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.

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%.

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 = \((0.0107 \times 0.9893) + (0.9893 \times 0.0107)\)

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 = \(0.0107 \times 0.0107\)

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: \({6 \choose 2} \times (0.0002^2 \times 0.9998^4) \approx 0.0000007\)