Problem 19 - Chemistry and Chemical Reactivity

Question

Silver (Ag) has two stable isotopes, \(107 \mathrm{Ag}\) and \(^{109} \mathrm{Ag}\). The isotopic mass of \(^{107} \mathrm{Ag}\) is \(106.9051,\) and the isotopic mass of \(^{109} \mathrm{Ag}\) is \(108.9047 .\) The atomic weight of Ag, from the periodic table, is \(107.868 .\) Estimate the percent of \(^{107} \mathrm{Ag}\) in a sample of the element. (a) \(0 \%\) (b) \(25 \%\) (c) \(50 \%\) (d) \(75 \%\)

Step-by-Step Solution

Answer

The answer is (c) 50%.

Step 1: Identify Known Values

We are given the isotopic masses and atomic weight of Silver, along with the need to find the percentage of \(^{107}\mathrm{Ag}\). Isotopic mass of \(^{107} \mathrm{Ag}\) is \(106.9051\), and isotopic mass of \(^{109} \mathrm{Ag}\) is \(108.9047\). The atomic weight of Silver is \(107.868\).

Step 2: Set Up the Equation

To calculate the percentage, we use the atomic weight formula: \[ \text{Atomic Weight} = (\text{Fraction of } ^{107}\mathrm{Ag}) \times 106.9051 + (\text{Fraction of } ^{109}\mathrm{Ag}) \times 108.9047 \] Let's denote the fraction of \(^{107}\mathrm{Ag}\) as \(x\), so the fraction of \(^{109}\mathrm{Ag}\) will be \(1-x\).

Step 3: Substitute Known Values into the Equation

Replace the atomic weight and isotopic masses in the equation:\[ 107.868 = (x) \times 106.9051 + (1-x) \times 108.9047 \]

Step 4: Simplify and Solve for x

Expand the equation:\[ 107.868 = 106.9051x + 108.9047 - 108.9047x \]Combine like terms:\[ 107.868 = -1.9996x + 108.9047 \]Rearrange to solve for \(x\):\[ 1.0369 = 1.9996x \]\[ x = \frac{1.0369}{1.9996} \]

Step 5: Calculate Percentage of \(^{107}\mathrm{Ag}\)

Solve for \(x\):\[ x \approx 0.5186 \]Convert this to a percentage by multiplying by 100:\[ 0.5186 \times 100 \approx 51.86\% \]

Step 6: Select Closest Option

The closest option to our calculated percentage of \(51.86\%\) is (c) 50\%.

Key Concepts

Isotopic Mass

Isotopic mass refers to the mass of a specific isotope of an element, measured in atomic mass units (amu). Each element can have different isotopes, which are atoms with the same number of protons but different numbers of neutrons. These isotopes have varying isotopic masses due to the differing number of neutrons.
For example, Silver (Ag) has two stable isotopes:

• ^{107}Ag, with an isotopic mass of 106.9051 amu
• ^{109}Ag, with an isotopic mass of 108.9047 amu

The actual mass of each isotope affects how they appear in nature and are used in various calculations involving elements. This concept is crucial in the fields of chemistry and physics, especially when calculating formulas involving isotopes.

Atomic Weight

Atomic weight, also known as atomic mass, is an average of the masses of the isotopes of an element, calculated based on their relative abundance. It provides a convenient weighted average mass for elements in atomic mass units. Atomic weight is what we usually see on the periodic table and is represented without any decimal places for simplicity.
The equation for calculating atomic weight is:\[\text{Atomic Weight} = (\text{Fraction of Isotope 1}) \times (\text{Isotopic Mass of Isotope 1}) + (\text{Fraction of Isotope 2}) \times (\text{Isotopic Mass of Isotope 2}) + \ldots\]In our example with Silver, the atomic weight of Ag is 107.868, calculated via the isotopic masses and fractions of \(^{107}Ag\) and \(^{109}Ag\). Atomic weight enables scientists to predict the behavior of an element in different chemical reactions.

Percent Composition

Percent composition refers to the percentage of each isotope in a naturally occurring sample of an element. It's quantified based on the contribution of each isotope to the atomic weight. To find this percentage, we use the fractions of the isotopes involved.
In the case of Silver, its atomic weight can be expressed by the formula:\[107.868 = (x) \times 106.9051 + (1-x) \times 108.9047\]where \(x\) is the fraction of \(^{107}Ag\). Solving this equation, we found that about \(51.86\%\) of naturally occurring silver is \(^{107}Ag\). This is translated into a percentage by multiplying the fraction by 100, helping in real-world calculations and understanding the composition of elements.