Question

Boron trifluoride is used as a catalyst in the synthesis of organic compounds. When this compound is analyzed by mass spectrometry, several different \(1+\) ions form, including ions representing the whole molecule as well as molecular fragments formed by the loss of one, two, and three \(\mathrm{F}\) atoms. Given that boron has two naturally occurring isotopes, \({ }^{10} \mathrm{~B}\) and \({ }^{11} \mathrm{~B},\) and fluorine has one, \({ }^{19} \mathrm{~F}\), calculate the masses of all possible \(1+\) ions.

Short answer

The masses of all possible 1+ ions are 67 amu, 68 amu, 48 amu, 49 amu, 29 amu, 30 amu, 10 amu, and 11 amu.

Step-by-step solution

Identify the isotopes and their masses

Boron (\text{B}) has two naturally occurring isotopes: \text{\({ }^{10} \text{B}\)} with a mass of 10 amu and \text{\({ }^{11} \text{B}\)} with a mass of 11 amu. Fluorine (\text{F}) has one naturally occurring isotope: \text{\({ }^{19} \text{F}\)} with a mass of 19 amu.

Calculate the mass of the whole molecule ions

The compound Boron trifluoride (\text{BF\(_3\)}) has three fluorine atoms and one boron atom. First, calculate the masses of the whole molecule ions using the two isotopes of boron: For \text{\({ }^{10} \text{B}\)}: \text{\({ }^{10} \text{B}\)} + 3\text{\({ }^{19} \text{F}\)} = 10 + 3(19) = 10 + 57 = 67 amu For \text{\({ }^{11} \text{B}\)}: \text{\({ }^{11} \text{B}\)} + 3\text{\({ }^{19} \text{F}\)} = 11 + 3(19) = 11 + 57 = 68 amu

Calculate the masses of molecular fragments with the loss of one fluorine atom

When a fluorine atom is lost, one \text{BF\(_2\)} ion is formed. Using both isotopes of boron: For \text{\({ }^{10} \text{B}\)}: \text{\({ }^{10} \text{B}\)} + 2\text{\({ }^{19} \text{F}\)} = 10 + 2(19) = 10 + 38 = 48 amu For \text{\({ }^{11} \text{B}\)}: \text{\({ }^{11} \text{B}\)} + 2\text{\({ }^{19} \text{F}\)} = 11 + 2(19) = 11 + 38 = 49 amu

Calculate the masses of molecular fragments with the loss of two fluorine atoms

When two fluorine atoms are lost, one \text{BF} ion is formed. Using both isotopes of boron: For \text{\({ }^{10} \text{B}\)}: \text{\({ }^{10} \text{B}\)} + \text{\({ }^{19} \text{F}\)} = 10 + 19 = 29 amu For \text{\({ }^{11} \text{B}\)}: \text{\({ }^{11} \text{B}\)} + \text{\({ }^{19} \text{F}\)} = 11 + 19 = 30 amu

Calculate the mass of the boron ion with the loss of all three fluorine atoms

When all three fluorine atoms are lost, the ion remaining is just the boron ion itself. Using both isotopes of boron: For \text{\({ }^{10} \text{B}\)}: \text{\({ }^{10} \text{B}\)} = 10 amu For \text{\({ }^{11} \text{B}\)}: \text{\({ }^{11} \text{B}\)} = 11 amu

Key concepts

Isotope Masses

Isotopes are variants of a particular chemical element that have the same number of protons but different numbers of neutrons. This difference in the neutron count changes the mass of the isotopes.

In the case of boron, it has two naturally occurring isotopes:

• ¹⁰B, which has a mass of 10 amu (atomic mass units).
• ¹¹B, which has a mass of 11 amu.

Fluorine, on the other hand, has only one naturally occurring isotope: ¹⁹F with a mass of 19 amu.
Understanding these isotope masses is crucial for calculating the overall mass of compounds and their fragments in mass spectrometry.

Molecular Fragments

In mass spectrometry, compounds are often broken down into smaller pieces called molecular fragments.

These fragments help in identifying the structure and composition of the molecule. For Boron trifluoride (BF₃), fragmentation can occur with the loss of one, two, or all three fluorine atoms. Let's look at these step-by-step:

• Losing one fluorine atom turns BF₃ into BF₂ with potential masses:
• ¹⁰BF₂: 10 amu (for B) + 2 x 19 amu (for F) = 10 + 38 = 48 amu.
• ¹¹BF₂: 11 amu (for B) + 2 x 19 amu = 11 + 38 = 49 amu.

• Losing two fluorine atoms gives BF:
• ¹⁰BF: 10 amu + 19 amu = 29 amu.
• ¹¹BF: 11 amu + 19 amu = 30 amu.

• Losing all three fluorine atoms leaves you only with the boron ion, either 10 or 11 amu depending on the isotope.

Boron Isotopes

Boron is unique as it exists naturally with two isotopes which play an essential role in mass spectrometry calculations.

These are:

• ¹⁰B (10 amu)
• ¹¹B (11 amu)

When analyzing boron-containing compounds, always consider the contributions from both isotopes.
In our example of Boron trifluoride (BF₃), the isotopic variation of boron leads to different mass readings for the entire molecule and its fragments.
Understanding this isotopic variation is key to interpreting mass spectrometry data accurately.

Fluorine Isotope

Fluorine is different from boron as it has only one naturally occurring isotope: ¹⁹F with a mass of 19 amu.

This simplifies calculations significantly as there is only one mass to consider for fluorine in mass spectrometry.

In the case of Boron trifluoride (BF₃), the mass contribution of fluorine is consistently:

• 3 x 19 = 57 amu for three fluorine atoms in BF₃
• 2 x 19 = 38 amu for two fluorine atoms in BF₂
• 1 x 19 = 19 amu for one fluorine atom in BF.

Given that fluorine doesn't have isotopic variation like boron, it makes the interpretation of mass spectrometry data more straightforward for fluorine-containing fragments.