Question

there are two naturally occurring isotopes of europium, $^{151}\text{eu}$ (151.0 amu) and $^{153}\text{eu}$ (153.0 amu). if the atomic mass of eu is 151.96 amu, what is the approximate natural abundance of $^{151}\text{eu}$? answer: a 20% b 40% c 50% d 60% e 80%

Explanation:

Step1: Let the abundance of \(^{151}Eu\) be \(x\), then the abundance of \(^{153}Eu\) is \(1 - x\).

The formula for average - atomic mass is \(A = m_1x_1+m_2x_2\), where \(A\) is the average atomic mass, \(m_1\) and \(m_2\) are the masses of the isotopes, and \(x_1\) and \(x_2\) are their abundances.
So, \(151.96=151.0x + 153.0(1 - x)\).

Step2: Expand and simplify the equation.

\[

$$\begin{align*} 151.96&=151.0x+153.0 - 153.0x\\ 151.96&=153.0-(153.0x - 151.0x)\\ 151.96&=153.0 - 2.0x \end{align*}$$

\]

Step3: Solve for \(x\).

First, move the terms around: \(2.0x=153.0 - 151.96\).
Then, \(2.0x = 1.04\).
Divide both sides by \(2.0\): \(x=\frac{1.04}{2.0}=0.52\) or \(52\%\approx50\%\).

Answer:

C. 50%