Natural rubidium has the average mass of \(85.4678 \mathrm{u}\) and is composed of isotopes \({ }^{85} \mathrm{Rb}\) (mass \(=84.9117 \mathrm{u}\) ) and \({ }^{\mathrm{s} 7} \mathrm{Rb}\). The ratio of atoms \({ }^{85} \mathrm{Rb} /{ }^{\mathrm{s} 7} \mathrm{Rb}\) in natural rubidium is \(2.591 .\) Calculate the mass of \(^{87} \mathrm{Rb}\).

In order to find the missing mass, we can use the weighted average formula. Since natural rubidium is a mixture of its two isotopes, we can set up an equation representing their weighted average: \[\text{average mass} =\frac{m_1 \cdot n_1 + m_2 \cdot n_2}{n_1 + n_2}\] where \(m_1\) and \(m_2\) are the masses of the isotopes, and \(n_1\) and \(n_2\) are their respective amounts in the mixture. In our case, the ratio \({}^{85}\mathrm{Rb} / {}^{87}\mathrm{Rb}\) is \(2.591\), so we have: \[\text{average mass} =\frac{m_{85}\cdot n_{85} + m_{87}\cdot n_{87}}{n_{85} + n_{87}}\] with \(m_{85} = 84.9117 \mathrm{u}\) being the mass of \(^{85}\mathrm{Rb}\) and \(m_{87}\) being the mass of \(^{87}\mathrm{Rb}\), which we want to find.

The ratio between the number of atoms of \(^{85}\mathrm{Rb}\) and \(^{87}\mathrm{Rb}\) is given as \(2.591\) (approximately \(2.59\)). We can rewrite this as: \[\frac{n_{85}}{n_{87}} = 2.591\] Now, let's set \(n_{87} = x\). Then, \(n_{85} = 2.591x\), and we can plug this relationship back into our weighted average formula.

Now, let's plug in the known values for the average mass of natural rubidium and the mass of \(^{85}\mathrm{Rb}\) into our weighted average equation and solve for \(m_{87}\): \[85.4678 = \frac{84.9117(2.591x) + m_{87} x}{2.591x + x}\] First, multiply both sides by \((2.591x + x)\) to get rid of the denominator: \[85.4678(2.591x + x) = 84.9117(2.591x) + m_{87} x\] Now, we can expand this equation and regroup to isolate \(m_{87}\): \[85.4678(3.591x) = 84.9117(2.591)x + m_{87} x\] \[85.4678(3.591x) - 84.9117(2.591)x = m_{87} x\] Now, we can divide both sides by \(x\) and simplify: \[85.4678(3.591) - 84.9117(2.591) = m_{87}\] \[307.064582 - 220.3584157 = m_{87}\] \[86.7061663 \approx m_{87}\]

So, the mass of \(^{87}\mathrm{Rb}\) is approximately \(86.706\mathrm{u}\).