AP Chemistry 1.2 Mass Spectroscopy of Elements Study Notes - New Syllabus 2024-2025
AP Chemistry 1.2 Mass Spectroscopy of Elements Study Notes- New syllabus
AP Chemistry 1.2 Mass Spectroscopy of Elements Study Notes – AP Chemistry – per latest AP Chemistry Syllabus.
LEARNING OBJECTIVE
- Explain the quantitative relationship between the mass spectrum of an element and the masses of the element’s isotopes
Key Concepts:
- Mass Spectra of Elements
Mass Spectra of Elements
A mass spectrum is a graph produced by a mass spectrometer that shows the isotopes of an element as peaks. The x-axis represents the mass-to-charge ratio (m/z), and the y-axis represents the relative abundance of each isotope.
For atoms that carry a +1 charge (most common in mass spectrometry), the m/z value is equal to the relative isotopic mass.
How a Mass Spectrum Relates to Isotopes
1. Peak positions: Each isotope of an element gives a distinct peak on the spectrum at a position corresponding to its mass number.
2. Peak heights (or intensities): These reflect the relative abundances of the isotopes present in the natural sample.
3. Multiple isotopes: The presence of more than one isotope produces multiple peaks.
4. Average atomic mass: The relative atomic mass seen in the periodic table is a weighted average of the isotopic masses and their abundances.
Quantitative Relationship
The quantitative relationship between the spectrum and the isotopes is expressed mathematically as:
\( A_r = \dfrac{\sum (\text{isotopic mass} \times \text{relative abundance})}{\sum (\text{relative abundance})} \)
This means that each isotope’s contribution to the average atomic mass is proportional to both its mass and its relative abundance.
Step-by-Step Approach
1. Identify the isotopes (from the m/z values).
2. Note their relative abundances (from peak heights).
3. Multiply each isotope’s mass by its abundance.
4. Add these values together.
5. Divide by the total abundance (usually 100 if percentages are used).
6. The result is the element’s relative atomic mass, which matches the value in the periodic table.
Conclusion
The quantitative relationship is that the mass spectrum directly reflects the isotope composition of an element. The peak positions give the isotopic masses, the peak heights give the isotopic abundances, and from these values the average atomic mass is calculated. This explains why many elements have non-integer relative atomic masses in the periodic table.
Example
The element chlorine has two main isotopes: \( ^{35}\text{Cl} \) with 75% abundance and \( ^{37}\text{Cl} \) with 25% abundance. Calculate the relative atomic mass of chlorine.
▶️ Answer/Explanation
\( A_r = \dfrac{(35 \times 75) + (37 \times 25)}{100} \)
\( A_r = \dfrac{2625 + 925}{100} \)
\( A_r = 35.5 \)
This explains why the relative atomic mass of chlorine is recorded as 35.5 on the periodic table.
Example
The mass spectrum of magnesium shows three isotopes: \( ^{24}\text{Mg} \) (79%), \( ^{25}\text{Mg} \) (10%), and \( ^{26}\text{Mg} \) (11%). Calculate the relative atomic mass of magnesium.
▶️ Answer/Explanation
\( A_r = \dfrac{(24 \times 79) + (25 \times 10) + (26 \times 11)}{100} \)
\( A_r = \dfrac{1896 + 250 + 286}{100} \)
\( A_r = 24.32 \)
The value (24.32) matches the relative atomic mass of magnesium given in the periodic table.
Example
The element boron has two isotopes: \( ^{10}\text{B} \) (20%) and \( ^{11}\text{B} \) (80%). Calculate the relative atomic mass of boron.
▶️ Answer/Explanation
\( A_r = \dfrac{(10 \times 20) + (11 \times 80)}{100} \)
\( A_r = \dfrac{200 + 880}{100} \)
\( A_r = 10.8 \)
This matches the average atomic mass of boron in the periodic table.