Question

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

Answer 1

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:

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