Question

The element rhenium (Re) has two naturally occurring isotopes, \({ }^{185} \mathrm{Re}\) and \({ }^{187} \mathrm{Re}\), with an average atomic mass of \(186.207\) amu. Rhenium is \(62.60 \%^{187} \mathrm{Re}\), and the atomic mass of \({ }^{187} \mathrm{Re}\) is \(186.956\) amu. Calculate the mass of \({ }^{185} \mathrm{Re}\).

Short answer

The mass of ${ }^{185} \mathrm{Re}$ is approximately \(185.46\) amu.

Step-by-step solution

Understand the weighted average equation

The weighted average equation we will use to solve for the mass of \({ }^{185} \mathrm{Re}\) is given by: Average atomic mass = (% of isotope 1) * (mass of isotope 1) + (% of isotope 2) * (mass of isotope 2) Let x be the mass of \({ }^{185} \mathrm{Re}\) Then we can rewrite the equation like this: \(186.207 = (62.60\% * 186.956) + ((100\%-62.60\%) * x)\)

Convert percentages

Now, convert the percentages to decimals, so that the equation can be solved easily: \(186.207 = (0.6260 * 186.956) + ((1-0.6260) * x)\)

Rearrange the equation

Now we need to rearrange the equation to isolate x on one side: \(186.207 = (0.6260 * 186.956) + (0.3740 * x)\) Now, subtract the value of the first term from both sides to isolate x: \(186.207 - (0.6260 * 186.956) = 0.3740 * x\)

Solve for x

Now we can solve for x to find the mass of \({ }^{185} \mathrm{Re}\): \( x=\frac{186.207 - (0.6260 * 186.956)}{0.3740} \) Calculating the value of x: \( x=\frac{186.207 - 116.91156}{0.3740} \) \( x \approx 185.45992 \) So, the mass of \({ }^{185} \mathrm{Re}\) is approximately \(185.46\) amu.

Key concepts

Natural Isotopes

Natural isotopes are variations of elements that have the same number of protons but different numbers of neutrons in their nuclei. This variance in neutron count leads to different masses among isotopes of the same element. A common element like carbon, for example, has isotopes such as Carbon-12 and Carbon-13, each named for their respective atomic mass numbers, which is the sum of protons and neutrons.

Elements found in nature usually exist as a mix of their isotopes, each with its own natural abundance, meaning the relative proportion at which each isotope appears. Understanding this concept is vital since it provides the base to comprehend exercises involving average atomic mass calculations and further, allows one to better appreciate how these slight differences in mass contribute to the diversity of physical properties among substances.

Weighted Average Atomic Mass

The weighted average atomic mass of an element is calculated by considering the natural abundances of its isotopes and their respective atomic masses. Rather than simply averaging the mass numbers, the weighted average ensures that more abundant isotopes have a greater effect on the overall atomic mass of the element.

It is a critical concept in chemistry since this value is what's listed for an element on the periodic table and is essential for a myriad of calculations in chemical reactions. Essentially, the weighted average takes into account the percentage (abundance) of each isotope and multiplies it by the isotope's mass, accounting for the fact that not all atoms of an element are identical in mass. The resulting values are then summed up to give the average atomic mass for that element.

Isotope Abundances

Isotope abundances refer to the percentage of each isotope present within a naturally occurring sample of an element. This concept is crucial when calculating the weighted average atomic mass of an element. Real-world samples of elements are rarely pure, containing a mixture of isotopes, which is why understanding abundances is important for accurate scientific work.

To begin a calculation using isotope abundances, these values must often be converted from their percentage form into decimal form; that is, 50% becomes 0.50, which makes calculations more straightforward. These decimal values will then play a key role when multiplied against isotopic masses, which can result in finding properties like the weighted average atomic mass.

Solving for Isotope Mass

The process of solving for an unknown isotope mass when given the weighted average atomic mass, and the masses and abundances of other isotopes, involves algebra. These exercises look to find a missing mass that completes the equation of the weighted average atomic mass.

To solve for the isotope's mass, a system of equations is used where the weighted average atomic mass is represented by the sum of the products of each isotope abundance (converted to decimal) and its respective mass. As shown in the example with rhenium (Re) isotopes, to find an unknown mass, you'll rearrange the equation to isolate the unknown mass term, then perform algebraic operations to solve for it. This process involves crucial thinking skills and an understanding of how to manipulate equations, reflecting the quantitative nature of chemistry.