Question

Determine the probability of the following in a natural sample of ethane. (a) One carbon in an ethane molecule is \({ }^{13} \mathrm{C}\). (b) Both carbons in an ethane molecule are \({ }^{13} \mathrm{C}\). (c) Two hydrogens in an ethane molecule are replaced by deuterium atoms.

Short answer

Answer: The probabilities of these events occurring in a natural sample of ethane are: (a) approximately 0.0214, (b) approximately 0.0001, and (c) approximately 0.0000007.

Step-by-step solution

Identify the composition of ethane and natural abundances of isotopes

Ethane is a molecule with the chemical formula C2H6, meaning it has two carbon atoms and six hydrogen atoms. There are two stable isotopes of carbon: \({ }^{12} \mathrm{C}\), which makes up about 98.93% of natural carbon, and \({ }^{13} \mathrm{C}\), which makes up about 1.07%. Similarly, hydrogen has two stable isotopes: \({ }^{1} \mathrm{H}\) (also called protium), which makes up about 99.98% of natural hydrogen, and \({ }^{2} \mathrm{H}\) (also called deuterium), which makes up about 0.02%.

Calculate the probability of one carbon in an ethane molecule being \({ }^{13} \mathrm{C}\)

For this event, we need the probability of one carbon atom being \({ }^{13} \mathrm{C}\) and the other carbon atom being \({ }^{12} \mathrm{C}\). The probability of choosing a \({ }^{13} \mathrm{C}\) atom is 1.07% (0.0107 in decimal form) and the probability of choosing a \({ }^{12} \mathrm{C}\) atom is 98.93% (0.9893 in decimal form). We can have either the first or the second carbon atom being \({ }^{13} \mathrm{C}\). So the probability for this event can be calculated as follows: Probability = \((P_{1st\:carbon\:being\:^{13}C} \times P_{2nd\:carbon\:being\:^{12}C}) + (P_{1st\:carbon\:being\:^{12}C} \times P_{2nd\:carbon\:being\:^{13}C})\) Probability = \((0.0107 \times 0.9893) + (0.9893 \times 0.0107)\)

Calculate the probability of both carbons in an ethane molecule being \({ }^{13} \mathrm{C}\)

For this event, we need the probability of both carbon atoms being \({ }^{13} \mathrm{C}\). The probability of choosing a \({ }^{13} \mathrm{C}\) atom is the same as before, 1.07% (0.0107 in decimal form). Probability = \(P_{1st\:carbon\:being\:^{13}C} \times P_{2nd\:carbon\:being\:^{13}C}\) Probability = \(0.0107 \times 0.0107\)

Calculate the probability of two hydrogens in an ethane molecule being replaced by deuterium atoms

For this event, we need the probability of two hydrogen atoms being \({ }^{2} \mathrm{H}\) (or deuterium) and the other four hydrogen atoms being \({ }^{1} \mathrm{H}\) (or protium). The probability of choosing a deuterium atom is 0.02% (0.0002 in decimal form) and the probability of choosing a protium atom is 99.98% (0.9998 in decimal form). There are \({6 \choose 2}\) ways to choose the positions of the two deuterium atoms. So the probability for this event can be calculated as follows: Probability = \({6 \choose 2} \times (P_{deuterium\:atoms\:chosen} \times P_{remaining\:protium\:atoms})\) Probability = \({6 \choose 2} \times (0.0002^2 \times 0.9998^4)\) Finally, let's calculate these probabilities: (a) Probability of one carbon in an ethane molecule being \({ }^{13} \mathrm{C}\): \((0.0107 \times 0.9893) + (0.9893 \times 0.0107) \approx 0.0214\) (b) Probability of both carbons in an ethane molecule being \({ }^{13} \mathrm{C}\): \(0.0107 \times 0.0107 \approx 0.0001\) (c) Probability of two hydrogens in an ethane molecule being replaced by deuterium atoms: \({6 \choose 2} \times (0.0002^2 \times 0.9998^4) \approx 0.0000007\)

Key concepts

Carbon Isotopes

Carbon isotopes, particularly \(^{12}C\) and \(^{13}C\), play a significant role in chemistry, especially when studying organic compounds like ethane. Isotopes of an element share the same number of protons but differ in neutrons, giving them different mass numbers. In nature, carbon exists primarily as two stable isotopes:

• \(^{12}C\): This is the most abundant isotope, making up approximately 98.93% of all carbon naturally found on Earth. Due to its greater abundance, it is often the carbon isotope encountered in organic compounds.
• \(^{13}C\): Comprising about 1.07% of natural carbon, this isotope is less common and has a slightly greater mass, which can affect the molecular weight calculations in isotopic studies and the analysis of natural samples of organic compounds.

In ethane \( C_{2}H_{6} \), each of the two carbon atoms can be either \(^{12}C\) or \(^{13}C\). Understanding the abundance and, consequently, the probability of each isotope forming part of ethane helps in fields such as geochemistry and organic chemistry to track processes like carbon cycling and chemical reactions.

Hydrogen Isotopes

Hydrogen, the lightest element, has two stable isotopes: \(^{1}H\) and \(^{2}H\), also known as protium and deuterium, respectively. Isotopic variation is key in understanding chemical reactions and processes in which ethane and other hydrocarbons participate. Here's the breakdown:

• \(^{1}H\) (Protium): The most prevalent hydrogen isotope, comprising about 99.98% of natural hydrogen. Its high abundance makes it the primary hydrogen isotope encountered in most reactions involving hydrogen.
• \(^{2}H\) (Deuterium): Though it only constitutes about 0.02% of natural hydrogen, deuterium is crucial in research due to its applications in tracer studies and understanding reaction mechanisms. Its greater mass compared to protium affects vibrational properties and bond strength in molecules.

In ethane, the presence of hydrogen isotopes affects both its molecular weight and properties. Replacing protium atoms with deuterium can be significant in studies that rely on fractional distillation or isotope-ratio mass spectrometry to trace biochemical and environmental pathways.

Probability Calculation in Chemistry

Probability in chemistry helps to predict the likelihood of various isotopes appearing in molecular compounds such as ethane. Calculating these probabilities involves understanding the natural abundances of isotopes and using combinatorial mathematics to determine the odds of different isotopic configurations occurring in a sample. Here’s a simplified approach:

• **Single Isotopic Event**: To find the probability of a single isotopic substitution (e.g., one \(^{13}C\) in ethane), use the abundance percentages. Multiply the likelihood of each isotopic event together for combined possibilities, considering that the substitution can occur in any order.
• **Multiple Isotopic Events**: For events requiring multiple isotopic substitutions (like both carbons being \(^{13}C\) or multiple deuterium atoms replacing hydrogen), calculate the probability for each instance and multiply them together. Afterward, apply combinatorial coefficients (such as binomial coefficients) to account for different possible arrangements.

In ethane, knowing these probabilities can aid in the study of mixtures in natural and experimental settings, providing insights into the proportions of isotopes and conditions affecting their distribution. These calculations are especially useful for making predictions in isotope geochemistry and organic synthesis.