Problem 36 - Chemistry: The Central Science

Question

Mass spectrometry is more often applied to molecules than to atoms. We will see in Chapter 3 that the molecular weight of a molecule is the sum of the atomic weights of the atoms in the molecule. The mass spectrum of \(\mathrm{H}_{2}\) is taken under conditions that prevent decomposition into \(\mathrm{H}\) atoms. The two naturally occurring isotopes of hydrogen are \({ }^{1} \mathrm{H}\) (atomic mass \(=1.00783\) amu; abundance \(99.9885 \%\) ) and \({ }^{2} \mathrm{H}\) (atomic mass \(=2.01410 \mathrm{amu}\); abundance \(0.0115 \%\). (a) How many peaks will the mass spectrum have? (b) Give the relative atomic masses of each of these peaks. (c) Which peak will be the largest, and which the smallest?

Step-by-Step Solution

Answer

(a) The mass spectrum of \(\mathrm{H}_{2}\) will have two peaks. (b) The relative atomic masses of these peaks are 2.01566 amu and 3.02193 amu. (c) The largest peak will be \({}^{1} \mathrm{H} - {}^{1} \mathrm{H}\) with an abundance of approximately 99.977%, and the smallest peak will be \({}^{1} \mathrm{H} - {}^{2} \mathrm{H}\) with an abundance of approximately 0.0115%.

Step 1: Identify the isotopes of hydrogen and their properties

We have two naturally occurring isotopes of hydrogen: 1. \({}^{1} \mathrm{H}\) with an atomic mass of 1.00783 amu and an abundance of 99.9885%. 2. \({}^{2} \mathrm{H}\) with an atomic mass of 2.01410 amu and an abundance of 0.0115%.

Step 2: Find all possible combinations of hydrogen isotopes to form \(\mathrm{H}_{2}\).

There are two ways to form a hydrogen molecule (\(\mathrm{H}_{2}\)) from available isotopes: 1. \({}^{1} \mathrm{H} - {}^{1} \mathrm{H}\) (two \({}^{1} \mathrm{H}\) atoms) 2. \({}^{1} \mathrm{H} - {}^{2} \mathrm{H}\) (one \({}^{1} \mathrm{H}\) and one \({}^{2} \mathrm{H}\) atom)

Step 3: Calculate the relative atomic masses for each peak.

We will now find the molecular weight for each combination in step 2. 1. For \({}^{1} \mathrm{H} - {}^{1} \mathrm{H}\): The molecular weight is the sum of the two atomic weights. Molecular weight = 1.00783 + 1.00783 = 2.01566 amu. 2. For \({}^{1} \mathrm{H} - {}^{2} \mathrm{H}\): Molecular weight = 1.00783 + 2.01410 = 3.02193 amu.

Step 4: Determine the largest and smallest peaks.

To determine the largest and smallest peaks, we will examine the abundances of the isotopes in each combination. 1. For \({}^{1} \mathrm{H} - {}^{1} \mathrm{H}\): The chance of this combination occurring is equal to the product of the abundances of the two \({}^{1} \mathrm{H}\) atoms. Abundance = 0.999885 * 0.999885 ≈ 0.99977 or 99.977%. 2. For \({}^{1} \mathrm{H} - {}^{2} \mathrm{H}\): The chance of this combination occurring is equal to the product of the abundances of one \({}^{1} \mathrm{H}\) atom and one \({}^{2} \mathrm{H}\) atom. Abundance = 0.999885 * 0.000115 ≈ 0.000115 or 0.0115%. Given these abundances, the \({}^{1} \mathrm{H} - {}^{1} \mathrm{H}\) peak will be the largest, and the \({}^{1} \mathrm{H} - {}^{2} \mathrm{H}\) peak will be the smallest. To answer the questions in the exercise: (a) The mass spectrum of \(\mathrm{H}_{2}\) will have two peaks. (b) The relative atomic masses of these peaks are 2.01566 amu and 3.02193 amu. (c) The largest peak will be \({}^{1} \mathrm{H} - {}^{1} \mathrm{H}\), and the smallest peak will be \({}^{1} \mathrm{H} - {}^{2} \mathrm{H}\).

Key Concepts

Hydrogen Isotopes

Hydrogen, the most abundant element in the universe, consists of isotopes that differ in neutron number. Isotopes have the same number of protons (which defines the element) but differ in the number of neutrons. This difference in neutron number causes variations in atomic mass.

• Protium (\({}^{1} \mathrm{H}\)): The most common isotope of hydrogen, it has one proton and no neutrons. Its atomic mass is 1.00783 amu, and it accounts for 99.9885% of natural hydrogen.
• Deuterium (\({}^{2} \mathrm{H}\)): This isotope contains one proton and one neutron, with an atomic mass of 2.01410 amu. It makes up about 0.0115% of natural hydrogen.

These isotopes do not significantly differ in chemical behavior since the number of electrons (and thus the electronic configuration) remains unchanged. However, they play crucial roles in nuclear reactions and fields like mass spectrometry, where small differences in mass can be measured and analyzed. By understanding the distribution and properties of these isotopes, one can predict the mass spectra of various substances like \(\mathrm{H}_{2}\).

Molecular Weight Calculation

Molecular weight is calculated by summing the atomic weights of all atoms within a molecule. For diatomic hydrogen (\(\mathrm{H}_{2}\)), we can form molecules with different isotope combinations, affecting the resulting molecular weight.

• \({}^{1} \mathrm{H} - {}^{1} \mathrm{H}\): This combination results in a molecular weight calculated as follows: 1.00783 amu (for each \({}^{1} \mathrm{H}\) atom) added together, yielding 2.01566 amu.
• \({}^{1} \mathrm{H} - {}^{2} \mathrm{H}\): Here, the molecular weight is obtained by adding 1.00783 amu (\({}^{1} \mathrm{H}\)) and 2.01410 amu (\({}^{2} \mathrm{H}\)), resulting in 3.02193 amu.

These calculations demonstrate the variance in molecular weight based on available isotopes. Calculating molecular weight precisely is crucial in determining the masses that appear in a mass spectrum.

Relative Atomic Mass

Relative atomic mass, not to be confused with atomic weight, is an average of the masses of the isotopes composing an element, weighted by their natural abundance. In the context of mass spectrometry, a mass spectrum will have peaks corresponding to the relative masses of an element's isotopes.
For diatomic hydrogen (\(\mathrm{H}_{2}\)), the mass spectrum reflects the relative atomic masses calculated for different isotope combinations. We observe two peaks:

• 2.01566 amu for the \({}^{1} \mathrm{H} - {}^{1} \mathrm{H}\) combination, reflecting the most abundant form.
• 3.02193 amu for the \({}^{1} \mathrm{H} - {}^{2} \mathrm{H}\) combination, which is much less common.

The abundance of \({}^{1} \mathrm{H}\) over \({}^{2} \mathrm{H}\) means that the peak at 2.01566 amu will be the most prominent, while the one at 3.02193 amu will be smaller. Understanding relative atomic mass helps in interpreting the dominance and presence of peaks within a spectrum, giving insights into isotopic compositions and chemical analysis.