Name: [tex]$\qquad$[/tex] Period: [tex]$\qquad$[/tex]

PROBLEM SET: SUBATOMIC PARTICLES & AVG ATOMIC MASS

Learning Objectives:
- I am learning to construct a mathematical model based on mass spectrum data to determine the average atomic mass of an element.
- Can I calculate the average atomic mass of an element from the relative abundances of its isotopes?
- Can I justify the average atomic mass using data from a mass spectrum?

Question 12:

A sample of metal was passed through a mass spectrometer, and the following plot and data were obtained:

\begin{tabular}{|c|c|c|}
\hline
Atomic Mass (amu) & Percent Abundance (\%) & Relative Abundance \\
\hline
58 & 68.25 & 0.6825 \\
\hline
60 & 26.07 & 0.2607 \\
\hline
61 & 1.6 & 0.016 \\
\hline
62 & 3.62 & 0.0362 \\
\hline
64 & 0.889 & 0.00889 \\
\hline
\end{tabular}

Calculate the atomic mass of the metal and identify which element it is using values from the periodic table.

Hint: Relative Abundance is just the decimal version of Percent Abundance. For example, [tex]$5.3 \% \rightarrow 0.053$[/tex]

DIFFERENTIATION BASED ON YOUR MATH LEVEL:



Answer :

Certainly! Let's proceed step-by-step to solve this problem.

### Step 1: Understanding Relative Abundance
Relative abundance is the percentage abundance of each isotope converted to a decimal. The data provided has already converted these percentages for us:

1. For 58 amu: [tex]\( 68.25\% \rightarrow 0.6825 \)[/tex]
2. For 60 amu: [tex]\( 26.07\% \rightarrow 0.2607 \)[/tex]
3. For 61 amu: [tex]\( 1.6\% \rightarrow 0.016 \)[/tex]
4. For 62 amu: [tex]\( 3.62\% \rightarrow 0.0362 \)[/tex]
5. For 64 amu: [tex]\( 0.889\% \rightarrow 0.00889 \)[/tex]

These are our relative abundances.

### Step 2: Weighted Average Atomic Mass Calculation
To find the average atomic mass, we calculate the weighted average using the atomic masses and their respective relative abundances. The formula to compute the weighted average atomic mass is:

[tex]\[ \text{Average Atomic Mass} = \sum (\text{atomic mass} \times \text{relative abundance}) \][/tex]

We will multiply each atomic mass by its relative abundance and then sum all these values.

#### Calculations:
1. [tex]\( 58 \times 0.6825 = 39.585 \)[/tex]
2. [tex]\( 60 \times 0.2607 = 15.642 \)[/tex]
3. [tex]\( 61 \times 0.016 = 0.976 \)[/tex]
4. [tex]\( 62 \times 0.0362 = 2.2444 \)[/tex]
5. [tex]\( 64 \times 0.00889 = 0.56896 \)[/tex]

Now, summing these results:

[tex]\[ 39.585 + 15.642 + 0.976 + 2.2444 + 0.56896 = 59.01636 \][/tex]

So, the average atomic mass is [tex]\( 59.01636 \)[/tex].

### Step 3: Identifying the Element
From the periodic table, we know the atomic masses that typically fall around 59 are closest to Nickel (Ni), which has atomic masses in its isotopes around these values. However, Nickel itself typically has an average atomic mass closer to 58.6934 amu. This result suggests that the average atomic mass of 59.01636 does not directly correspond with the most common isotopic signature of Nickel around its standard atomic weight.

Thus, based on the calculated average atomic mass of [tex]\( 59.01636 \)[/tex], it does not match exactly with Nickel or any other element's weighted atomic mass naturally observed in the periodic table within an error boundary.

### Conclusion
- The calculated average atomic mass of the sample is [tex]\( 59.01636 \)[/tex].
- The closest known element to this atomic mass in the periodic table is Nickel; however, the exact match is not visible from the provided data.

Thus, with an average atomic mass of [tex]\( 59.01636 \)[/tex], the element remains unidentified based on common isotopic data, leading us to label it as 'Unknown' until further detailed analysis or comparison with additional spectral data can be performed.