Question

Boron consists of two isotopes, \({ }^{10} \mathrm{~B}\) and \({ }^{11} \mathrm{~B}\). Chlorine also has two isotopes, \({ }^{35} \mathrm{Cl}\) and \({ }^{37} \mathrm{Cl}\). Consider the mass spectrum of \(\mathrm{BCl}_{3}\). How many peaks would be present, and what approximate mass would each peak correspond to in the \(\mathrm{BCl}_{3}\) mass spectrum?

Short answer

The mass spectrum of BCl₃ will have 8 peaks corresponding to the following approximate masses: 115, 116, 117, 118, 119, 120, 121, and 122.

Step-by-step solution

List all possible combinations of isotopes

To determine all the unique BCl₃ molecules that can be formed from the given isotopes, we need to combine \({ }^{10} \mathrm{B}\) and \({ }^{11} \mathrm{B}\) with \({ }^{35} \mathrm{Cl}\) and \({ }^{37} \mathrm{Cl}\) in all possible ways, keeping in mind that there are three Cl atoms in each BCl₃ molecule.

Calculate the mass of each unique BCl₃ molecule

Next, we will calculate the mass of each unique BCl₃ molecule formed using the atomic mass of each isotope. 1. B(10)Cl(35)Cl(35)Cl(35): \(10 + 3 \times 35 = 10 + 105 = 115 \) 2. B(10)Cl(35)Cl(35)Cl(37): \(10 + 2 \times 35 + 37 = 10 + 70 + 37 = 117 \) 3. B(10)Cl(35)Cl(37)Cl(37): \(10 + 35 + 2 \times 37 = 10 + 35 + 74 = 119 \) 4. B(10)Cl(37)Cl(37)Cl(37): \(10 + 3 \times 37 = 10 + 111 = 121 \) 5. B(11)Cl(35)Cl(35)Cl(35): \(11 + 3 \times 35 = 11 + 105 = 116 \) 6. B(11)Cl(35)Cl(35)Cl(37): \(11 + 2 \times 35 + 37 = 11 + 70 + 37 = 118 \) 7. B(11)Cl(35)Cl(37)Cl(37): \(11 + 35 + 2 \times 37 = 11 + 35 + 74 = 120 \) 8. B(11)Cl(37)Cl(37)Cl(37): \(11 + 3 \times 37 = 11 + 111 = 122 \)

Count the number of different masses and list the corresponding peaks

Now, let's count the number of different masses calculated in step 2 and list the corresponding peaks: 1. Mass 115: B(10)Cl(35)Cl(35)Cl(35) 2. Mass 116: B(11)Cl(35)Cl(35)Cl(35) 3. Mass 117: B(10)Cl(35)Cl(35)Cl(37) 4. Mass 118: B(11)Cl(35)Cl(35)Cl(37) 5. Mass 119: B(10)Cl(35)Cl(37)Cl(37) 6. Mass 120: B(11)Cl(35)Cl(37)Cl(37) 7. Mass 121: B(10)Cl(37)Cl(37)Cl(37) 8. Mass 122: B(11)Cl(37)Cl(37)Cl(37) There are 8 unique BCl₃ molecules and 8 different masses. This means there will be 8 peaks in the BCl₃ mass spectrum. These peaks will correspond to the masses 115, 116, 117, 118, 119, 120, 121, and 122.

Key concepts

Isotopes of Boron

Isotopes are atoms of the same element that have the same number of protons but a different number of neutrons. This variation causes them to have different atomic masses. Boron has two naturally occurring isotopes: boron-10 and boron-11. The numbers (10 and 11) represent the total count of protons and neutrons in the nucleus, with 5 protons being constant as they define the element boron. Therefore, boron-10 has 5 neutrons, while boron-11 has 6 neutrons. These isotopes can be identified through mass spectrometry, which is a technique that measures the masses of atoms or molecules.

In the context of the exercise, the presence of two isotopes of boron impacts the resulting mass spectrum of BCl3 because each isotope contributes to a different molecular mass. As we consider possible molecule combinations, it's important to take into account that these two isotopes will lead to molecules with slightly varying weights, hence the multiple peaks in the mass spectrum.

Isotopes of Chlorine

Much like boron, chlorine also has isotopes, specifically chlorine-35 and chlorine-37. The most abundant is chlorine-35, which accounts for about 75% of naturally occurring chlorine, while chlorine-37 makes up approximately 25%. The mass spectrum of a molecule containing chlorine will show peaks that correspond to these isotopes.

In our BCl3 example, since each molecule of BCl3 contains three chlorine atoms, the combination of chlorine isotopes will drastically increase the number of potential unique molecular masses. For instance, a molecule with three chlorine-35 atoms will be lighter than a molecule with two chlorine-35 atoms and one chlorine-37 atom. This is the reason we witness multiple peaks for the BCl3 molecule in the mass spectrum, as it reflects all the possible isotopic combinations of chlorine within the molecule.

Molecular Mass Calculation

The molecular mass of a compound can be found by adding up the atomic masses of its constituent atoms. However, when dealing with isotopes, it's essential to consider the specific mass of each isotope present in the molecule.

In our exercise, we see the detailed calculation for every possible combination of boron and chlorine isotopes within a BCl3 molecule. By using the mass of the isotopes (10 or 11 for boron and 35 or 37 for chlorine), we multiply the atomic mass of the chlorine by three (since there are three chlorine atoms) and add it to the mass of boron to get the total molecular mass for each possible BCl3 molecule. This is a straightforward arithmetic process, yet the presence of different isotopes makes it a bit more complex since it produces multiple unique masses. It's these unique masses that are observed as different peaks on a mass spectrum, allowing scientists and students to deduce the isotopic composition of the sample in question.