Problem 13 - Introductory Chemistry Atoms First

Question

Chlorine exists as two isotopes in nature, \({ }_{1} 5 \mathrm{Cl}\) (atomic mass \(34.969 \mathrm{amu}\), abundance \(75.77 \%\) ) and \({ }_{17}^{37} \mathrm{Cl}\) (atomic mass \(36.966\) amu). (a) What is the percent abundance of the \({ }_{1}^{37} \mathrm{Cl}\) isotope? (b) Calculate the weighted average of the atomic mass of naturally occurring chlorine. (c) How many times more massive is \({ }_{1}^{37} \mathrm{Cl}\) than \({ }_{17}^{35} \mathrm{Cl}\) ?

Step-by-Step Solution

Answer

(a) The percent abundance of the \(_{1}^{37} \mathrm{Cl}\) isotope is 24.23%. (b) The weighted average of the atomic mass of naturally occurring chlorine is 35.453 amu. (c) \(_{1}^{37} \mathrm{Cl}\) is 1.057 times more massive than \(_{17}^{35} \mathrm{Cl}\).

Step 1: Find the percent abundance of 37Cl

To find the percent abundance of 37Cl, we will subtract the given percent abundance of 35Cl from 100%: Percent_abundance_37Cl = 100% - Percent_abundance_35Cl Percent_abundance_37Cl is the percent abundance of 37Cl isotope that we want to find out, and Percent_abundance_35Cl is the percent abundance of 35Cl isotope, which is given as 75.77%. Now, the calculation is straightforward: Percent_abundance_37Cl = 100% - 75.77% = 24.23% The percent abundance of the 37Cl isotope is 24.23%.

Step 2: Calculate the weighted average of the atomic mass of naturally occurring chlorine

To find the weighted average of the atomic mass, we will use the given percent abundance values of the isotopes and their atomic masses as follows: Weighted_average = (Percent_abundance_35Cl * Atomic_mass_35Cl + Percent_abundance_37Cl * Atomic_mass_37Cl)/100 Here, Atomic_mass_35Cl is the atomic mass of 35Cl isotope, which is 34.969 amu, and Atomic_mass_37Cl is the atomic mass of 37Cl isotope, which is 36.966 amu. Now, plugging the values in the formula and calculating: Weighted_average = (75.77 * 34.969 + 24.23 * 36.966)/100 = 35.453 amu The weighted average of the atomic mass of naturally occurring chlorine is 35.453 amu.

Step 3: Find how many times more massive is 37Cl than 35Cl

To determine how many times more massive 37Cl is than 35Cl, we will divide the atomic mass of 37Cl by the atomic mass of 35Cl: Mass_ratio = Atomic_mass_37Cl / Atomic_mass_35Cl Here, Atomic_mass_35Cl = 34.969 amu and Atomic_mass_37Cl = 36.966 amu. Now, substituting the values in the formula: Mass_ratio = 36.966 / 34.969 = 1.057 37Cl is 1.057 times more massive than 35Cl.

Key Concepts

Percent Abundance of Isotopes

When we talk about the percent abundance of isotopes, we are referring to the proportion of each isotope present in a natural sample of an element. Since isotopes of an element have the same number of protons but a different number of neutrons, they can vary slightly in mass. The percent abundance is crucial for understanding the overall atomic mass of the element as found in nature. For example, in the case of chlorine, we have two stable isotopes: Chlorine-35 and Chlorine-37.

To find the percent abundance of Chlorine-37, we subtract the percent abundance of Chlorine-35 from 100%, assuming only these two isotopes exist. This is based on the principle that the sum of the percent abundances of all isotopes of an element must equal 100%. If Chlorine-35 has an abundance of 75.77%, then Chlorine-37, which makes up the remainder, has an abundance of 24.23%. Understanding this concept is critical when calculating the average atomic mass of an element.

Atomic Mass Calculation

Atomic mass calculation involves determining the mass of a single atom of an element. It's usually expressed in atomic mass units (amu), where one amu is defined as one twelfth the mass of a carbon-12 atom. For isotopes, these masses are measured using mass spectrometry and are crucial for scientific fields ranging from chemistry to geology. For instance, Chlorine-35 has an atomic mass of approximately 34.969 amu, and Chlorine-37 has a mass of approximately 36.966 amu. These values are constants for the isotopes and are used in calculations involving isotopic composition, such as computing the weighted average atomic mass.

Weighted Average Atomic Mass

The weighted average atomic mass of an element takes into account the masses of its isotopes and their percent abundances in nature. It's like the average grade in a class where some assignments count more towards the final grade than others. In this analogy, the 'grades' are the isotopic masses, and their 'weight' is the percent abundance of each isotope.

To calculate the weighted average atomic mass for chlorine, each isotope's mass is multiplied by its percent abundance, and the products are then summed up and divided by 100. This gives us the average mass of a chlorine atom as it typically occurs in nature, considering the relative abundance of Chlorine-35 and Chlorine-37. It's this weighted average that is listed on the periodic table for elements with naturally occurring isotopes.

Isotope Mass Ratio

The isotope mass ratio is a comparison of the masses of different isotopes of the same element. It informs us about the relative heaviness of isotopes compared to each other. This ratio is calculated by simply dividing the atomic mass of one isotope by the atomic mass of another. In our example, to find out how much heavier Chlorine-37 is compared to Chlorine-35, we divide the atomic mass of Chlorine-37 by that of Chlorine-35, yielding a ratio of 1.057.

This means that Chlorine-37 is approximately 1.057 times more massive than Chlorine-35. Knowing the isotope mass ratio is particularly useful in disciplines such as isotope geochemistry and radiometric dating, where isotopic differences play a significant role.