= Isotopic abundance
{title2=$f_i$}
An <isotopic abundance> is an atom fraction within a specified population of one <chemical element>.<Compendium of Chemical Terminology (Gold Book), second edition/isotopic abundance>{ref} If $n_i$ counts its <chemical atoms> of <isotope> $i$, use
$$
f_i=\frac{n_i}{\sum_j n_j},\qquad f_i\geq0,\qquad\sum_i f_i=1.
$$
The 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 mean>
$$
\begin{aligned}
\bar m&=\sum_i f_i m_i\\
&=\bigl(0.9048\times19.9924401762+0.0027\times20.993846685+0.0925\times21.991385114\bigr)\,\mathrm u\\
&=20.18004638052026\,\mathrm u\approx20.1800\,\mathrm u.
\end{aligned}
$$
These are rounded reference inputs with uncertainties, not exact universal <isotopic abundances>; the long decimal shows their arithmetic result, not measurement precision.<Atomic Weights and Isotopic Compositions for Neon>{ref}
The population convention matters. If instead $w_i$ is the fraction of total neutral-atom <mass> supplied by <isotope> $i$, then $w_i=f_i m_i/\bar m$ and
$$
\bar m=\left(\sum_i\frac{w_i}{m_i}\right)^{-1}.
$$
This 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>.
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