Natural abundances of `""^(12)C and ""^(13)C` isotopes of carbon are 99% and 1%, respectively. Assuming they only contributes to the mol. wt. of `C_2 F_4`, the percentage of `C_2 F_4` having a molecular mass of 101 is

Options

A. 1.98
B. 98
C. 0.198
D. 99

Explanation

To solve the problem, we need to determine the percentage of the compound \( C_2F_4 \) that has a molecular mass of 101, considering the natural abundances of the isotopes \( ^{12}C \) and \( ^{13}C \).

### Step 1: Calculate the molecular weight of \( C_2F_4 \) using \( ^{12}C \)
The molecular formula of \( C_2F_4 \) consists of 2 carbon atoms and 4 fluorine atoms. The atomic weights are:
- \( ^{12}C \): 12 g/mol
- Fluorine (F): 19 g/mol

Calculating the molecular weight using \( ^{12}C \):
\[ \text{Molecular weight} = 2 \times 12 + 4 \times 19 = 24 + 76 = 100 \text{ g/mol} \]

### Step 2: Calculate the molecular weight of \( C_2F_4 \) using \( ^{13}C \)
If both carbon atoms are \( ^{13}C \):
\[ \text{Molecular weight} = 2 \times 13 + 4 \times 19 = 26 + 76 = 102 \text{ g/mol} \]

### Step 3: Determine the contribution of each isotope to the molecular weight
Given the natural abundances:
- \( ^{12}C \): 99%
- \( ^{13}C \): 1%

### Step 4: Calculate the percentage of \( C_2F_4 \) with molecular weight 101
The molecular weight of 101 can occur when one carbon atom is \( ^{12}C \) and the other is \( ^{13}C \):
\[ \text{Molecular weight} = 12 + 13 + 76 = 101 \text{ g/mol} \]

### Step 5: Calculate the probability of having one \( ^{12}C \) and one \( ^{13}C \)
The probability of having one \( ^{12}C \) and one \( ^{13}C \) in \( C_2F_4 \) can be calculated as follows:
\[ \text{Probability} = \text{(Probability of } ^{12}C) \times \text{(Probability of } ^{13}C) + \text{(Probability of } ^{13}C) \times \text{(Probability of } ^{12}C) \]
\[ = (0.99 \times 0.01) + (0.01 \times 0.99) = 0.0198 + 0.0198 = 0.0396 \]

### Step 6: Convert the probability to percentage
To convert the probability to a percentage:
\[ \text{Percentage} = 0.0396 \times 100 = 3.96\% \]

### Final Calculation
Since we are looking for the percentage of \( C_2F_4 \) that has a molecular mass of 101, we conclude that:
\[ \text{Percentage of } C_2F_4 \text{ with molecular mass 101} = 3.96\% \]