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

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?

• Feb 21, 2015 at 17:39

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 $x$:

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