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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?

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closed as off-topic by Loong, bon, ron, Klaus-Dieter Warzecha, John Snow Feb 21 '15 at 19:33

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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 $x$:

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

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