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Obtain the fractional abundances for the two naturally occurring isotopes of europium. The masses of the isotopes are \({ }^{151} \mathrm{Eu}, 150.9196 \mathrm{amu} ;{ }^{153} \mathrm{Eu}, 152.9209 \mathrm{amu} .\) The atomic weight is 151.9641 amu.

Short Answer

Expert verified
The fractional abundances are 0.4786 for \(^{151}\text{Eu}\) and 0.5214 for \(^{153}\text{Eu}.\)

Step by step solution

01

Define Variables

Let the fractional abundance of \(^{151}\text{Eu}\) be \(x\) and the fractional abundance of \(^{153}\text{Eu}\) be \(y\). We know that the total fractional abundance of the isotopes must be equal to 1, so \(x + y = 1\).
02

Set Up the Weighted Average Equation

The atomic weight is a weighted average of the weights of the isotopes, which can be represented as:\[ 150.9196x + 152.9209y = 151.9641 \]
03

Solve the System of Equations

We have the system of equations:1. \(x + y = 1\)2. \(150.9196x + 152.9209y = 151.9641\)From equation 1, express \(y\) in terms of \(x\):\[ y = 1 - x \]Substitute into equation 2:\[ 150.9196x + 152.9209(1-x) = 151.9641 \]
04

Simplify and Solve for \(x\)

Substituting \(y = 1 - x\) into the weighted average equation yields:\[ 150.9196x + 152.9209 - 152.9209x = 151.9641 \]Simplify it:\[ -2.0013x + 152.9209 = 151.9641 \]Rearrange to solve for \(x\):\[ -2.0013x = 151.9641 - 152.9209 \]\[ -2.0013x = -0.9568 \]\[ x = \frac{-0.9568}{-2.0013} \approx 0.4786 \]
05

Solve for \(y\)

Substituting \(x = 0.4786\) into \(y = 1 - x\):\[ y = 1 - 0.4786 \approx 0.5214 \]
06

Verify the Solution

Ensure that the fractional abundances add up to 1 and check the weighted average:\[ x + y = 0.4786 + 0.5214 = 1 \]Check the weighted average calculation:\[ 150.9196 \times 0.4786 + 152.9209 \times 0.5214 \approx 151.9641 \]The values check out, confirming the solution.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Fractional Abundance
The concept of fractional abundance is central to understanding isotopic composition. Fractional abundance refers to the proportion of each isotope of an element in a natural sample. It is expressed as a fraction, with the sum of all isotopic abundances equaling 1.
In the case of europium's isotopes, this means:
  • The fractional abundance for \(^{151}\text{Eu}\) is denoted as \(x\).
  • The fractional abundance for \(^{153}\text{Eu}\) is denoted as \(y\).
Since europium has two naturally occurring isotopes, the sum of \(x\) and \(y\) must equal 1, giving us the equation \(x + y = 1\).
This basic equation helps to determine the relative amounts of each isotope in a given sample of europium. Understanding how to express these abundances in fractional terms is foundational for calculating atomic weight and other related chemical properties.
Weighted Average
A weighted average takes into account both the magnitude of the items being averaged and their relative importance or abundance. In the context of isotopic abundances, the weighted average is used to determine the atomic weight of an element by considering the masses and fractional abundances of its isotopes.
For europium, the atomic weight is 151.9641 amu, achieved by weighting the isotopic masses of \(^{151}\text{Eu}\) and \(^{153}\text{Eu}\). The equation representing this relationship is:
  • \(150.9196x + 152.9209y = 151.9641\)
Here, \(150.9196\) represents the mass of \(^{151}\text{Eu}\), \(152.9209\) is the mass of \(^{153}\text{Eu}\), \(x\) is the fractional abundance of \(^{151}\text{Eu}\), and \(y\) is the fractional abundance of \(^{153}\text{Eu}\).
The concept of weighted average enables us to fairly account for each isotope's contribution to the overall atomic weight, reflecting how often each isotope appears in natural samples.
System of Equations
A system of equations is a set of equations with multiple variables that can be solved using methods such as substitution or elimination. In the case of isotopic abundance calculations, we use a system of equations to find the fractional abundances of isotopes.

The system of equations set for europium’s isotopes is:
  • \(x + y = 1\)
  • \(150.9196x + 152.9209y = 151.9641\)
To solve this, we express \(y\) in terms of \(x\) from the first equation: \(y = 1 - x\).
Substituting \(y\) in the second equation allows us to solve for \(x\). This method untangles the variables, making it possible to calculate the specific fractional abundance for each isotope of europium. By manipulating these equations, we can ascertain the distribution of isotopes within a given sample, enhancing our understanding of the element's properties in nature.

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Most popular questions from this chapter

An element has three naturally occurring isotopes with the following masses and abundances: $$ \begin{array}{ll} \text { Isotopic Mass (amu) } & \text { Fractional Abundance } \\ 27.977 & 0.9221 \\ 28.976 & 0.0470 \\ 29.974 & 0.0309 \end{array} $$ Calculate the atomic weight of this element. What is the identity of the element?

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