Question

Natural rubidium has the average mass of \(85.4678 \mathrm{u}\) and is composed of isotopes \({ }^{85} \mathrm{Rb}\) (mass \(=84.9117 \mathrm{u}\) ) and \({ }^{\mathrm{s} 7} \mathrm{Rb}\). The ratio of atoms \({ }^{85} \mathrm{Rb} /{ }^{\mathrm{s} 7} \mathrm{Rb}\) in natural rubidium is \(2.591 .\) Calculate the mass of \(^{87} \mathrm{Rb}\).

Short answer

The mass of \(^{87}\mathrm{Rb}\) is approximately \(86.706\mathrm{u}\).

Step-by-step solution

Set up an equation using the weighted average formula

In order to find the missing mass, we can use the weighted average formula. Since natural rubidium is a mixture of its two isotopes, we can set up an equation representing their weighted average: \[\text{average mass} =\frac{m_1 \cdot n_1 + m_2 \cdot n_2}{n_1 + n_2}\] where \(m_1\) and \(m_2\) are the masses of the isotopes, and \(n_1\) and \(n_2\) are their respective amounts in the mixture. In our case, the ratio \({}^{85}\mathrm{Rb} / {}^{87}\mathrm{Rb}\) is \(2.591\), so we have: \[\text{average mass} =\frac{m_{85}\cdot n_{85} + m_{87}\cdot n_{87}}{n_{85} + n_{87}}\] with \(m_{85} = 84.9117 \mathrm{u}\) being the mass of \(^{85}\mathrm{Rb}\) and \(m_{87}\) being the mass of \(^{87}\mathrm{Rb}\), which we want to find.

Set up the ratio

The ratio between the number of atoms of \(^{85}\mathrm{Rb}\) and \(^{87}\mathrm{Rb}\) is given as \(2.591\) (approximately \(2.59\)). We can rewrite this as: \[\frac{n_{85}}{n_{87}} = 2.591\] Now, let's set \(n_{87} = x\). Then, \(n_{85} = 2.591x\), and we can plug this relationship back into our weighted average formula.

Plug in the values and solve for \(m_{87}\)

Now, let's plug in the known values for the average mass of natural rubidium and the mass of \(^{85}\mathrm{Rb}\) into our weighted average equation and solve for \(m_{87}\): \[85.4678 = \frac{84.9117(2.591x) + m_{87} x}{2.591x + x}\] First, multiply both sides by \((2.591x + x)\) to get rid of the denominator: \[85.4678(2.591x + x) = 84.9117(2.591x) + m_{87} x\] Now, we can expand this equation and regroup to isolate \(m_{87}\): \[85.4678(3.591x) = 84.9117(2.591)x + m_{87} x\] \[85.4678(3.591x) - 84.9117(2.591)x = m_{87} x\] Now, we can divide both sides by \(x\) and simplify: \[85.4678(3.591) - 84.9117(2.591) = m_{87}\] \[307.064582 - 220.3584157 = m_{87}\] \[86.7061663 \approx m_{87}\]

Final answer

So, the mass of \(^{87}\mathrm{Rb}\) is approximately \(86.706\mathrm{u}\).

Key concepts

Isotopes

Isotopes are atoms that have the same number of protons but differ in the number of neutrons. This means isotopes of an element have the same atomic number but different mass numbers. Isotopes are important because they help us understand the atomic mass, which is the average of all isotopes of an element as they occur naturally. Here are some key aspects of isotopes:

• The same element can have multiple isotopes, each with a different mass. For instance, carbon has isotopes like carbon-12 and carbon-14, where the numbers denote the mass number or the total number of protons and neutrons.
• Isotopes can be classified as stable or radioactive. Stable isotopes do not change over time, while radioactive isotopes decay and transform into other elements or isotopes.
• Understanding isotopes is crucial in fields like chemistry, physics, and even environmental science where isotopes are used to date archaeological finds or to study climate changes in ice cores.

Rubidium Isotopes

Rubidium is an element that naturally occurs with two main isotopes. These are Rb-85 and Rb-87. They each contribute to the overall atomic mass of rubidium, but in different ratios:
The symbol for rubidium is Rb, and it is found on the periodic table with an atomic number of 37. Here's more about the isotopes:

• **Rb-85**: This isotope has a mass of approximately 84.9117 u. It is more abundant in nature compared to its counterpart, accounting for about 72% of natural rubidium.
• **Rb-87**: Though less common, with a natural abundance of around 28%, Rb-87 still significantly influences the average atomic mass. It has a mass that is higher than Rb-85 and is crucial for calculations involving the element's average mass.

When studying rubidium, one often needs to understand how these isotopes contribute to measurements in atomic physics and chemistry. Calculating the average atomic mass involves understanding and using these isotopic masses and their relative abundances.

Weighted Average Formula

The weighted average formula is an essential tool in chemistry for calculating the average atomic mass of an element. This formula takes into account the mass and relative abundance of each isotope to find an overall average, which reflects what is found naturally. Here's how it works:

You start with the formula for weighted average mass: \[\text{average mass} = \frac{m_1 \cdot n_1 + m_2 \cdot n_2}{n_1 + n_2}\] Where:

• \(m_1\) and \(m_2\) represent the masses of the isotopes.
• \(n_1\) and \(n_2\) represent the amounts (or proportions) of each isotope in the mixture.

To solve real-world problems like calculating the mass of an isotope (as in rubidium), you would iterate through:
- Establishing known values (mass and abundance).- Setting and solving equations to find unknown values, often using ratios to represent the relative amounts of isotopes.- Solving for missing isotope masses involves substituting known values into the formula and algebraically rearranging to find the unknown. Understanding this concept allows students to explore chemistry problems concerning isotopic compositions and average atomic masses effectively.