Question

part a
on a mission to a newly discovered planet, an astronaut finds chlorine abundances of 13.85% for $^{35}cl$ and 86.15% for $^{37}cl$. what is the atomic mass of chlorine for this location?
the mass of $^{35}cl$ is 34.9700 amu. the mass of $^{37}cl$ is 36.9700 amu.
express your answer to two decimal places, and include the appropriate units.
view available hint(s)

Explanation:

Step1: Convert percentages to decimals

For \(^{35}\text{Cl}\), the abundance as a decimal is \(0.1385\) (since \(13.85\%=\frac{13.85}{100} = 0.1385\)). For \(^{37}\text{Cl}\), the abundance as a decimal is \(0.8615\) (since \(86.15\%=\frac{86.15}{100}=0.8615\)).

Step2: Calculate the contribution of each isotope to the atomic mass

The contribution of \(^{35}\text{Cl}\) is \(0.1385\times34.9700\) amu. Using the formula \(a\times b\), where \(a = 0.1385\) and \(b=34.9700\), we get \(0.1385\times34.9700=4.843345\) amu.
The contribution of \(^{37}\text{Cl}\) is \(0.8615\times36.9700\) amu. Using the formula \(a\times b\), where \(a = 0.8615\) and \(b = 36.9700\), we have \(0.8615\times36.9700=31.849655\) amu.

Step3: Sum the contributions

The atomic mass \(M\) is the sum of the contributions of each isotope. So \(M=(0.1385\times34.9700)+(0.8615\times36.9700)=4.843345 + 31.849655\) amu. Using the formula \(a + b\), where \(a=4.843345\) and \(b = 31.849655\), we get \(M=36.693\) amu. Rounding to two decimal places, we use the rule: if the third - decimal digit is less than 5, we keep the second - decimal digit as it is. Since the third - decimal digit of \(36.693\) is \(3\lt5\), the rounded value is \(36.69\) amu.

Answer:

\(36.69\) amu