In Average Atomic Mass you weighted each isotope's mass by how common it is. Where do those abundances come from? From a mass spectrometer, a machine that sorts atoms by mass and counts them.
Inside, atoms of the sample are turned into positive ions, sped up by an electric field and sent through a magnetic field. The field bends their paths, and lighter ions bend more, so ions of each mass land in a different place on the detector. The result is a mass spectrum: one peak for each isotope.
Key idea
A peak's position is the mass of one isotope (in u). Its height tells how many atoms have that mass, so it gives the relative abundance. The average atomic mass is the abundance-weighted average: $$\bar{m} = \sum (\text{mass of isotope}) \times (\text{fraction of atoms}).$$
Why it works
Picture 100 chlorine atoms taken at random: about 76 have mass 34.97 u and 24 have mass 36.97 u. Their total mass is $76(34.97) + 24(36.97)$, and dividing by 100 gives the mass of an "average" atom. Dividing each count by 100 first gives the same thing, which is why fractions multiply the masses. The heavier isotope pulls the average toward itself only as much as its share of atoms allows.
Worked example: Chlorine
Chlorine's spectrum has peaks at $34.97\ \text{u}$ (75.76% of the atoms) and $36.97\ \text{u}$ (24.24%). Find the average atomic mass.
- Change percents to fractions: $0.7576$ and $0.2424$.
- $$\bar{m} = 34.97(0.7576) + 36.97(0.2424) = 26.49 + 8.96$$
- $\bar{m} = 35.45\ \text{u}$, the value on the periodic table.
Answer: $35.45\ \text{u}$
Worked example: Peaks measured against the tallest
Magnesium's spectrum shows peaks at $23.99$, $24.99$ and $25.98\ \text{u}$ with heights $100$, $12.66$ and $13.94$ (the tallest set to 100). Find the average atomic mass.
- Heights are proportional to the number of atoms. The total is $100 + 12.66 + 13.94 = 126.60$.
- Fractions: $100/126.60 = 0.7899$, $12.66/126.60 = 0.1000$, $13.94/126.60 = 0.1101$.
- $$\bar{m} = 23.99(0.7899) + 24.99(0.1000) + 25.98(0.1101) = 18.95 + 2.50 + 2.86 = 24.31\ \text{u}$$
Answer: $24.31\ \text{u}$
Worked example: Working backward to an abundance
Copper's average atomic mass is $63.55\ \text{u}$. Its two isotopes have masses $62.93\ \text{u}$ and $64.93\ \text{u}$. What percent of copper atoms are copper-63?
- Let $x$ be the fraction of copper-63. The rest, $1 - x$, is copper-65.
- $62.93x + 64.93(1 - x) = 63.55$, so $64.93 - 2.00x = 63.55$.
- $x = \dfrac{64.93 - 63.55}{2.00} = 0.690$, or $69.0\%$. Check: the average lies closer to 62.93, so the lighter isotope should be the more common one.
Answer: About $69.0\%$ copper-63.
Common mistake
Reading the mass of the tallest peak as the average atomic mass, or adding the peak masses and dividing by the number of peaks. For copper, the plain average of 62.93 and 64.93 is 63.93, but the real average is 63.55, because 69% of the atoms are the lighter kind. Check: the weighted average always lies between the lightest and heaviest peaks, closest to the tallest.
Spectra of molecules show extra patterns. A sample of $\mathrm{Cl_2}$ gives three peaks near 70, 72 and 74 u, for $^{35}\mathrm{Cl}\!-\!{}^{35}\mathrm{Cl}$, $^{35}\mathrm{Cl}\!-\!{}^{37}\mathrm{Cl}$ and $^{37}\mathrm{Cl}\!-\!{}^{37}\mathrm{Cl}$, in heights of about $9.8 : 6.3 : 1$ (from $0.7576^2$, $2(0.7576)(0.2424)$ and $0.2424^2$). Chemists use such fingerprints to spot chlorine in an unknown compound.