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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 35.444 amu.

Step by step solution

01

Understand the Concept of Average Atomic Mass

The average atomic mass of an element is calculated using the isotopes of that element. Each isotope contributes to the average atomic mass based on its relative abundance. The formula for average atomic mass is: \[ \text{Average Atomic Mass} = (\text{Mass of Isotope 1} \times \text{Abundance of Isotope 1}) + (\text{Mass of Isotope 2} \times \text{Abundance of Isotope 2}) + ... \] The abundance must be in decimal form.
02

Convert Percentage Abundances to Decimal Form

Convert the percentage abundances into decimals by dividing each percentage by 100. For \({ }_{17}^{35} \text{Cl}\): \(75.53\% = 0.7553\) For \({ }_{17}^{37} \text{Cl}\): \(24.47\% = 0.2447\)
03

Multiply Isotope Masses by Their Abundances

Multiply each isotope's mass by its corresponding abundance in decimal form. For \({ }_{17}^{35} \text{Cl}\): Mass: \(34.968 \text{ amu}\) × Abundance: \(0.7553\) = \(26.398\) amu For \({ }_{17}^{37} \text{Cl}\): Mass: \(36.956 \text{ amu}\) × Abundance: \(0.2447\) = \(9.046\) amu
04

Sum the Contributions of Each Isotope

Add the results from Step 3 to find the average atomic mass. \[ 26.398 \text{ amu} + 9.046 \text{ amu} = 35.444 \text{ amu} \]
05

State the Average Atomic Mass

The calculated average atomic mass of chlorine is the result from Step 4. Write down this value as the final answer.

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Key Concepts

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

Isotopes
Atoms of the same element can have different numbers of neutrons. This causes them to have different masses, even though they have the same number of protons. These different forms are called isotopes.
For example, chlorine has two main isotopes:
  • extsuperscript{35}Cl, which has 18 neutrons.
  • extsuperscript{37}Cl, which has 20 neutrons.
Each isotope has a unique atomic mass, because the total number of protons and neutrons in its nucleus differs. Importantly, they still behave the same way chemically because they have the same number of protons and electrons. Understanding isotopes is essential to comprehending how average atomic mass is determined based on their masses and abundances.
Relative Abundance
Relative abundance refers to how common each isotope of an element is relative to the total amount of the element found in nature. It is usually expressed as a percentage. For instance, for chlorine:
  • extsuperscript{35}Cl has a relative abundance of 75.53%.
  • extsuperscript{37}Cl has a relative abundance of 24.47%.
Relative abundance must be converted into decimal form when calculating the average atomic mass. You do this by dividing the percentage by 100. For example, 75.53% becomes 0.7553.
When calculating average atomic mass, each isotope's mass is multiplied by its relative abundance, and the results are summed. This helps effectively weigh each isotope's contribution.
Understanding relative abundance ensures that students accurately represent how much of each isotope naturally contributes to an element's overall atomic structure in their calculations.
Atomic Mass Unit (amu)
The atomic mass unit (amu) is the standard unit for expressing atomic mass. It's used to express the mass of atoms and molecules in a way that is easier to manage, given the incredibly small scale at which atoms operate.
One amu is defined as one-twelfth the mass of a carbon-12 atom. Effectively, it serves as a baseline for determining atomic mass on a relative scale.
When calculating average atomic masses, the masses of isotopes are given in amu. This allows for a consistent and comparable way to express the mass contributions of isotopes. For example, the isotope extsuperscript{35}Cl has a mass of 34.968 amu.
Remember, the final average mass of an element, calculated from its isotopes and relative abundances, is also expressed in amu. This consistency simplifies understanding and communicating atomic masses.

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Most popular questions from this chapter

A sample containing \(\mathrm{NaCl}, \mathrm{Na}_{2} \mathrm{SO}_{4},\) and \(\mathrm{NaNO}_{3}\) gives the following elemental analysis: Na: 32.08 percent; O: 36.01 percent; Cl: 19.51 percent. Calculate the mass percent of each compound in the sample.

A mixture of methane \(\left(\mathrm{CH}_{4}\right)\) and ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) of mass \(13.43 \mathrm{~g}\) is completely burned in oxygen. If the total mass of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) produced is \(64.84 \mathrm{~g},\) calculate the fraction of \(\mathrm{CH}_{4}\) in the mixture.

One of the reactions that occurs in a blast furnace, where iron ore is converted to cast iron, is $$ \mathrm{Fe}_{2} \mathrm{O}_{3}+3 \mathrm{CO} \longrightarrow 2 \mathrm{Fe}+3 \mathrm{CO}_{2} $$ Suppose that \(1.64 \times 10^{3} \mathrm{~kg}\) of Fe are obtained from a \(2.62 \times 10^{3}-\mathrm{kg}\) sample of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\). Assuming that the reaction goes to completion, what is the percent purity of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) in the original sample?

Silicon tetrachloride \(\left(\mathrm{SiCl}_{4}\right)\) can be prepared by heating Si in chlorine gas: $$ \mathrm{Si}(s)+2 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SiCl}_{4}(l) $$ In one reaction, 0.507 mole of \(\mathrm{SiCl}_{4}\) is produced. How many moles of molecular chlorine were used in the reaction?

A certain metal M forms a bromide containing 53.79 percent Br by mass. What is the chemical formula of the compound?

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