass}) \times (\text{its percentage}) =$ _x

   Each mass counts in proportion to its abundance.

2. Weight the heavier isotope.

   $(\text{heavier mass}) \times (\text{its percentage}) =$ _y

   The same rule for the second peak.

3. Add and divide by a hundred.

   $(\text{sum}) \div \text{a hundred} =$ _r

   The percentages make up the whole sample.

19. Guided practice

The mass spectrum of a sample of copper is shown. The bars stand in order of increasing mass from left to right. Put each mass number on the bar it belongs to.

This task has no paper form; do it on a device.

20. Practice

A mass spectrum of copper shows two peaks: mass number $63$ at $75\%$ and mass number $65$ at $25\%$. What is the relative atomic mass of copper? A relative atomic mass is a comparison and carries no unit.

Answer:

21. Practice

A battery plant in Nevada checks a batch of lithium on a mass spectrometer. The batch is $8\%$ lithium-6 and $92\%$ lithium-7. What is the batch's relative atomic mass?

The answer: _a.

22. Somewhere new

A works that grows crystals for the electronics industry checks a batch of silicon on a mass spectrometer before it is used. The spectrum comes back with three peaks: mass number $28$ at $90\%$, mass number $29$ at $5\%$ and mass number $30$ at $5\%$. What relative atomic mass does this batch have?

Answer:

23. Lesson test

Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.

24. Test question

A sample of rubidium is $72\%$ of the isotope with mass number $85$ and $28\%$ of the isotope with mass number $87$. Lay out the weighted mean: for each isotope give its mass number multiplied by its percentage, and that product divided by 100.

25. What you can do now

You can read a spectrum as evidence and turn it into a weighted mean. Say out loud why chlorine's relative atomic mass is 35.5 when no chlorine atom has that mass, and what would be wrong with averaging 35 and 37. Next: how the electrons arrange themselves outside the nucleus, and what that arrangement predicts.

Working for the steps left to you

16. Your turn: a spectrum of gallium shows mass $69$ at $60\%$ and mass $71$ at $40\%$. What is its relative atomic mass?, step 3