Chlorine consists of the two isotopes `""_(17)Cl^(35)` and `"'_(17)Cl^(37)` of masses 34.99 amu and 36.97 amu, respectively. Calculate the relative abundance of each isotope if atomic mass of chlorine is 35.46 amu.

Answer

To calculate the relative abundance of the isotopes \( \text{Cl}^{35} \) and \( \text{Cl}^{37} \), we can follow these steps:

### Step 1: Define Variables
Let:
- \( x \) = the relative abundance (percentage) of \( \text{Cl}^{35} \)
- \( 100 - x \) = the relative abundance (percentage) of \( \text{Cl}^{37} \)

### Step 2: Set Up the Equation
The weighted average atomic mass of chlorine can be expressed as:
\[ \text{Atomic mass} = \frac{(x \cdot 34.99) + ((100 - x) \cdot 36.97)}{100} \]
Given that the atomic mass of chlorine is 35.46 amu, we can set up the equation:
\[ 35.46 = \frac{(x \cdot 34.99) + ((100 - x) \cdot 36.97)}{100} \]

### Step 3: Eliminate the Denominator
Multiply both sides of the equation by 100 to eliminate the denominator:
\[ 3546 = (x \cdot 34.99) + ((100 - x) \cdot 36.97) \]

### Step 4: Expand the Equation
Expand the right side:
\[ 3546 = 34.99x + 3697 - 36.97x \]

### Step 5: Combine Like Terms
Combine the terms involving \( x \):
\[ 3546 = 3697 - 1.98x \]

### Step 6: Isolate \( x \)
Rearranging gives:
\[ 1.98x = 3697 - 3546 \]
\[ 1.98x = 151 \]
Now, divide both sides by 1.98:
\[ x = \frac{151}{1.98} \approx 76.26 \]

### Step 7: Calculate Abundance of \( \text{Cl}^{37} \)
Now, calculate the abundance of \( \text{Cl}^{37} \):
\[ 100 - x = 100 - 76.26 \approx 23.74 \]

### Final Results
- The relative abundance of \( \text{Cl}^{35} \) is approximately **76.26%**.
- The relative abundance of \( \text{Cl}^{37} \) is approximately **23.74%**.