Question

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

Short answer

The mass of \({ }^{109} \mathrm{Ag}\) is approximately \(108.905 \mathrm{amu}\).

Step-by-step solution

Gather the given information

We are given the following information: 1. Silver consists of \(51.82\% \:^{107} \mathrm{Ag}\) 2. The average atomic mass of silver is \(107.868 \mathrm{amu}\) 3. The mass of \({ }^{107} \mathrm{Ag}\) is \(106.905 \mathrm{amu}\)

Calculate the proportion of \({ }^{109} \mathrm{Ag}\) in natural silver

Since there are only two isotopes in natural silver, we can calculate the proportion of \({ }^{109} \mathrm{Ag}\) using the proportion of \({ }^{107} \mathrm{Ag}\). Percentage of \({ }^{109} \mathrm{Ag} = 100 - 51.82\) Percentage of \({ }^{109} \mathrm{Ag} = 48.18\%\)

Write down the equation for calculating average atomic mass

We can use the equation for average atomic mass which takes into account the proportion and mass of each isotope. The equation is: Average Atomic Mass = (Mass of isotope 1 × Proportion of isotope 1) + (Mass of isotope 2 × Proportion of isotope 2)

Substitute the given values into the equation

Substitute the given values into the equation: \(107.868 \mathrm{amu} = (106.905 \mathrm{amu}\times 51.82\%) + (m_{109} \times 48.18\%)\) Here, \(m_{109}\) is the mass of \({ }^{109} \mathrm{Ag}\).

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

Now we can solve for the mass of \({ }^{109} \mathrm{Ag}\) by first rearranging the equation: \(m_{109} \times 48.18\% = 107.868 \mathrm{amu} - (106.905 \mathrm{amu}\times 51.82\%)\) Next, divide both sides by 48.18\%: \(m_{109} = \frac{107.868 \mathrm{amu} - (106.905 \mathrm{amu} \times 51.82\%)}{48.18\%}\) Now, calculate the mass of \({ }^{109} \mathrm{Ag}\): \(m_{109} = \frac{107.868 \mathrm{amu} - (106.905 \mathrm{amu} \times 0.5182)}{0.4818}\) \(m_{109} \approx 108.905\ \mathrm{amu}\) The mass of \({ }^{109} \mathrm{Ag}\) is approximately \(108.905 \mathrm{amu}\).

Key concepts

Average Atomic Mass

The concept of average atomic mass is crucial for understanding isotopes. An element can have different isotopes, which are atoms with the same number of protons but different numbers of neutrons. As a result, each isotope has a different mass number. The average atomic mass of an element reflects the weighted average of the masses of its naturally occurring isotopes.
This calculation considers both the mass of each isotope and its relative abundance in nature. For example, if an element has two isotopes, the average atomic mass can be calculated using the formula:

• \( \text{Average Atomic Mass} = (\text{mass of isotope 1} \times \text{percent abundance of isotope 1}) + (\text{mass of isotope 2} \times \text{percent abundance of isotope 2}) \)

This way, the average atomic mass provides a more accurate picture of the atomic mass expected in a sample of the element.

Silver Isotopes

Silver, denoted by the chemical symbol Ag, primarily consists of two naturally occurring isotopes:

• \(^{107}\mathrm{Ag} \)
• \(^{109}\mathrm{Ag} \)

These isotopes of silver differ in terms of neutron numbers, which contributes to their different individual atomic masses. Despite these differences, both isotopes share the same chemical characteristics because they have the same number of protons.
The uniqueness of these isotopes in the calculation of silver’s average atomic mass is based on their natural abundance. Typically, you will find silver in nature as a mixture of these isotopes, and knowing their exact percentages allows scientists to calculate its average atomic mass accurately.

Mass of Isotopes

Each isotope of an element has its own distinct atomic mass due to varying numbers of neutrons. The mass of a specific isotope is measured in atomic mass units (amu). For silver, we have the isotope

• \(^{107}\mathrm{Ag} \)
  with a known mass of \(106.905 \) amu.

To find the mass of another silver isotope,

• \(^{109}\mathrm{Ag} \)

we utilize the values of average atomic mass and the percent abundance of each isotope to solve for it with the formula provided in the concept of average atomic mass. Through calculation,

• \(^{109}\mathrm{Ag} \)

is found to have a mass of approximately \(108.905 \) amu.

Percent Abundance

The percent abundance of an isotope refers to the percentage of that particular isotope found in a natural sample of the element. This value is crucial because it influences the calculation of the average atomic mass. For silver,

• \(^{107}\mathrm{Ag} \)

composes approximately 51.82% of naturally occurring silver, while the balance,

• \(^{109}\mathrm{Ag} \)

makes up 48.18%. Knowing these percentages helps us determine how much each isotope weighs into the average atomic mass calculation.
Thus, percent abundance not only helps in identifying the predominance of an isotope but also plays a vital role in scientific calculations involving isotopic distributions.