Question

Write molecular formulas for the five possible molecular ions of \(\mathrm{m} / z 100\) containing only the elements \(\mathrm{C}, \mathrm{H}, \mathrm{N}\), and \(\mathrm{O}\).

Short answer

Question: Determine the five possible molecular formulas for molecular ions that have an m/z value of 100, considering they contain only the elements C, H, N, and O. Answer: The five possible molecular formulas are: 1. C8H4 2. C7H12 3. C6H8N2 4. C6H4O 5. C5H16N

Step-by-step solution

List the atomic masses of the elements

We have the following atomic masses for each element: C: 12 H: 1 N: 14 O: 16

Start with the highest number of C atoms possible and decrease

We will first try to see if we can create a molecule with the highest possible number of carbon atoms and still have an m/z value of 100. We can divide 100 by 12 to find the maximum number of carbon atoms: \(\frac{100}{12} \approx 8.33\), so the maximum number of carbon atoms is 8.

Find possible molecular formulas and check mass

We will now try different combinations of the other elements with 8 carbon atoms and calculate the mass. Then, we will decrease the number of carbon atoms and repeat the process until we find five possible molecular formulas with m/z value of 100. 1. C8H4 Mass = 8(12) + 4(1) = 100 2. C7H12 Mass = 7(12) + 12(1) = 100 3. C6H8N2 Mass = 6(12) + 8(1) + 2(14) = 100 4. C6H4O Mass = 6(12) + 4(1) + 1(16) = 100 5. C5H16N Mass = 5(12) + 16(1) + 1(14) = 100 The five possible molecular formulas are: 1. C8H4 2. C7H12 3. C6H8N2 4. C6H4O 5. C5H16N

Key concepts

Atomic Masses

Understanding atomic masses is crucial for a foundational grasp of chemistry. Every element on the periodic table has a unique atomic mass, typically expressed in atomic mass units (amu). These atomic masses are averaged values taking into account the different isotopes of an element and their abundances.

In our exercise, we used the atomic masses of carbon (C), hydrogen (H), nitrogen (N), and oxygen (O) — 12 amu, 1 amu, 14 amu, and 16 amu respectively. These masses allow us to calculate the total mass of a molecule by adding up the atomic masses of the individual atoms contained within the molecule.

For example, to calculate the mass of a molecule of C8H4, we multiply the number of carbon atoms (8) by the atomic mass of carbon (12 amu) and the number of hydrogen atoms (4) by the atomic mass of hydrogen (1 amu) to get a total mass of 100 amu. It's like putting together pieces of a puzzle; each piece has a set weight, and when combined correctly, they give us the full picture – or in this case, the total mass of the molecular ion.

Mass-to-Charge Ratio

The mass-to-charge ratio (m/z) is a key concept in mass spectrometry, which is a technique used to determine the masses of particles, identify compounds within a sample, and elucidate the chemical structures of molecules. The m/z ratio represents the mass of a particle (such as a molecular ion) divided by its electrical charge.

In this exercise, we're considering molecular ions with an m/z value of 100. This means the mass of the ions we need to create should be 100 amu if they carry a single positive charge. The mass-to-charge ratio is particularly helpful as it simplifies our search for potential molecular formulas—since we're dealing with singly charged ions, the mass itself must be 100 amu.

Keep in mind that the charge of molecular ions can affect the m/z ratio significantly. For example, a molecular ion with a mass of 200 amu and a charge of +2 would also have an m/z of 100 (200/2=100). However, in this exercise, we're focused on ions with a single positive charge.

Molecular Formula Calculation

Calculating molecular formulas involves determining the number of atoms of each element in a compound. It's like solving a mathematical equation where the sum of the atomic masses times their respective number of atoms equals the molar mass of the compound.

In the exercise, we began by attempting to fit as many carbon atoms as possible into the mass limit of 100 amu set by the m/z ratio. Using trial and error, we gradually adjusted the number of other atoms—hydrogen, nitrogen, and oxygen—to first reach and then fine-tune the total mass to meet our target. This systematic approach is effective because it considers the heaviest elements first, adjusting for lighter ones later.

For instance, after starting with eight carbon atoms, the remaining mass for hydrogen atoms was calculated, which is how we derived the molecular formula C8H4. We continued to adjust the number of each type of atom, balancing them until we derived five distinct molecular formulas, all with an exact mass of 100 amu. Each formula represents a unique compound, showcasing the beauty of molecular diversity within strict numeric constraints.