Chlorine has 2 isotopes 35 Cl and 37 Cl with percentage abundances of 75.78% and 24.22% calculate its R.A.M

1 answer

To calculate the relative atomic mass (R.A.M.) of chlorine using its isotopes and their percentage abundances, you can use the following formula:

\[ \text{R.A.M.} = \left( \frac{(\text{Isotope mass}_1 \times \text{Abundance}_1) + (\text{Isotope mass}_2 \times \text{Abundance}_2)}{100} \right) \]

Where:

  • Isotope mass of \( ^{35}\text{Cl} = 35 , \text{u} \)
  • Isotope mass of \( ^{37}\text{Cl} = 37 , \text{u} \)
  • Abundance of \( ^{35}\text{Cl} = 75.78% \)
  • Abundance of \( ^{37}\text{Cl} = 24.22% \)

Now substitute these values into the formula:

\[ \text{R.A.M.} = \left( \frac{(35 \times 75.78) + (37 \times 24.22)}{100} \right) \]

Calculating each term:

  1. For \( ^{35}\text{Cl} \): \[ 35 \times 75.78 = 2652.3 \]

  2. For \( ^{37}\text{Cl} \): \[ 37 \times 24.22 = 896.14 \]

Now add these two results:

\[ 2652.3 + 896.14 = 3548.44 \]

Now, divide by 100 to find the R.A.M.:

\[ \text{R.A.M.} = \frac{3548.44}{100} = 35.4844 \]

Thus, the relative atomic mass of chlorine, rounded to two decimal places, is approximately:

\[ \text{R.A.M.} \approx 35.48 , \text{u} \]