Mass of Mole of Isotope of Element

Theorem

Let $S$ be a substance made up entirely of a particular isotope $Q$ of a particular element.

Let one atom of $Q$ contain $n$ neutrons and $p$ protons.

Then one mole of $S$ has a mass of approximately $n + p$ grams.

Proof

Let $m_Q$ grams be the mass of one atom of $Q$.

Let $m_C$ grams be the mass of one atom of carbon-12.

Let $M_Q$ grams be the mass of one mole of $Q$.

Let $M_C$ grams be the mass of one mole of carbon-12.

The mass of a proton and the mass of a neutron are approximately equal.

Together, the protons and neutrons make up the vast majority of the mass of the atom they compose.

Hence the mass of one atom is proportional to the total number of its protons and neutrons.

There are $6$ protons and $6$ neutrons in one atom of carbon-12.

Thus we have:

\(\ds m_Q : m_C\)

\(=\)

\(\ds \paren {n + p} : {6 + 6}\)

\(\ds \leadsto \ \ \)

\(\ds \dfrac {m_Q} {m_C}\)

\(=\)

\(\ds \dfrac {n + p} {12}\)

\(\ds \leadsto \ \ \)

\(\ds \dfrac {M_Q} {M_C}\)

\(=\)

\(\ds \dfrac {n + p} {12}\)

\(\text {(1)}: \quad\)

\(\ds \leadsto \ \ \)

\(\ds M_Q\)

\(=\)

\(\ds \paren {n + p} \dfrac {M_C} {12}\)

Before $2018$, the mole was defined as the number of atoms in $12$ grams of carbon-12.

While the mole is no longer so defined, the mass of $1$ mole of carbon-12 is still $12$ grams, for all practical purposes.

Thus we have:

\(\ds M_C\)

\(=\)

\(\ds 12 \ \mathrm g\)

\(\ds \leadsto \ \ \)

\(\ds M_Q\)

\(=\)

\(\ds \paren {n + p} \dfrac {12 \ \mathrm g} {12}\)

substituting for $M_C$ in $(1)$

\(\ds \leadsto \ \ \)

\(\ds M_Q\)

\(=\)

\(\ds \paren {n + p} \ \mathrm g\)

$\blacksquare$