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?


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

This question appears to be off-topic. The users who voted to close gave this specific reason:

If this question can be reworded to fit the rules in the help center, please edit the question.


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


Solve for $x$:

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


Not the answer you're looking for? Browse other questions tagged or ask your own question.