How to find the relative abundances of isotopes of an unknown element? [closed]








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An unknown element $\ce{Q}$ has two unknown isotopes: $\ce{^60Q}$ and $\ce{^63Q}$. If the average atomic mass is $\pu{61.5 u}$, what are the relative percentages of the isotopes?




You can reverse engineer the formula used to calculate the average atomic mass of all isotopes.


For example, carbon has two naturally occurring isotopes:


\begin{array}{lrr}
\text{Isotope} & 
\text{Isotopic Mass $A$} & 
\text{Abundance $p$} \\\hline
\ce{^{12}C}: & \pu{12.000000 u} & 0.98892 \\
\ce{^{13}C}: & \pu{13.003354 u} & 0.01108 
\end{array}


The formula to get a weighted average is the sum of the product of the abundances and the isotope mass:
$$A = \sum\limits\_{i=1}^n p\_i A\_i$$


For carbon this is:  

$$0.989 \times 12.000 + 0.0111 \times 13.003 = 12.011$$


As you can see, we can set the abundance of one isotope to $x$, and the other to $1 - x$.


If $x = 0.989$, then $1 - x = 0.0111$, *OR* if $x = 0.0111$, then $1 - x = 0.989$.
Therefore, we can simply set up an algebraic equation:
$$A\_1(x\_1) + A\_2(1 - x\_1) = A$$


We know $A\_1$, $A\_2$, and $A$ in your example, so:


\begin{align}
60 x + 63(1 - x) &= 61.5\\
63 - 3 x &= 61.5\\
x &= \frac{-1.5}{-3} = 0.5
\end{align}


Therefore, the element with a mass of $\pu{60 u}$ has a $50\%$ abundance, and the element with the mass of $\pu{63 u}$ has also a $50\%$ abundance.


Proof: $\pu{60 u}(0.5) + \ce{63 u}(0.5) = \pu{61.5 u}$


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Solve for $x$:


$$60x + 63(1-x)=61.5$$


 0