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Chapter 5: Problem 162

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

Short Answer

Expert verified
The mass of \(^{87}\mathrm{Rb}\) is approximately \(86.938\mathrm{u}\).

Step by step solution

01

Define the mass fraction of each isotope

We are given the ratio of atoms \(^{85}\mathrm{Rb}/^{87}\mathrm{Rb} = 2.591\). To calculate the mass fraction of each isotope (the percentage of each isotope in the mixture), let's first denote the mass fraction of \(^{85}\mathrm{Rb}\) as x and the mass fraction of \(^{87}\mathrm{Rb}\) as y. Since there are only two isotopes of rubidium, we know that their mass fractions must sum up to 1 or 100% : x + y = 1.
02

Determine the mass fraction of \(^{85}\mathrm{Rb}\) and \(^{87}\mathrm{Rb}\)

We are given the ratio of atoms as \(^{85}\mathrm{Rb}/^{87}\mathrm{Rb} = 2.591\). We can rewrite this as \(x/y = 2.591\). Since x + y = 1, we can also write y = 1 - x. Now we substitute y with 1 - x in the \(x/y = 2.591\) equation: $$\frac{x}{1-x} = 2.591$$ Solving this equation for x, we get: $$x = \frac{2.591}{1+2.591} = \frac{2.591}{3.591} \approx 0.721$$ Now we can find the mass fraction of \(^{87}\mathrm{Rb}\) (y) by substituting the value of x: $$y = 1 - x = 1 - 0.721 = 0.279$$
03

Calculate the mass of \(^{87}\mathrm{Rb}\)

We can now use the average mass of natural rubidium and the mass fractions of both isotopes to find the mass of \(^{87}\mathrm{Rb}\). We know the average mass (\(m_{avg}\)) is related to the mass fractions and the individual masses as follows: $$m_{avg}= x \cdot m(^{85}\mathrm{Rb}) + y \cdot m(^{87}\mathrm{Rb})$$ where \(m(^{85}\mathrm{Rb}) = 84.9117\mathrm{u}\) and \(m_{avg} = 85.4678\mathrm{u}\). We already know x and y, so now we solve for the mass of \(^{87}\mathrm{Rb}\) (denoted as \(m(^{87}\mathrm{Rb})\)): $$85.4678\mathrm{u} = (0.721 \cdot 84.9117\mathrm{u}) + (0.279 \cdot m(^{87}\mathrm{Rb}))$$
04

Solve for the mass of \(^{87}\mathrm{Rb}\)

Rearrange the equation and solve for \(m(^{87}\mathrm{Rb})\): $$m(^{87}\mathrm{Rb}) = \frac{85.4678\mathrm{u} - (0.721 \cdot 84.9117\mathrm{u})}{0.279}$$ Now, we just need to simplify: $$m(^{87}\mathrm{Rb}) \approx \frac{85.4678\mathrm{u} - 61.1430\mathrm{u}}{0.279} \approx 86.938\mathrm{u}$$ So, the mass of \(^{87}\mathrm{Rb}\) is approximately \(86.938\mathrm{u}\).

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Most popular questions from this chapter

Bornite \(\left(\mathrm{Cu}_{3} \mathrm{FeS}_{3}\right)\) is a copper ore used in the production of copper. When heated, the following reaction occurs: $$2 \mathrm{Cu}_{3} \mathrm{FeS}_{3}(s)+7 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{Cu}(s)+2 \mathrm{FeO}(s)+6 \mathrm{SO}_{2}(g)$$ If 2.50 metric tons of bornite is reacted with excess \(\mathrm{O}_{2}\) and the process has an \(86.3 \%\) yield of copper, what mass of copper is produced?

A compound with molar mass \(180.1 \mathrm{g} / \mathrm{mol}\) has the following composition by mass: $$\begin{array}{|ll|}\hline C & 40.0 \% \\\H & 6.70 \% \\\O & 53.3 \% \\\\\hline\end{array}$$ Determine the empirical and molecular formulas of the compound.

A compound containing only sulfur and nitrogen is \(69.6 \% \mathrm{S}\) by mass; the molar mass is \(184 \mathrm{g} / \mathrm{mol}\). What are the empirical and molecular formulas of the compound?

Consider samples of phosphine \(\left(\mathrm{PH}_{3}\right),\) water \(\left(\mathrm{H}_{2} \mathrm{O}\right),\) hydrogen sulfide \(\left(\mathrm{H}_{2} \mathrm{S}\right),\) and hydrogen fluoride (HF), each with a mass of \(119 \mathrm{g} .\) Rank the compounds from the least to the greatest number of hydrogen atoms contained in the samples.

A 0.755-g sample of hydrated copper(II) sulfate $$\mathrm{CuSO}_{4} \cdot x \mathrm{H}_{2} \mathrm{O}$$ was heated carefully until it had changed completely to anhydrous copper(II) sulfate (CuSO_) with a mass of 0.483 g. Determine the value of \(x .\) [This number is called the number of waters of hydration of copper(II) sulfate. It specifies the number of water molecules per formula unit of \(\mathrm{CuSO}_{4}\) in the hydrated crystal.]
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