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Chapter 3: Problem 13

Chlorine exists mainly as two isotopes, \({ }^{37} \mathrm{Cl}\) and \({ }^{35} \mathrm{Cl}\). Which is more abundant? How do you know?

Short Answer

Expert verified
The \({ }^{35}\mathrm{Cl}\) isotope is more abundant than the \({ }^{37}\mathrm{Cl}\) isotope. By considering the atomic mass of chlorine from the periodic table (35.5 amu) and the masses of its two isotopes, we can calculate their relative abundances (A for \({ }^{37}\mathrm{Cl}\) and B for \({ }^{35}\mathrm{Cl}\)). Solving the equation, we find that A = 0.25 and B = 0.75, which indicates that \({ }^{35}\mathrm{Cl}\) is more abundant (0.75 > 0.25).

Step by step solution

01

Find the atomic mass of chlorine from the periodic table

Look for chlorine on the periodic table and find its atomic mass, which is around 35.5 amu (atomic mass units).
02

Consider the atomic masses of the two isotopes and their relative abundances

The two isotopes are \({ }^{37}\mathrm{Cl}\), with a mass of 37 amu, and \({ }^{35}\mathrm{Cl}\), with a mass of 35 amu. Let's assume that \({ }^{37}\mathrm{Cl}\) has an abundance of A, while \({ }^{35}\mathrm{Cl}\) has an abundance of B. Since only these two isotopes contribute to the atomic mass, their relative abundance sum is 1: A+B = 1
03

Calculate the weighted sum of the isotopic masses

To get the atomic mass from the isotopes, we need to multiply the isotopic mass by its abundance and sum it up to compare it to the atomic mass found on the periodic table. The weighted sum is: \[(37 \times A) + (35 \times B) = 35.5\]
04

Solve for the relative abundances

We can substitute the sum from step 2 into the weighted sum (step 3) and solve for the relative abundances of the isotopes: \[(37 \times A) + (35 \times (1 - A)) = 35.5\] Expanding and rearranging, we get: \[2A = 0.5\] Solving for A, we find: \[A = 0.25\] From this, we can calculate B: \[B = 1 - A = 1 - 0.25 = 0.75\]
05

Compare the relative abundances

Since the relative abundance of \({ }^{35}\mathrm{Cl}\) (represented by B) is greater than the relative abundance of \({ }^{37}\mathrm{Cl}\) (represented by A): B > A or 0.75 > 0.25 Therefore, the \({ }^{35}\mathrm{Cl}\) isotope is more abundant than the \({ }^{37}\mathrm{Cl}\) isotope.

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Most popular questions from this chapter

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