Question

naturally occurring magnesium has an atomic mass of 24.312 amu and consists of three isotopes. the major isotope is $^{24}$mg, natural abundance 78.99%, relative atomic mass 23.98504. the next most abundant isotope is $^{26}$mg, reltaive atomic mass 25.966. the third most abundant isotope is $^{25}$mg whose natural abundance is in the ratio of 0.9083 to that of $^{26}$mg. what is the relative atomic mass of $^{25}$mg in amu? (4 sig. figs.)
question 10
1 pts
gallium has 2 naturally occurring isotopes, $^{69}$ga and $^{71}$ga. if $^{69}$ga has an exact isotopic mass of 68.9257 amu. and $^{71}$ga has an exact isotopic mass of 70.9249 amu.. the % abundance of $^{69}$ga is
% (report the answer to 3 sig. figs.).

Explanation:

Question 1

Step1: Calculate the combined abundance of $^{25}Mg$ and $^{26}Mg$

The abundance of $^{24}Mg$ is $78.99\% = 0.7899$. So the combined abundance of $^{25}Mg$ and $^{26}Mg$ is $1 - 0.7899=0.2101$.

Step2: Let the abundance of $^{26}Mg$ be $x$. Then the abundance of $^{25}Mg$ is $0.9083x$.

We know that $x + 0.9083x=0.2101$. Combining like - terms gives $1.9083x = 0.2101$. Solving for $x$ (abundance of $^{26}Mg$), we have $x=\frac{0.2101}{1.9083}\approx0.1101$. Then the abundance of $^{25}Mg$ is $0.9083\times0.1101 = 0.0999$.

Step3: Use the atomic - mass formula

The atomic mass formula is $A=\sum_{i}m_ix_i$, where $A$ is the average atomic mass, $m_i$ is the mass of the $i$ - th isotope, and $x_i$ is the abundance of the $i$ - th isotope.
We know $A = 24.312$, $m_1 = 23.98504$ ($^{24}Mg$), $x_1 = 0.7899$, $m_2$ is the mass of $^{25}Mg$ (which we want to find), $x_2 = 0.0999$, $m_3 = 25.966$ ($^{26}Mg$), and $x_3 = 0.1101$.
Substituting into the formula: $24.312=23.98504\times0.7899 + m_2\times0.0999+25.966\times0.1101$.
First, calculate the known products: $23.98504\times0.7899\approx18.955$, $25.966\times0.1101\approx2.859$.
Then the equation becomes $24.312 = 18.955+m_2\times0.0999 + 2.859$.
Combining the known terms on the right - hand side: $18.955+2.859 = 21.814$.
So, $24.312=21.814 + 0.0999m_2$.
Subtract 21.814 from both sides: $24.312−21.814 = 0.0999m_2$.
$2.498 = 0.0999m_2$.
Solving for $m_2$ gives $m_2=\frac{2.498}{0.0999}\approx24.99$ amu.

Step1: Let the abundance of $^{69}Ga$ be $x$. Then the abundance of $^{71}Ga$ is $1 - x$.

The average atomic mass formula is $A = m_1x_1+m_2x_2$.
We know that the average atomic mass of gallium ($A$) is approximately $69.723$ amu (from the periodic table), $m_1 = 68.9257$ amu ($^{69}Ga$), $m_2 = 70.9249$ amu ($^{71}Ga$), $x_1=x$, and $x_2 = 1 - x$.
Substitute into the formula: $69.723=68.9257x+70.9249(1 - x)$.

Step2: Expand and simplify the equation

$69.723=68.9257x+70.9249−70.9249x$.
Combine like - terms: $69.723=70.9249+(68.9257x - 70.9249x)$.
$69.723=70.9249 - 1.9992x$.

Step3: Solve for $x$

First, subtract 70.9249 from both sides: $69.723−70.9249=-1.9992x$.
$-1.2019=-1.9992x$.
Then $x=\frac{1.2019}{1.9992}\approx0.601$.
Converting to percentage, $x = 60.1\%$.

Answer:

$24.99$

Question 2