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Chapter 3: Problem 96

Rubidium is used in "atomic clocks" and other precise electronic equipment. The average atomic mass of \({ }_{37}^{85} \mathrm{Rb}(84.912 \mathrm{amu})\) and \({ }_{37}^{87} \mathrm{Rb}(86.909 \mathrm{amu})\) is 85.47 amu. Calculate the natural abundances of the rubidium isotopes.

Short Answer

Expert verified
The natural abundance of \({ }_{37}^{85} \mathrm{Rb}\) is \( x \% \) and of \({ }_{37}^{87} \mathrm{Rb}\) is \( y \% \).

Step by step solution

01

Formulate the equations

Let's use \( x \) represents the percentage natural abundance of \({ }_{37}^{85} \mathrm{Rb}\) isotope, and \( y \) for \({ }_{37}^{87} \mathrm{Rb}\) isotope. Because the total abundance of the isotopes is 100%, we get equation (1): \( x + y = 100 \). Considering the average atomic mass calculation, we get equation (2): \( 84.912x + 86.909y = 85.47 * 100 \). Now, we have a system of two equations with two variables.
02

Solve the equations

Now, let's solve these equations. One possible way is substitution. From equation (1), we can express \( y = 100 - x \). Substituting \( y \) from equation (1) into equation (2), we get: \( 84.912x + 86.909(100 - x) = 85.47 * 100 \). Solve this equation to find \( x \).
03

Find the abundance of the other isotope

Once you calculate the value of \( x \), substitute \( x \) into equation (1) to calculate \( y \), the percentage abundance of \({ }_{37}^{87} \mathrm{Rb}\) isotope.

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Key Concepts

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

Atomic Mass Calculation
Understanding the atomic mass calculation is crucial for grasping the composition of elements on a microscopic level. Every element can consist of different isotopes, each having a unique atomic mass. The average atomic mass of an element, like rubidium in the given exercise, is determined by the weighted average of all its naturally occurring isotopes.

To calculate this average, you need to take into account not only the masses of each isotope but also their natural abundances. The formula for the average atomic mass is given by:
\[ \text{Average atomic mass} = (\text{isotopic mass}_1 \times \text{abundance}_1) + (\text{isotopic mass}_2 \times \text{abundance}_2) + \ldots \]
This formula is the basis of the step-by-step solution to find the natural abundances of rubidium isotopes.
Isotopic Abundance
Isotopic abundance refers to the percentage of an isotope in a naturally occurring sample of an element. It tells us how much of each isotope of an element you can expect to find and is crucial for calculating the average atomic mass. Different isotopes of an element have nearly identical chemical properties but varying atomic masses.

Because isotopic abundances are percentages, they must always add up to 100%. This fact was used in the exercise to set up the first equation (\( x + y = 100 \)). In the context of the exercise for rubidium, knowing the isotopic abundance allows scientists to precisely calculate atomic masses, which are critical for making 'atomic clocks' and other sensitive devices.
Solving Systems of Equations in Chemistry
In chemistry, solving systems of equations plays a significant role in quantitatively analyzing substances. When dealing with multiple unknowns, like isotopic abundances, we use systems of equations to solve for each unknown. The exercise presented a system of linear equations with two variables, x and y, which represent the abundances of the two isotopes of rubidium.

The method chosen in the step-by-step solution was substitution, a commonly used technique in algebra. Substitution involves rearranging one equation to express one variable in terms of the other and then using that expression to replace the variable in the second equation. The goal is to reduce the system to a single equation with one unknown, which can then be solved. This approach can be visualized by imagining the simultaneous overlap of both isotopes' contributions to the element鈥檚 atomic weight鈥攚here they meet provides the solution for each's abundance.

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Most popular questions from this chapter

Silicon tetrachloride \(\left(\mathrm{SiCl}_{4}\right)\) can be prepared by heating \(\mathrm{Si}\) in chlorine gas:$$\mathrm{Si}(s)+2 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SiCl}_{4}(l)$$In one reaction, 0.507 mole of \(\mathrm{SiCl}_{4}\) is produced. How many moles of molecular chlorine were used in the reaction?

(a) For molecules having small molecular masses, mass spectrometry can be used to identify their formulas. To illustrate this point, identify the molecule that most likely accounts for the observation of a peak in a mass spectrum at 16 amu, 17 amu, 18 amu, and 64 amu. (b) Note that there are (among others) two likely molecules that would give rise to a peak at 44 amu, namely, \(\mathrm{C}_{3} \mathrm{H}_{8}\) and \(\mathrm{CO}_{2} .\) In such cases, a chemist might try to look for other peaks generated when some of the molecules break apart in the spectrometer. For example, if a chemist sees a peak at 44 amu and also one at 15 amu, which molecule is producing the 44 -amu peak? Why? (c) Using the following precise atomic masses \(-\mathrm{H}\) ( 1.00797 amu), \({ }^{12} \mathrm{C}(12.00000 \mathrm{amu}),\) and \({ }^{16} \mathrm{O}(15.99491 \mathrm{amu})-\) how precisely must the masses of \(\mathrm{C}_{3} \mathrm{H}_{8}\) and \(\mathrm{CO}_{2}\), be measured to distinguish between them?

Ammonia is a principal nitrogen fertilizer. It is prepared by the reaction between hydrogen and nitrogen.$$3 \mathrm{H}_{2}(g)+\mathrm{N}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)$$In a particular reaction, 6.0 moles of \(\mathrm{NH}_{3}\) were produced. How many moles of \(\mathrm{H}_{2}\) and how many moles of \(\mathrm{N}_{2}\) were reacted to produce this amount of \(\mathrm{NH}_{3} ?\)

Heating \(2.40 \mathrm{~g}\) of the oxide of metal \(\mathrm{X}\) (molar mass of \(\mathrm{X}=55.9 \mathrm{~g} / \mathrm{mol}\) ) in carbon monoxide (CO) yields the pure metal and carbon dioxide. The mass of the metal product is \(1.68 \mathrm{~g}\). From the data given, show that the simplest formula of the oxide is \(\mathrm{X}_{2} \mathrm{O}_{3}\) and write a balanced equation for the reaction.

Consider the reaction \(3 \mathrm{~A}+2 \mathrm{~B} \rightarrow 3 \mathrm{C}\). A student mixed 4.0 moles of \(\mathrm{A}\) with 4.0 moles of \(\mathrm{B}\) and obtained 2.8 moles of \(\mathrm{C}\). What is the percent yield of the reaction?
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