Problem 162 - Chemistry An Atoms First Approach

Question

Natural rubidium has the average mass of 85.4678 u and is composed of isotopes \(^{85} \mathrm{Rb}\) (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 \(^{87} \mathrm{Rb}\).

Step-by-Step Solution

Answer

The mass of \(^{87}\mathrm{Rb}\) is approximately \(86.938\mathrm{u}\).

Step 1: Define the mass fraction of each isotope

We are given the ratio of atoms \(^{85}\mathrm{Rb}/^{87}\mathrm{Rb} = 2.591\). To calculate the mass fraction of each isotope (the percentage of each isotope in the mixture), let's first denote the mass fraction of \(^{85}\mathrm{Rb}\) as x and the mass fraction of \(^{87}\mathrm{Rb}\) as y. Since there are only two isotopes of rubidium, we know that their mass fractions must sum up to 1 or 100% : x + y = 1.

Step 2: Determine the mass fraction of \(^{85}\mathrm{Rb}\) and \(^{87}\mathrm{Rb}\)

We are given the ratio of atoms as \(^{85}\mathrm{Rb}/^{87}\mathrm{Rb} = 2.591\). We can rewrite this as \(x/y = 2.591\). Since x + y = 1, we can also write y = 1 - x. Now we substitute y with 1 - x in the \(x/y = 2.591\) equation: $$\frac{x}{1-x} = 2.591$$ Solving this equation for x, we get: $$x = \frac{2.591}{1+2.591} = \frac{2.591}{3.591} \approx 0.721$$ Now we can find the mass fraction of \(^{87}\mathrm{Rb}\) (y) by substituting the value of x: $$y = 1 - x = 1 - 0.721 = 0.279$$

Step 3: Calculate the mass of \(^{87}\mathrm{Rb}\)

We can now use the average mass of natural rubidium and the mass fractions of both isotopes to find the mass of \(^{87}\mathrm{Rb}\). We know the average mass (\(m_{avg}\)) is related to the mass fractions and the individual masses as follows: $$m_{avg}= x \cdot m(^{85}\mathrm{Rb}) + y \cdot m(^{87}\mathrm{Rb})$$ where \(m(^{85}\mathrm{Rb}) = 84.9117\mathrm{u}\) and \(m_{avg} = 85.4678\mathrm{u}\). We already know x and y, so now we solve for the mass of \(^{87}\mathrm{Rb}\) (denoted as \(m(^{87}\mathrm{Rb})\)): $$85.4678\mathrm{u} = (0.721 \cdot 84.9117\mathrm{u}) + (0.279 \cdot m(^{87}\mathrm{Rb}))$$

Step 4: Solve for the mass of \(^{87}\mathrm{Rb}\)

Rearrange the equation and solve for \(m(^{87}\mathrm{Rb})\): $$m(^{87}\mathrm{Rb}) = \frac{85.4678\mathrm{u} - (0.721 \cdot 84.9117\mathrm{u})}{0.279}$$ Now, we just need to simplify: $$m(^{87}\mathrm{Rb}) \approx \frac{85.4678\mathrm{u} - 61.1430\mathrm{u}}{0.279} \approx 86.938\mathrm{u}$$ So, the mass of \(^{87}\mathrm{Rb}\) is approximately \(86.938\mathrm{u}\).

Key Concepts

rubidium isotopes

Rubidium is a chemical element symbolized by Rb, appearing in the periodic table with the atomic number 37. This element exists in nature primarily as two isotopes, which are different forms of the same element that have the same number of protons but a different number of neutrons. The isotopes of rubidium found naturally are

• \(^{85}\mathrm{Rb}\): This isotope comprises the bulk of natural rubidium.
• \(^{87}\mathrm{Rb}\): Although less abundant compared to \(^{85}\mathrm{Rb}\), this isotope also occurs naturally.

Rubidium isotopes behave chemically the same because they have the same number of electrons and thus participate in reactions in the same way. However, differences arise in their physical properties due to the difference in mass. Understanding the natural occurrence of these isotopes helps in numerous scientific calculations, including calculating atomic masses.

atomic mass calculation

Atomic mass is essentially the weighted average of all the isotope masses of an element, considering their relative abundance in nature. It offers an average value that you might use to calculate the mass of a sample containing various isotopes. Let's imagine we have natural rubidium with an average atomic mass of 85.4678 u. This average is a result of the differing masses and abundances of its isotopes, \(^{85}\mathrm{Rb}\) and \(^{87}\mathrm{Rb}\). The calculation involves:

• Multiplying the mass of each isotope by its relative abundance.
• Adding these values together to get the average mass.

For example, \[m_{avg} = (0.721 \cdot 84.9117 \mathrm{u}) + (0.279 \cdot 86.938 \mathrm{u})\] This depicts a simplified way of calculating the atomic mass of elements based on their isotopic composition. Understanding atomic mass calculation allows scientists to predict the behavior of elements in reactions and natural processes.

mass fractions

Mass fraction is a way to express the concentration of each component in a mixture. It's particularly useful in chemistry for determining the ratio of various isotopes of an element in a sample. In the case of rubidium, you often look at the mass fraction of its two primary isotopes, \(^{85}\mathrm{Rb}\) and \(^{87}\mathrm{Rb}\). When given the ratio of atoms, such as \(^{85}\mathrm{Rb}/^{87}\mathrm{Rb} = 2.591\), you can use this to calculate their mass fractions.

• First, express the atom ratio as a fraction of the total mass.
• For \(x\), the mass fraction of \(^{85}\mathrm{Rb}\), solve the equation \(x/(1-x) = 2.591\) to find \(x = 0.721\).
• Then, determine \(y = 1 - x = 0.279\), the mass fraction of \(^{87}\mathrm{Rb}\).

These fractions help in further calculations regarding the mass and behavior of isotopic mixes.

chemical calculations

Chemical calculations encompass various mathematical processes used to determine quantitative aspects of chemistry, including isotope ratios and atomic masses. In rubidium's case, we start by understanding the isotope composition and any given ratios, like \(^{85}\mathrm{Rb}/^{87}\mathrm{Rb} = 2.591\). With such ratios, you can find mass fractions, and from there, calculate unknowns like the mass of a particular isotope.To find the mass of \(^{87}\mathrm{Rb}\):

• Identify known values, such as the average atomic mass and mass fractions.
• Set up an equation relating the average atomic mass to these fractions and solve for unknown variables using algebra.

For example:
\[85.4678\mathrm{u} = (0.721 \cdot 84.9117\mathrm{u}) + (0.279 \cdot m(^{87}\mathrm{Rb}))\]Solving will give you the mass of \(^{87}\mathrm{Rb}\) as approximately \(86.938\mathrm{u}\). These calculations are crucial for predicting reactions and understanding the composition of natural elements.