Question

The element europium exists in nature as two isotopes: ${}^{151}\mathrm{Eu}$ has a mass of 150.9196 u and ${}^{153}\mathrm{Eu}$ has a mass of 152.9209 u. The average atomic mass of europium is 151.96 u. Calculate the relative abundance of the two europium isotopes.

Short answer

The relative abundances of the two europium isotopes are approximately 49.17% for ${}^{151}\mathrm{Eu}$ and 50.83% for ${}^{153}\mathrm{Eu}$.

Step-by-step solution

Assign variables to the unknowns

Let x be the relative abundance of ${}^{151}\mathrm{Eu}$ and y be the relative abundance of ${}^{153}\mathrm{Eu}$.

Set up the equations

Since we are given the average atomic mass of europium, we can set up an equation using the weighted average formula: $$151.96=150.9196x+152.9209y$$ Additionally, the sum of the relative abundances of both isotopes should equal 100%, or 1 in decimal form: $$x+y=1$$

Solve the system of equations

We have a system of two linear equations with two unknowns, x and y: $$\{\begin{array}{l}151.96=150.9196x+152.9209y\\ x+y=1\end{array}$$ To solve this system, we can first solve the second equation for x: $$x=1-y$$ Next, we can substitute this expression for x into the first equation: $$151.96=150.9196(1-y)+152.9209y$$ Now, solve for y: $$151.96=150.9196-150.9196y+152.9209y$$ Simplify the equation and solve for y: $$151.96=150.9196+(152.9209-150.9196)y$$ $$y=\frac{151.96-150.9196}{152.9209-150.9196}\approx 0.5083$$ Now that we have the value for y, we can substitute it back into the equation for x: $$x=1-y=1-0.5083\approx 0.4917$$

Convert the relative abundances to percentages

To convert the relative abundances (in decimal form) to percentages, multiply each by 100: $$\begin{gathered} {}^{151}\mathrm{Eu}:0.4917\times 100=49.17\mathrm{\%}{ \\ }^{153}\mathrm{Eu}:0.5083\times 100=50.83\mathrm{\%} \end{gathered}$$ Therefore, the relative abundances of the two europium isotopes are approximately 49.17% for ${}^{151}\mathrm{Eu}$ and 50.83% for ${}^{153}\mathrm{Eu}$.

Key concepts

Average Atomic Mass

The term average atomic mass is a weighted average that takes into account the masses of all isotopes of an element, according to their natural abundance. This value is what we typically find on the periodic table under each element.

When calculating the average atomic mass, we multiply the mass of each isotope by its relative abundance (the fraction of that isotope that can be found in a natural sample), and then sum these products. Mathematically, the formula is expressed as:$$\text{Average atomic mass}=\sum (\text{isotope mass}\times \text{relative abundance})$$In the exercise given, the two isotopes of europium have different masses and relative abundances, which contribute to the overall average atomic mass of the element. The question, therefore, seeks to deduce the individual relative abundances given the average atomic mass.

Relative Abundance

The concept of relative abundance pertains to how much of each isotope is present in a natural sample of an element as a percentage of the total amount of that element. It is essentially the isotope's share of the entire elemental pie. For chemists, this is crucial information, as it allows them to accurately determine the average atomic mass of the element.

One of the main challenges students face when dealing with relative abundance is understanding that it must always sum up to 100%, or 1 if working with decimal fractions. In the provided solution, we applied this rule to find out the proportion of each europium isotope by setting up an equation with the total of the relative abundances equal to 1.

Chemical Isotopes

Chemical isotopes are variants of a particular chemical element that share the same number of protons but differ in the number of neutrons. This variation in neutron count leads to differences in mass and sometimes differences in radioactive properties, although chemically, isotopes of the same element behave identically.

For example, in the case of europium, the isotopes are $${}^{151}\mathrm{Eu}$$and$${}^{153}\mathrm{Eu}$$,where the superscript indicates the mass number (sum of protons and neutrons). Because isotopes have different mass numbers, their percentage in nature (relative abundance) directly influences the average atomic mass of the element.

System of Linear Equations

A system of linear equations consists of two or more linear equations that share common variables. The goal is to find the value of the variables that make all the equations true at the same time. Such systems can be solved using various methods, including substitution, elimination, or matrix operations.

In the context of the europium isotope exercise, we work with a simple system of two linear equations with two variables representing the relative abundances. The first equation comes from the expectation that the weighted average of the isotopes' masses yields the average atomic mass of europium, while the second ensures that the sum of the relative abundances equals 100%. Doing so allows for the determination of each isotope's relative abundance in the elemental composition of europium.