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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 $\ce{Cl}$.

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    $\begingroup$ Your teacher is wrong. You know precisely that 24% are $\ce{^37 Cl}$, as this is given in the question. Also not that amu is a deprecated unit and it should be replaced by u, the unified atomic mass unit. $\endgroup$ – Martin - マーチン Sep 22 '16 at 5:39
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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})} &= \pu{10.2 g} \ce{^37Cl} \\ \implies \frac{\pu{10.2 g}~ \ce{^37Cl}}{\pu{36.97 g//mol}\ce{^37Cl}} &= \pu{0.276 mol} \end{align}

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