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Isotopes of a chemical element have the same atomic number $Z$, hence the same number of protons, but different neutron numbers $N$ and therefore different mass numbers $A=Z+N$. For example, every neon atomic nucleus contains ten protons: neon-20, neon-21, and neon-22 contain ten, eleven, and twelve neutrons, respectively. They occupy one chemical element box in the periodic table, rather than three boxes.

Three labels answer different questions. The chemical element fixes $Z$; the isotope specifies which neutron number accompanies that $Z$; the ion label specifies the net electric charge. Neutral ${}_{10}^{22}\mathrm{Ne}$ and the ion ${}_{10}^{22}{\mathrm{Ne}}^{+}$ both contain ten protons and twelve neutrons; they contain ten and nine Electrons, respectively. Ordinary ionization changes the Electron count, leaving the isotope unchanged. Changing the neutron number can change nuclear binding energy and the possibility of radioactive decay. Isotopic abundances can help distinguish material sources, while radioactive decay can additionally supply chronological information.

A mass spectrum makes the distinction experimentally visible: a single chemical element can give separate ion peaks at different mass-to-charge ratios. Its bulk atomic weight is a weighted mean governed by isotopic abundances, rather than the location of one compulsory peak. The historical mass spectrograph experiment and the synthetic neon mass spectrum below illustrate these complementary observations.

An isotopic abundance is an atom fraction within a specified population of one chemical element.^[ref] If $n_i$ counts its chemical atoms of isotope $i$, useThe denominator includes every isotope of that chemical element included in the stated population, not every chemical atom of every chemical element in the sample. Thus an oxygen atom fraction among all chemical elements is a different quantity.

For an illustrative neutral neon population, take the displayed central isotopic abundances $f_{20}=0.9048$, $f_{21}=0.0027$, and $f_{22}=0.0925$ from the reference table, so $0.9048+0.0027+0.0925=1$. Its mean atomic mass is the weighted arithmetic meanThese are rounded reference inputs with uncertainties, not exact universal isotopic abundances; the long decimal shows their arithmetic result, not measurement precision.^[ref]

The population convention matters. If instead $w_i$ is the fraction of total neutral-atom mass supplied by isotope $i$, then $w_i=f_i{m}_i/\bar{m}$ andThis is a weighted harmonic mean in the mass fractions. Using ${\sum }_i{w}_i{m}_i$ with those mass fractions would answer a different question. The conversion follows directly by counting chemical atoms and adding their masses.

The relative isotopic abundances in a sample.

A mass spectrum displays detected signal against mass-to-charge ratio. Peak positions identify candidate ions; peak areas give isotopic abundances only after the response and overlapping contributions have been accounted for.

Broadening can hide a weak component without changing its area in this ideal model. An actual mass spectrum can also contain molecular ions, different electric charge states, background, and unequal detection responses. A missing visible peak therefore cannot by itself prove the absence of an isotope. Conversely, a fractional atomic weight is a property of a population: there is no requirement for an ion peak at the weighted mean.

Molar mass turns a weighed sample into a count of specified particles. For a pure chemical substance with specified chemical formula and isotopic abundances, its molar mass iswhere $m$ is the sample mass and $n$ its amount of substance. The SI unit is $\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$; the common laboratory unit is $\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$, with $1\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}=10^{-3}\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$. Equal values of amount of substance correspond to equal counts of specified entities, but heavier entities give greater masses.

Combining this definition with molecular count and amount of substance givesThe Avogadro constant is exact under the modern definition of the mole. The mass $M/N_{\mathrm{A}}$ is the mean mass per specified entity; if isotopes differ within the sample, individual molecules need not all have that mean mass.^[ref]

Specify what $n$ counts: water molecules, hydrogen chemical atoms, and oxygen chemical atoms give different values of amount of substance for the same sample. For a mixture, dividing total mass by total amount of substance of its chosen entities gives a composition-dependent mean. A polymer sample can contain many chain molar masses; its number-average molar mass and weight-average molar mass answer different averaging questions. Relating either to degree of polymerization also requires a unit-mass convention and end-group accounting.

For a chemical element in a specified population, its relative atomic mass or atomic weight is the dimensionless ratio $A_{\mathrm{r}}=\bar{m}/\mathrm{u}$, with $\bar{m}$ the mean neutral atomic mass weighted by isotopic abundance. For the illustrative neon population, $A_{\mathrm{r}}\approx 20.1800$, whereas $\bar{m}\approx 20.1800\mathrm{u}$ has mass units.^[ref]

A recommended atomic weight applicable to normal materials, rather than the result for every possible enriched sample. The tabulated neon value $20.1797(6)$ and the arithmetic result above need not agree in every last displayed digit. Some chemical elements have intervals of standard atomic weight because their isotopic abundances vary across normal materials.^[ref]

Choose approximate atomic weights $A_{\mathrm{r}}(\mathrm{H})=1.008$ and $A_{\mathrm{r}}(\mathrm{O})=15.999$, and use $M_{\mathrm{u}}\approx 1\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$. The chemical formula ${\mathrm{H}}_2\mathrm{O}$ then gives water molar massThese are deliberately chosen rounded inputs, not exact atomic weights or a universal value for every water sample. Different isotopic abundances, notably those in heavy water, change the mean molar mass; chemical formula alone does not specify them. Adding atomic weights also neglects the tiny mass change associated with chemical bond energies, immaterial at this rounding precision.

For an illustrative pure water sample of mass $4.50375\mathrm{g}$, this convention givesThe count refers to water molecules. Their hydrogen chemical atom count is approximately $2N=3.01107038\times 10^{23}$, and their oxygen chemical atom count is approximately $N$. The mean mass per water molecule is $M/N_{\mathrm{A}}\approx 2.99146\times 10^{-26}\mathrm{kg}$. Extra displayed digits expose the arithmetic; they do not imply that the sample composition or mass was measured that precisely.

The same conversion makes stoichiometry a mass calculation. For the chemical equation $2{\mathrm{H}}_2+{\mathrm{O}}_2\to 2{\mathrm{H}}_2\mathrm{O}$, complete conversion of $0.250000\mathrm{mol}$ of molecular hydrogen and $0.125000\mathrm{mol}$ of molecular oxygen gives $0.250000\mathrm{mol}$ of water. With the chosen approximate atomic weights, the reactant masses are $0.50400\mathrm{g}$ and $3.99975\mathrm{g}$, summing to $4.50375\mathrm{g}$. The $2:1:2$ ratio belongs to molecule counts and amount of substance, not to masses; this bookkeeping neglects the much smaller reaction energy contribution to mass.