Calculate the amount of substance of an isotope based on the percent composition Calculate the amount of substance of $\ce{^37Cl}$ in a $\pu{42.3 g}$ sample of $\ce{Cl}$ containing $24.00\%~\ce{^37Cl}$ and $76.00\%~ \ce{^35Cl}.$
  Given the atomic masses $m(\ce{Cl}) =\pu{35.45 amu}$, $m(\ce{^35Cl}) =\pu{34.97 amu}$, and $m(\ce{^37Cl}) =\pu{36.97 amu}$.


I know you would multiply $\pu{42.3 g}$ by $0.240$ to get $\pu{10.2 g}$ of $\ce{^37Cl}$, but my teacher says you then divide by the atomic mass of $\ce{Cl}$ rather than $\ce{^37Cl}$. That is the part that confuses me, she says it is because 


  You don't know what those $\pu{10.2 g}$ are made of, so you need to use the atomic mass of $\ce{Cl}$.
If you have given an accurate account then your teacher is mistaken. We've all gotten confused while trying to talk and think at the same time. 
\begin{align}
\pu{42.3 g} \ce{Cl}\cdot 0.24\frac{m(\ce{^37Cl})}{m(\ce{Cl})} 
&amp;= \pu{10.2 g} \ce{^37Cl} \\
\implies
\frac{\pu{10.2 g}~ \ce{^37Cl}}{\pu{36.97 g//mol}\ce{^37Cl}} &amp;= \pu{0.276 mol}
\end{align}