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Chapter 4: Problem 42

Two isotopes of rubidium occur naturally, with \({ }_{37}^{85} \mathrm{Rb}\) at \(72.17 \%\) \((84.91 \mathrm{amu})\) and \({ }_{37}^{87} \mathrm{Rb}\) at \(27.83 \%(86.91 \mathrm{amu}) .\) Calculate the atomic mass for rubidium using the weighted average mass method.

Short Answer

Expert verified
85.48 amu

Step by step solution

01

- Identify the Isotopes and Their Properties

Two naturally occurring isotopes of rubidium are \({ }_{37}^{85} \mathrm{Rb}\) and \({ }_{37}^{87} \mathrm{Rb}\). The isotope \({ }_{37}^{85} \mathrm{Rb}\) has a mass of 84.91 amu and an abundance of 72.17%, and the isotope \({ }_{37}^{87} \mathrm{Rb}\) has a mass of 86.91 amu and an abundance of 27.83%.
02

- Convert Percentages to Decimals

Convert the percentages to decimal form by dividing by 100. \(72.17 \% = 0.7217\) and \(27.83 \% = 0.2783\).
03

- Multiply Each Isotope's Mass by Its Relative Abundance

Multiply the atomic mass of each isotope by its decimal abundance: \[84.91 \mathrm{amu} \times 0.7217 = 61.29 \mathrm{amu} \] \[86.91 \mathrm{amu} \times 0.2783 = 24.19 \mathrm{amu} \]
04

- Add the Results Together

Add the results from the previous step to find the weighted average atomic mass: \[61.29 \mathrm{amu} + 24.19 \mathrm{amu} = 85.48 \mathrm{amu} \]

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Key Concepts

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

Isotopes
Isotopes are variants of a particular chemical element that have the same number of protons but different numbers of neutrons. This means they have the same atomic number but different mass numbers. An element can have multiple isotopes that are naturally occurring or artificially made.
In the context of our rubidium example, there are two isotopes: \({ }_{37}^{85} \text{Rb}\) and \({ }_{37}^{87} \text{Rb}\). Both isotopes have 37 protons, but \({ }_{37}^{85} \text{Rb}\) has 48 neutrons, while \({ }_{37}^{87} \text{Rb}\) has 50 neutrons.
Understanding isotopes is crucial because they contribute differently to the atomic mass of an element, depending on their relative abundance.
Relative Abundance
Relative abundance refers to the percentage of a particular isotope that occurs naturally compared to the total amount of the element. It helps us understand how common or rare a specific isotope is.
In the rubidium example, \({ }_{37}^{85} \text{Rb}\) has a relative abundance of 72.17%, meaning it makes up about 72.17% of all rubidium atoms found in nature. On the other hand, \({ }_{37}^{87} \text{Rb}\) has a relative abundance of 27.83%.
To use these percentages in calculations, we convert them to decimal form by dividing by 100. This gives us 0.7217 for \({ }_{37}^{85} \text{Rb}\) and 0.2783 for \({ }_{37}^{87} \text{Rb}\).
Weighted Average
The weighted average is a method used to calculate an average that takes into account the relative importance or frequency of some values. It is essential when dealing with isotopes because not all isotopes contribute equally to the atomic mass of an element.
For rubidium, calculate the weighted average atomic mass by multiplying each isotope's mass by its relative abundance and then summing the results:
  • For \({ }_{37}^{85} \text{Rb}\): \84.91 \text{ amu} \times 0.7217 = 61.29 \text{ amu}\
  • For \({ }_{37}^{87} \text{Rb}\): \86.91 \text{ amu} \times 0.2783 = 24.19 \text{ amu}\
Add these two numbers together:
\61.29 \text{ amu} + 24.19 \text{ amu} = 85.48 \text{ amu}\
This weighted average gives us the atomic mass of rubidium (85.48 amu).
Atomic Mass Unit (amu)
The atomic mass unit (amu) is a standard unit of mass used to express atomic and molecular weights. One amu is defined as one-twelfth of the mass of a carbon-12 atom.
Using amu simplifies calculations involving atomic masses, as it provides a consistent unit of measure that's easy to work with.
For our rubidium isotopes:
  • \({ }_{37}^{85} \text{Rb}\) has a mass of 84.91 amu
  • \({ }_{37}^{87} \text{Rb}\) has a mass of 86.91 amu
By using amu, we ensure our calculations are precise and manageable, making it easier to find the atomic mass of elements like rubidium.

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Most popular questions from this chapter

With the help of the periodic table, write the numbers of protons in and the group positions of the following elements: (4.4) a. Bromine b. Rhenium c. Ruthenium d. Zinc e. Gallium f. Lead

Argon has three naturally occurring isotopes, with mass numbers 36,38 , and 40 . a. Write the atomic symbol for each of these atoms. b. How are these isotopes alike? c. How are they different? d. Why is the atomic mass of argon listed on the periodic table not a whole number? e. Which isotope is the most prevalent in a sample of argon?

The most prevalent isotope of gold is Au- \(197 .(4.5)\) a. How many protons, neutrons, and electrons are in this isotope? b. What is the atomic symbol of another isotope of gold with 116 neutrons? c. What is the atomic symbol of an atom with an atomic number of 78 and 116 neutrons?

The most abundant isotope of nickel is \({ }_{28}^{58} \mathrm{Ni}\). (4.4) a. How many protons, neutrons, and electrons are in \(\frac{58}{28} \mathrm{Ni}\) ? b. What is the atomic symbol of another isotope of nickel with 35 neutrons? c. What is the name and symbol of an atom with the same mass number as in part \(\mathbf{b}\) and 34 neutrons?

Consider the following atoms in which \(\mathrm{X}\) represents the chemical symbol of the element: \((4.4,4.5)\) \({ }_{11}^{22} \mathrm{X},{ }_{10}^{22} \mathrm{X},{ }_{10}^{19} \mathrm{X},{ }_{92}^{238} \mathrm{X},{ }_{92}^{236} \mathrm{X}\) a. Which atoms have the same number of protons? b. Which atoms are isotopes? Of what element? c. Which atoms have the same mass number?
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