Question

Nitrogen has two naturally occurring isotopes: nitrogen-14 (mass = 14.00 amu) and nitrogen-15 (mass \(=15.00 \mathrm{amu}\) ). Bromine also has two naturally occurring isotopes: bromine-79 (mass \(=78.92 \mathrm{amu}\) ) and bromine-81 (mass \(=80.92\) amu). How many types of \(\mathrm{NBr}_{3}\) molecules of different masses can exist? Determine the mass (in amu) of each of them.

Short answer

There can be 8 different types of \(NBr_{3}\) molecules, with masses of 251.76 amu, 253.76 amu, 255.76 amu, 257.76 amu, 252.76 amu, 254.76 amu, 256.76 amu, and 258.76 amu.

Step-by-step solution

Identify Isotopes and Molecular Formula

The molecular formula for the compound in question is \(\mathrm{NBr}_{3}\), which indicates that there is one nitrogen atom (\(\mathrm{N}\)) and three bromine atoms (\(\mathrm{Br}\)) in each molecule. Nitrogen has two isotopes, \(\mathrm{^{14}N}\) and \(\mathrm{^{15}N}\), while bromine has two isotopes, \(\mathrm{^{79}Br}\) and \(\mathrm{^{81}Br}\).

Calculate Possible Molecule Masses

We should consider each isotope combination of nitrogen and bromine that could form an \(NBr_{3}\) molecule: 1) \(\mathrm{^{14}N}\mathrm{Br}_{3}\) with \(3 \times \mathrm{^{79}Br}\), 2) \(\mathrm{^{14}N}\mathrm{Br}_{3}\) with \(2 \times \mathrm{^{79}Br}\) and \(1 \times \mathrm{^{81}Br}\), 3) \(\mathrm{^{14}N}\mathrm{Br}_{3}\) with \(1 \times \mathrm{^{79}Br}\) and \(2 \times \mathrm{^{81}Br}\), 4) \(\mathrm{^{14}N}\mathrm{Br}_{3}\) with \(3 \times \mathrm{^{81}Br}\), 5) \(\mathrm{^{15}N}\mathrm{Br}_{3}\) with \(3 \times \mathrm{^{79}Br}\), 6) \(\mathrm{^{15}N}\mathrm{Br}_{3}\) with \(2 \times \mathrm{^{79}Br}\) and \(1 \times \mathrm{^{81}Br}\), 7) \(\mathrm{^{15}N}\mathrm{Br}_{3}\) with \(1 \times \mathrm{^{79}Br}\) and \(2 \times \mathrm{^{81}Br}\), 8) \(\mathrm{^{15}N}\mathrm{Br}_{3}\) with \(3 \times \mathrm{^{81}Br}\).

Calculate Masses for Each Molecule Type

Mass of each molecule type: 1) \(14.00 + 3 \times 78.92 = 251.76 \mathrm{amu}\), 2) \(14.00 + 2 \times 78.92 + 80.92 = 253.76 \mathrm{amu}\), 3) \(14.00 + 78.92 + 2 \times 80.92 = 255.76 \mathrm{amu}\), 4) \(14.00 + 3 \times 80.92 = 257.76 \mathrm{amu}\), 5) \(15.00 + 3 \times 78.92 = 252.76 \mathrm{amu}\), 6) \(15.00 + 2 \times 78.92 + 80.92 = 254.76 \mathrm{amu}\), 7) \(15.00 + 78.92 + 2 \times 80.92 = 256.76 \mathrm{amu}\), 8) \(15.00 + 3 \times 80.92 = 258.76 \mathrm{amu}\). Each mass corresponds to a different molecule type, resulting in 8 types of \(NBr_{3}\) molecules.

Key concepts

Isotopes and Their Role in Molecules

Isotopes are atoms of the same element that have the same number of protons but different numbers of neutrons, resulting in different atomic masses. For example, nitrogen has two stable isotopes, nitrogen-14 (), with 7 protons and 7 neutrons, and nitrogen-15 (), with 7 protons and 8 neutrons. Despite having different masses, isotopes exhibit very similar chemical behavior.

Understanding isotopes is crucial when analyzing molecular composition. In molecules, the presence of different isotopes results in molecules of the same type but with slightly varying masses. This is especially important in fields like chemistry, physics, and medicine, where isotopic composition can influence molecular properties, reaction rates, and even diagnostic techniques such as Magnetic Resonance Imaging (MRI).

In our exercise, we examine how isotopes of nitrogen () and bromine ( and ) result in varying masses for the molecules. Each isotope combination represents a unique molecular mass, affecting the overall physical properties of the substance.

Molecular Mass Calculation

Calculating the molecular mass of a compound is a fundamental task in chemistry that adds up the atomic masses of all atoms in a molecule. The unit for atomic mass is the atomic mass unit (amu), which is based on the mass of a carbon-12 atom. One amu is defined as the mass of a carbon-12 atom.

To calculate the molecular mass, we tally the mass of each isotope present in the molecule. For molecules with isotopic variations such as , we account for each possible isotope combination. This leads to different calculated molecular masses for the different isotope-based molecule types.

By systematically considering all the combinations of isotopes in our exercise, we're able to discern that there are multiple possible masses for the molecule. Each unique mass found in the calculation represents a different molecular species with potentially differing physical and chemical properties.

Chemical Formulas and Molecular Diversity

Chemical formulas provide a representation of the composition of molecules in a compact form, indicating both the types and quantities of atoms involved. The chemical formula encapsulates vital information required to understand a substance's structure and, consequently, its behavior in chemical reactions.

When we look at the chemical formula of the molecule, it signifies one nitrogen atom bonded to three bromine atoms. While the formula itself is simple, the isotopic diversity adds complexity, leading to multiple unique molecules. The process laid out in the exercise showcases how even with a single chemical formula, the existence of isotopes allows for several different molecules that vary in mass but share the same chemical structure. This variability is a fundamental aspect of chemistry that highlights the diversity hidden within seemingly simple molecules.

Each potential isotope combination of is central to understanding the molecule's overall mass distribution and can impact properties like boiling point, density, and behavior in chemical reactions. This insight is essential for students learning how to predict and interpret the outcomes of experiments involving isotopic variants.