Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

A periodic table might list the average atomic mass of magnesium as being 24.312u, which the result of weighting the atomic masses of the magnesium isotopes is according to their natural abundances on Earth. The three isotopes and their masses are $^{24} Mg (23.98504 u)$ , $^{25} Mg (23.98584 u)$ , and $^{26} Mg (23.9859 u)$ . The natural abundance of $^{24} Mg$ is 78.99% by mass (that is, 78.99% of the mass of a naturally occurring sample of magnesium is due to the presence of $^{24} Mg$ ).What is the abundance of (a) $^{25} M g$ and (b) $^{26} M g$ ?

The abundance of ${}^{25}M g$ is 9.303%.
The abundance of ${}^{26}M g$ is 11.71%.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Write the given data

Average atomic mass of magnesium, $M_{m g} = 24.312 u$
The mass of the isotope $^{24} Mg$ , $M_{1} = 23.98504 u$
The mass of the isotope $^{25} Mg$ , $M_{2} = 24.98584 u$
The mass of the isotope $^{26} Mg$ , $M_{3} = 24.98259 u$
The natural abundance of the isotope $^{24} Mg$ is 78.99% of its mass.

Step 2: Understanding the concept of natural abundance

The abundance of the three isotopes represents their percent of mass percent in the environment. Considering that there are three isotopes of magnesium, we can get an equation of the total abundance of the three isotopes to be 100% of the natural magnesium substance. Thus, using this concept, we can calculate the abundance of the two other isotopes that are unknown.

Step 3: a) Calculate the abundance of isotope 25Mg

Let $f_{24}$ be the abundance of $^{24} Mg$ , let $f_{25}$ be the abundance of $^{25} Mg$ , and let $f_{26}$ be the abundance of $^{26} Mg$ . Then, the entry in the periodic table for Mg is:

$24.312 = 23.98504 f_{24} + 24.98584 f_{25} + 25.98259 f_{26}$ ….. (a)

Since, there are only three isotopes $f_{24} + f_{25} + f_{26} = 1$ . Solving for $f_{25}$ and $f_{26}$ , we have the above equation as follows:

$f_{26} = 1 - f_{24} - f_{25}$ …… (b)

Now, substituting this above value and $f_{24} = 0.7899$ in equation (a), determine the abundance of isotope ${}^{25}{Mg}$ as follows:

role="math" localid="1661585620014" $24.312 = 23.98504 (0.7899) + 24.98259 - 24.98259 (0.7899) - 25.98259 (f_{25}) f_{25} = 0.09303 f_{25} = 9.303 %$

Hence, the value of abundance is 9.303%.

Step 4: b) Calculate the abundance of isotope 26Mg

Now, substitute the values $f_{24} = 0.7899$ and role="math" localid="1661585674376" $f_{25} = 0.09303$ in equation (b), solve to obtain the abundance of isotope ${}^{26}M g$ as follows:

$\begin{array}{l} f_{26} = 1 - 0.7899 - 0.09303 \\ = 0.1171 \\ = 11.71 % \end{array}$

Hence, the value of abundance is 11.71%.

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Fundamentals Of Physics

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Short Answer

Step by Step Solution

Step 1: Write the given data

Step 2: Understanding the concept of natural abundance

Step 3: a) Calculate the abundance of isotope 25Mg

Step 4: b) Calculate the abundance of isotope 26Mg

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Fundamentals Of Physics

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