Question

the element rhenium (re) has two naturally - occurring isotopes, ¹⁸⁵re and ¹⁸⁷re, with an average atomic mass of 186.2 u. rhenium is 62.60% ¹⁸⁷re, and the atomic mass of ¹⁸⁵re is 184.956 u. calculate the atomic mass of ¹⁸⁷re.

Explanation:

Step1: Recall average - atomic - mass formula

The average atomic mass ($A_{avg}$) of an element with two isotopes $A_1$ and $A_2$ and their relative abundances $x_1$ and $x_2$ is given by $A_{avg}=x_1A_1 + x_2A_2$, where $x_1 + x_2=1$. Given $A_{avg}=186.956$ u, $A_1$ (atomic mass of $^{185}$Re) is unknown, $A_2$ (atomic mass of $^{187}$Re) is to be found, $x_1 = 0.474$ (since $^{185}$Re is $47.4\%$ abundant as $^{186}$Re is $52.6\%$ abundant, so $x_1=1 - 0.526$), and $x_2 = 0.526$.

Step2: Substitute known values into the formula

Let the atomic mass of $^{185}$Re be $m_1$ and of $^{187}$Re be $m_2$. We know $186.956=0.474m_1+0.526m_2$. Also, assume the atomic mass of $^{185}$Re is approximately 185 u (since it is named $^{185}$Re). Then we have $186.956 = 0.474\times185+0.526m_2$.

Step3: Solve for $m_2$

First, calculate $0.474\times185 = 87.69$. Then the equation becomes $186.956=87.69 + 0.526m_2$. Rearrange to get $0.526m_2=186.956 - 87.69$. So $0.526m_2=99.266$. Then $m_2=\frac{99.266}{0.526}\approx188.72$ u. But if we consider the more accurate way, let the average - atomic - mass formula $A_{avg}=x_1A_1+x_2A_2$. We know $A_{avg} = 186.956$, $x_1=0.474$, $x_2 = 0.526$. Let $A_1$ be the atomic mass of $^{185}$Re and $A_2$ be the atomic mass of $^{187}$Re.

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We know that the average atomic mass is given, and we can also use the fact that the isotopes contribute to the average. Since the average atomic mass is $186.956$ u and the abundance of $^{185}$Re is $47.4\%$ and of $^{187}$Re is $52.6\%$.

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Answer:

The atomic mass of $^{187}$Re is approximately $188.53$ u.