Question

The element silver (Ag) has two naturally occurring isotopes: \(^{109} \mathrm{Ag}\) and \(^{107} \mathrm{Ag}\) with a mass of 106.905 u. Silver consists of \(51.82 \%^{107} \mathrm{Ag}\) and has an average atomic mass of \(107.868 \mathrm{u}\) Calculate the mass of \(^{109} \mathrm{Ag.}\)

Short answer

The mass of the isotope \(^{109}\mathrm{Ag}\) is approximately 108.833 u.

Step-by-step solution

Find the percentage of \(^{109}\mathrm{Ag}\)

First, we need to find the percentage of the other isotope, \(^{109}\mathrm{Ag}\). Since there are only two isotopes, we can simply subtract the given percentage of \(^{107}\mathrm{Ag}\) from 100%. Percentage of \(^{109}\mathrm{Ag} = 100 \% - 51.82 \% = 48.18 \%\)

Set up the weighted average equation

We know that the average atomic mass of silver is a weighted average based on the masses and percentages of each isotope. We'll denote the mass of \(^{109}\mathrm{Ag}\) as \(m_{109}\). The equation will look like: \(107.868 \mathrm{u} = 0.5182 \times 106.905 \mathrm{u} + 0.4818 \times m_{109}\) Here, 0.5182 and 0.4818 are the decimal percentages of the isotopes \(^{107}\mathrm{Ag}\) and \(^{109}\mathrm{Ag}\) respectively.

Solve for the mass of \(^{109}\mathrm{Ag}\)

Now, we simply solve for \(m_{109}\): \(107.868 = 0.5182 \times 106.905 + 0.4818 \times m_{109}\) Rearrange the equation to isolate \(m_{109}\): \(m_{109} = \frac{107.868 - 0.5182 \times 106.905}{0.4818}\) Now, we plug in the values and calculate the result: \(m_{109} = \frac{107.868 - 0.5182 \times 106.905}{0.4818} \approx 108.833 \mathrm{u}\) So, the mass of the isotope \(^{109}\mathrm{Ag}\) is approximately 108.833 u.

Key concepts

Isotopes

Isotopes are variations of an element that have the same number of protons but different numbers of neutrons. This difference in neutron count doesn't affect the chemical behavior of the isotopes, but it does alter their mass slightly. For example, Silver (Ag) has isotopes like \(^{107}\mathrm{Ag}\) and \(^{109}\mathrm{Ag}\). Although both are isotopes of silver, they have 107 and 109 nucleons (protons plus neutrons) respectively. Understanding isotopes is crucial when discussing atomic mass because the atomic mass we commonly see is a weighted average of the masses of all the isotopes of that element as they occur in nature.
Elements in nature typically exist as a mixture of their isotopes, and each isotope contributes to the element's overall atomic mass based on its natural abundance. That's why knowing the percentage of each isotope present in a sample is essential for calculating the average atomic mass of the element.

Weighted Average

A weighted average takes into account not only the individual values but also their relative significance, which is often represented as their proportions in the overall mix. When calculating the average atomic mass of an element, we use a weighted average because some isotopes are simply more abundant than others.

In practice, this means multiplying the mass of each isotope by its fractional abundance (or percentage) and then summing these products to get the average atomic mass. This approach reflects the fact that more abundant isotopes have a bigger 'weight' in determining the overall atomic mass. To solve exercises involving the calculation of atomic mass, mastery of setting up and solving a weighted average equation is key, as demonstrated in the solution section with the silver isotopes. Applying this concept correctly will ensure an accurate representation of the isotopic composition of an element.

Atomic Mass Unit

The atomic mass unit, abbreviated as u, is a standard unit of mass that quantifies the atomic and molecular mass scale. It is defined as one twelfth of the mass of a carbon-12 atom, which is approximately \(1.66053906660 \times 10^{-27}\) kilograms.

This small unit is incredibly useful in chemistry and physics because it allows us to compare the masses of different atoms and molecules on a scale suitable for very small entities. For instance, when we say that the mass of the isotope \(^{107}\mathrm{Ag}\) is 106.905 u, we're saying it is 106.905 times the mass of 1/12 of a carbon-12 atom. It's essential to understand the atomic mass unit when delving into atomic mass calculations because all atomic masses and the calculations involving them use this unit.