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Chapter 3: Problem 172

Natural rubidium has the average mass of 85.4678 \(\mathrm{u}\) and is composed of isotopes \(^{85} \mathrm{Rb}(\mathrm{mass}=84.9117 \mathrm{u})\) and $^{87} \mathrm{Rb}\( . The ratio of atoms \)^{85} \mathrm{Rb} /^{87} \mathrm{Rb}$ in natural rubidium is \(2.591 .\) Calculate the mass of \(^{87} \mathrm{Rb}\) .

Short Answer

Expert verified
The mass of \(^{87}\mathrm{Rb}\) is approximately \(86.9090\mathrm{u}\).

Step by step solution

01

Write down the formula for the average mass of the element

The average mass (M) of an element can be computed as the sum of the product of the mass of each isotope (m_i) and their respective abundances (a_i). The formula is: \[M = \sum m_i a_i\] In this case, there are only two isotopes, \(^{85}\mathrm{Rb}\) and \(^{87}\mathrm{Rb}\). So the formula becomes: \[M = m_{85}a_{85} + m_{87}a_{87}\]
02

Express the abundance of \(^{87}\mathrm{Rb}\) in terms of the abundance of the \(^{85}\mathrm{Rb}\)

We know that the ratio of atoms \(^{85}\mathrm{Rb}\) and \(^{87}\mathrm{Rb}\) is \(2.591\). So, we can write: \[\frac{a_{85}}{a_{87}} = 2.591\] Now, let \(x\) be the abundance of \(^{87}\mathrm{Rb}\), then the abundance of \(^{85}\mathrm{Rb}\) will be \(2.591x\). Since the sum of abundances of all isotopes must be 1, we can write: \[2.591x + x = 1\] Solving this equation for \(x\): \[x = \frac{1}{3.591} \approx 0.27842\] Thus, the abundance of \(^{87}\mathrm{Rb}\) is approximately \(0.27842\) and the abundance of \(^{85}\mathrm{Rb}\) is approximately \(0.72158\).
03

Insert the known values into the formula

We are given that the average mass of natural rubidium is \(85.4678\mathrm{u}\) and the mass of \(^{85}\mathrm{Rb}\) is \(84.9117\mathrm{u}\). Let \(m_{87}\) be the mass of \(^{87}\mathrm{Rb}\). We now have all the values required to solve for \(m_{87}\) using the formula: \[85.4678 = 84.9117(0.72158) + m_{87}(0.27842)\]
04

Solve for the mass of \(^{87}\mathrm{Rb}\)

Rearrange the equation to isolate \(m_{87}\): \[m_{87} = \frac{85.4678 - 84.9117(0.72158)}{0.27842}\] Now, solve for \(m_{87}\): \[m_{87} \approx 86.9090\mathrm{u}\] Thus, the mass of \(^{87}\mathrm{Rb}\) is approximately \(86.9090\mathrm{u}\).

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