Question

silver has two naturally occurring isotopes, ag - 107 (106.905 amu) and ag - 109 (108.904 amu) and the average atomic mass of silver is 107.87 amu. what are the percent abundances of the two naturally occurring isotopes? you must show your work to receive full credit.

Explanation:

Step1: Let the percent abundance of Ag - 107 be $x$.

Then the percent abundance of Ag - 109 is $1 - x$.

Step2: Set up the average - atomic - mass equation.

The average atomic mass formula is $\text{Average atomic mass}=\sum_{i}(m_i\times a_i)$, where $m_i$ is the mass of the isotope and $a_i$ is its abundance. So, $107.87 = 106.905x+108.904(1 - x)$.

Step3: Expand the equation.

$107.87=106.905x + 108.904-108.904x$.

Step4: Combine like terms.

$107.87-108.904=106.905x-108.904x$.
$-1.034=-1.999x$.

Step5: Solve for $x$.

$x=\frac{1.034}{1.999}\approx0.517$.

Step6: Find the percent abundances.

The percent abundance of Ag - 107 is $x\times100 = 51.7\%$.
The percent abundance of Ag - 109 is $(1 - x)\times100=(1 - 0.517)\times100 = 48.3\%$.

Answer:

The percent abundance of Ag - 107 is $51.7\%$ and the percent abundance of Ag - 109 is $48.3\%$.