Question

In using a mass spectrometer, a chemist sees a peak at a mass of \(30.0106\). Of the choices \({ }^{12} \mathrm{C}_{2}{ }^{1} \mathrm{H}_{6},{ }^{12} \mathrm{C}^{1} \mathrm{H}_{2}{ }^{16} \mathrm{O}\), and \({ }^{14} \mathrm{~N}^{16} \mathrm{O}\), which is responsible for this peak? Pertinent masses are \({ }^{1} \mathrm{H}, 1.007825\); \({ }^{16} \mathrm{O}, 15.994915 ;\) and \({ }^{14} \mathrm{~N}, 14.003074 .\)

Short answer

The molecule responsible for the observed peak in the mass spectrometer at 30.0106 units is \({ }^{12} \mathrm{C}^{1} \mathrm{H}_{2}{ }^{16} \mathrm{O}\), as its total mass of 30.010565 is the closest match to the given mass in the peak.

Step-by-step solution

First, find the total mass of this molecule, considering that carbon has a mass of 12 units: Total mass = 2 (mass of carbon) + 6 (mass of hydrogen) = 2 * 12 + 6 * 1.007825 = 24 + 6.04695 = 30.04695 #Step 2: Calculate mass of \({ }^{12} \mathrm{C}^{1} \mathrm{H}_{2}{ }^{16} \mathrm{O}\)

Now find the total mass of this molecule: Total mass = 1 (mass of carbon) + 2 (mass of hydrogen) + 1 (mass of oxygen) = 12 + 2 * 1.007825 + 15.994915 = 12 + 2.01565 + 15.994915 = 30.010565 #Step 3: Calculate mass of \({ }^{14} \mathrm{~N}^{16} \mathrm{O}\)

Find the total mass of this last molecule: Total mass = 1 (mass of nitrogen) + 1 (mass of oxygen) = 14.003074 + 15.994915 = 29.997989 #Step 4: Identify the molecule responsible for the peak

Compare the total masses calculated in steps 1-3 with the observed peak of the spectrometer at 30.0106: The total mass of the molecules: * \({ }^{12} \mathrm{C}_{2}{ }^{1} \mathrm{H}_{6}:\) 30.04695 * \({ }^{12} \mathrm{C}^{1} \mathrm{H}_{2}{ }^{16} \mathrm{O}:\) 30.010565 * \({ }^{14} \mathrm{~N}^{16} \mathrm{O}:\) 29.997989 The mass that is closest to the peak is 30.010565 (for the molecule \({ }^{12} \mathrm{C}^{1} \mathrm{H}_{2}{ }^{16} \mathrm{O}\)). Therefore, this molecule is responsible for the observed peak.

Key concepts

Molecular Mass Calculation

Molecular mass calculation is a fundamental process in mass spectrometry. It involves determining the total mass of a molecule by adding up the masses of its constituent elements. Each element has a characteristic atomic mass, typically measured in atomic mass units (amu).

To calculate the molecular mass, you need to know the type and quantity of each atom in the molecule. The atomic masses you use are based on the most common isotopes of each element. For instance, carbon usually has an atomic mass of 12 amu because the isotope \(^{12}\mathrm{C}\) is the most prevalent.

A common example includes calculating the mass of a hydrocarbon such as \(^{12}\mathrm{C}_{2}^{1}\mathrm{H}_{6}\). You multiply the number of atoms by their respective atomic masses: - \(2 \times 12\) amu for carbon- \(6 \times 1.007825\) amu for hydrogen

Summing these gives the total molecular mass of about 30.04695 amu. Understanding these calculations helps in identifying compounds by comparing the calculated molecular masses to experimental data collected from tools like mass spectrometers.

Isotope Analysis

In mass spectrometry, isotope analysis plays a crucial role in understanding the composition of molecules. Isotopes are variants of elements that have the same number of protons but different numbers of neutrons. This leads to different atomic masses for isotopes of the same element.

In the context of the exercise, choosing the correct isotope mass is vital for accurate molecular mass calculation. For example, \(^{16}O\) has an atomic mass of 15.994915 amu, which needs to be used when calculating molecules containing oxygen.

Knowing which isotopes to use can be derived from natural abundance, where certain isotopes are more common and hence have a greater influence on the average atomic mass listed in periodic tables. Using precise isotope masses assists in narrowing down the potential candidates in a mass spectrum, making it easier to identify unknown compounds.

Mass spectrometry utilizes isotope analysis to differentiate and validate the molecular compositions of samples, especially when a molecule's mass is very close to potential alternatives.

Spectral Peak Identification

Identifying spectral peaks is an essential aspect of utilizing mass spectrometry data. When a sample is analyzed, the resulting spectrum shows peaks corresponding to the different molecules or fragments present, each peak represented by an accurate mass measurement.

The process involves comparing the observed mass of peaks to computed molecular masses to identify possible compounds. For example, if you have a spectral peak at a mass of 30.0106 amu, you compare this to the calculated molecular masses from known potential compounds like - \(30.04695\) amu for \(^{12}\mathrm{C}_{2}^{1}\mathrm{H}_{6}\)- \(30.010565\) amu for \(^{12}\mathrm{C}^{1}\mathrm{H}_{2}^{16}\mathrm{O}\)- \(29.997989\) amu for \(^{14}\mathrm{N}^{16}\mathrm{O}\)

The closest matching mass guides you to the most likely molecular candidate causing the peak. In this case, \(^{12}\mathrm{C}^{1}\mathrm{H}_{2}^{16}\mathrm{O}\) with a calculated mass of 30.010565 amu aligns best with the peak at 30.0106, indicating it as the responsible molecule. This technique allows chemists to deduce the structure and composition of unknown substances.