Question

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

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

Step-by-step solution

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: \[M = m_{85}a_{85} + m_{87}a_{87}\]

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 \(0.72158\).

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: \[85.4678 = 84.9117(0.72158) + m_{87}(0.27842)\]

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 \(86.9090\mathrm{u}\).

Key concepts

Abundance Ratio

The idea of abundance ratio refers to the relative proportion of isotopes found in a natural sample of an element. In the case of rubidium, we have two isotopes: \(^{85}\text{Rb}\) and \(^{87}\text{Rb}\). The abundance ratio is simply the ratio of the number of atoms of one isotope to another. For example, the abundance ratio of \(^{85}\text{Rb}\) to \(^{87}\text{Rb}\) is given as 2.591. This means, for every 2.591 atoms of \(^{85}\text{Rb}\), there is 1 atom of \(^{87}\text{Rb}\).

• The abundance ratio helps us understand the composition of isotopes in a sample.
• It is crucial when calculating the average atomic mass of an element with multiple isotopes.
• Knowing the ratio allows us to express the abundance of one isotope in terms of the other.

In our calculation, we can express the abundances of \(^{85}\text{Rb}\) and \(^{87}\text{Rb}\) as \(2.591x\) and \(x\), respectively. Since the total abundance is always 1, this relationship forms the basis for further calculations.

Average Atomic Mass

The average atomic mass of an element is the weighted average of the masses of its isotopes, considering their relative abundances. Natural elements often exist as a mixture of isotopes, each contributing to the overall atomic mass based on how common they are in nature. The formula to find the average atomic mass \( M \) is:

\[ M = \sum m_i a_i \] where \(m_i\) is the mass of each isotope and \(a_i\) is its abundance.

• This calculation considers both the mass and the abundance of each isotope.
• It results in an average that can be compared to the standard atomic mass seen on periodic tables.
• The key is to sum the products of each isotope's mass and its abundance.

In the rubidium example, the average atomic mass is given as \(85.4678 \text{u}\), which is calculated by combining \(^{85}\text{Rb}\) and \(^{87}\text{Rb}\) with their respective abundances. Solving the equation \[ 85.4678 = 84.9117 \times 0.72158 + m_{87} \times 0.27842 \] gives us the mass of the \(^{87}\text{Rb}\) isotope, showing how significant each isotope's contribution is to the total atomic mass.

Rubidium Isotopes

Rubidium is a fascinating element found in nature as two main isotopes: \(^{85}\text{Rb}\) and \(^{87}\text{Rb}\). Each isotope has a different atomic mass, resulting in a slightly different chemical behavior and mass contribution.

• The isotope \(^{85}\text{Rb}\) has a mass of \(84.9117 \text{u}\) and is more abundant in natural rubidium.
• The isotope \(^{87}\text{Rb}\) is present to a lesser extent but plays a significant role in dating geological formations due to its radioactivity.
• The isotopes differ in neutron number while sharing the same number of protons (37, which is unique to rubidium).

Understanding these isotopes helps in applications ranging from isotopic dating to materials science. Their individual properties collectively influence the behavior of rubidium in both chemical reactions and geological formations. The calculated mass of \(^{87}\text{Rb}\), approximately \(86.9090 \text{u}\), is essential for precision in scientific calculations and understanding the element's role in various processes.