Problem 5 - Chemistry

Question

The atomic masses of \({ }_{17}^{35} \mathrm{Cl}(75.53\) percent) and \({ }_{17}^{37} \mathrm{Cl}(24.47\) percent) are 34.968 amu and 36.956 amu, respectively. Calculate the average atomic mass of chlorine. The percentages in parentheses denote the relative abundances.

Step-by-Step Solution

Answer

The average atomic mass of chlorine is \(35.4647\) amu.

Step 1: Convert the percentage abundances to decimal

The percentage abundance needs to be converted to a decimal to be used in calculations. This can be done by dividing the percentage by 100. So for \({ }_{17}^{35} \mathrm{Cl}\), it would be \(75.53/100=0.7553\). Similarly for ${}_{17}^{37} \mathrm{Cl}\), it would be \(24.47/100=0.2447\)

Step 2: Multiply the atomic masses with the respective decimal abundance

For each isotope, multiply the atomic mass by its decimal abundance: For \({ }_{17}^{35} \mathrm{Cl}\): \(34.968 * 0.7553 = 26.4248\). For ${}_{17}^{37} \mathrm{Cl}\), \(36.956 * 0.2447 = 9.0399\)

Step 3: Calculate the average atomic mass

The average atomic mass of chlorine is found by adding these multiplied values together. So, average atomic mass = \(26.4248 + 9.0399 = 35.4647\) amu

Key Concepts

Isotopic Abundance

Isotopic abundance refers to the percentage of a particular isotope present in a sample containing various isotopes of an element. Since elements naturally occur as mixtures of isotopes, understanding their abundance is crucial for calculating the average atomic mass. This abundance is typically expressed as a percentage but is converted to a decimal form for calculations. This conversion is simply done by dividing the given percentage by 100. For instance, if Cl-35 has an isotopic abundance of 75.53%, we convert this to 0.7553 to use it in further calculations. It is this decimal form that represents the proportion of each isotope in the natural mixture, essential for accurately determining the average atomic mass of an element.

Remember that the isotopic abundance can vary depending on the source material, which can lead to slight variations in average atomic mass from one sample to another. This variation is why the concept of isotopic abundance is key to understanding not just theoretical calculations, but also the practical aspects of chemistry.

Atomic Mass Unit (amu)

The atomic mass unit (amu) is a standardized unit of mass that specifically serves the field of chemistry and physics. It is defined as one-twelfth of the mass of a carbon-12 atom in its ground state and unbound. The carbon-12 isotope is chosen as the standard reference because of its stable nature and abundance. One atomic mass unit is therefore very small - approximately equal to 1.66053906660 × 10^-24 grams.

When we talk about the mass of an atom or a chemical isotope, we refer to it in amu to maintain a standardized scale for comparison. For instance, the atomic masses of Cl-35 and Cl-37 isotopes are given as 34.968 amu and 36.956 amu, respectively, which aligns them with the atomic mass of carbon-12 for easy comparison.

Chemical Isotopes

Chemical isotopes are different forms of the same element that have the same number of protons but a different number of neutrons within their nuclei. This difference in neutron count leads to variations in atomic mass while retaining identical chemical properties. For chlorine, the isotopes Cl-35 and Cl-37 differ in their mass because Cl-37 has two more neutrons than Cl-35.

Isotopes play a pivotal role in various fields - from medicine, where radioactive isotopes are used for treatment and diagnostics, to archaeology, where isotope ratios in samples help determine the age of artifacts. In the context of our exercise, it's the isotopes Cl-35 and Cl-37 that must be considered to calculate the average atomic mass of chlorine.

Relative Atomic Mass

The relative atomic mass of an element, often simply referred to as its atomic weight, is the weighted average mass of the atoms in a naturally occurring sample of the element. It is dimensionless and takes into account the atomic masses of the individual isotopes, as well as their isotopic abundance. The calculation is akin to finding the