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.

Short Answer

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

Step by step solution

01

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 \(24.47/100=0.2447\)
02

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}\), \(36.956 * 0.2447 = 9.0399\)
03

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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

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Most popular questions from this chapter

What mole ratio of molecular chlorine \(\left(\mathrm{Cl}_{2}\right)\) to molecular oxygen \(\left(\mathrm{O}_{2}\right)\) would result from the breakup of the compound \(\mathrm{Cl}_{2} \mathrm{O}_{7}\) into its constituent elements?

Write balanced equations for the following reactions described in words. (a) Pentane burns in oxygen to form carbon dioxide and water. (b) Sodium bicarbonate reacts with hydrochloric acid to form carbon dioxide, sodium chloride, and water. (c) When heated in an atmosphere of nitrogen, lithium forms lithium nitride. (d) Phosphorus trichloride reacts with water to form phosphorus acid and hydrogen chloride. (e) Copper(II) oxide heated with ammonia will form copper, nitrogen gas, and water.

Consider the reaction $$\mathrm{MnO}_{2}+4 \mathrm{HCl} \longrightarrow \mathrm{MnCl}_{2}+\mathrm{Cl}_{2}+2 \mathrm{H}_{2} \mathrm{O} $$If 0.86 mole of \(\mathrm{MnO}_{2}\) and \(48.2 \mathrm{~g}\) of \(\mathrm{HCl}\) react, which reactant will be used up first? How many grams of \(\mathrm{Cl}_{2}\) will be produced?

Define limiting reactant and excess reactant. What is the significance of the limiting reactant in predicting the amount of the product obtained in a reaction? Can there be a limiting reactant if only one reactant is present?

A certain metal oxide has the formula \(\mathrm{MO},\) where \(\mathrm{M}\) denotes the metal. A \(39.46-\mathrm{g}\) sample of the compound is strongly heated in an atmosphere of hydrogen to remove oxygen as water molecules. At the end, \(31.70 \mathrm{~g}\) of the metal is left over. If \(\mathrm{O}\) has an atomic mass of 16.00 amu, calculate the atomic mass of \(\mathrm{M}\) and identify the element.

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