The natural abundances of chlorine’s two major isotopes are 35Cl (75.5%) and 37Cl (24.5%). If the mass spectrum of chlorobenzene contains a peak at m/z 114 with an abundance of 5.0%, calculate the percent abundance of the peak at m/z 12.
The Correct Answer and Explanation is:
To calculate the percent abundance of the peak at m/z = 112 (which you likely meant instead of “12”, as 12 is too low for chlorobenzene), we consider the mass spectrum of chlorobenzene (C₆H₅Cl), which arises from combinations of chlorine isotopes:
Step 1: Isotopes and Molar Mass
- Chlorine has two main isotopes:
- ³⁵Cl with a mass of ~34.97 amu and abundance 75.5% (0.755)
- ³⁷Cl with a mass of ~36.97 amu and abundance 24.5% (0.245)
- Benzene (C₆H₅) has a mass of approximately 77 amu.
So:
- Chlorobenzene with ³⁵Cl: m/z ≈ 77 + 35 = 112
- Chlorobenzene with ³⁷Cl: m/z ≈ 77 + 37 = 114
Given:
- m/z 114 peak (³⁷Cl) has relative abundance = 5.0%
- Need to find the abundance of m/z 112 (³⁵Cl)
Step 2: Use Isotopic Ratio
The abundance ratio of the two peaks is directly proportional to the isotopic abundances of ³⁵Cl and ³⁷Cl:Ratio=Abundance of 37ClAbundance of 35Cl=0.2450.755=5.0%x\text{Ratio} = \frac{\text{Abundance of }^{37}\text{Cl}}{\text{Abundance of }^{35}\text{Cl}} = \frac{0.245}{0.755} = \frac{5.0\%}{x}Ratio=Abundance of 35ClAbundance of 37Cl=0.7550.245=x5.0%
Solving:0.2450.755=5.0x\frac{0.245}{0.755} = \frac{5.0}{x}0.7550.245=x5.0x=5.0×0.7550.245=3.7750.245≈15.4%x = \frac{5.0 \times 0.755}{0.245} = \frac{3.775}{0.245} \approx 15.4\%x=0.2455.0×0.755=0.2453.775≈15.4%
✅ Final Answer:
The percent abundance of the peak at m/z 112 is approximately 15.4%.
📘 Textbook-Style Explanation
The mass spectrum of chlorobenzene (C₆H₅Cl) shows distinct peaks due to the presence of chlorine isotopes. Chlorine naturally exists primarily as two isotopes: ³⁵Cl and ³⁷Cl. Their respective abundances are approximately 75.5% and 24.5%. In mass spectrometry, these isotopic variations cause the molecular ion peak of chlorobenzene to appear at different m/z values depending on the isotope present.
The molecular mass of the C₆H₅ (phenyl) group is about 77 amu. When bonded with ³⁵Cl, the molecular ion has a mass of roughly 112 amu (77 + 35). When bonded with ³⁷Cl, the ion has a mass of 114 amu (77 + 37). Therefore, peaks in the mass spectrum are expected at m/z = 112 and m/z = 114, corresponding to ³⁵Cl and ³⁷Cl variants, respectively.
Since isotopic abundance affects peak intensity, the relative peak heights mirror the abundance ratio. The ratio of ³⁷Cl to ³⁵Cl is:24.575.5≈0.3245\frac{24.5}{75.5} \approx 0.324575.524.5≈0.3245
If the less abundant isotope peak at m/z 114 (³⁷Cl) has a reported abundance of 5.0%, we apply this ratio to determine the peak at m/z 112:5.0%x=0.2450.755⇒x≈15.4%\frac{5.0\%}{x} = \frac{0.245}{0.755} \Rightarrow x \approx 15.4\%x5.0%=0.7550.245⇒x≈15.4%
This reflects the greater natural abundance of ³⁵Cl. Hence, the m/z 112 peak is expected to be more intense, consistent with the isotope’s prevalence in nature.
