College

A certain element consists of two stable isotopes:

- The first has a mass of 138 amu and a percent natural abundance of [tex]8.90 \times 10^{-2} \%[/tex].
- The second has a mass of 139 amu and a percent natural abundance of [tex]99.9 \%[/tex].

What is the atomic weight of the element? [tex]\square[/tex] amu

Answer :

Sure! Let's solve the problem step-by-step. We're given two stable isotopes of an element, including their masses and natural abundances. We need to find the atomic weight of the element.

### Step-by-Step Solution:

1. Identify the masses and natural abundances of the isotopes:

- The mass of the first isotope ([tex]\( \text{mass}_1 \)[/tex]) is 138 amu with a natural abundance ([tex]\( \text{abundance}_1 \)[/tex]) of [tex]\( 8.90 \times 10^{-2} \% \)[/tex].
- The mass of the second isotope ([tex]\( \text{mass}_2 \)[/tex]) is 139 amu with a natural abundance ([tex]\( \text{abundance}_2 \)[/tex]) of [tex]\( 99.9 \% \)[/tex].

2. Convert the natural abundances from percentages to decimal form:

- [tex]\( \text{abundance}_1 \)[/tex] is [tex]\( 8.90 \times 10^{-2} \% \)[/tex]:
[tex]\[
\text{abundance}_1 = \frac{8.90 \times 10^{-2}}{100} = 0.00089
\][/tex]

- [tex]\( \text{abundance}_2 \)[/tex] is [tex]\( 99.9 \% \)[/tex]:
[tex]\[
\text{abundance}_2 = \frac{99.9}{100} = 0.9990
\][/tex]

3. Use the formula for the atomic weight:

The atomic weight of the element is the weighted average of the masses of its isotopes, calculated using their natural abundances. The formula is:
[tex]\[
\text{atomic weight} = (\text{mass}_1 \times \text{abundance}_1) + (\text{mass}_2 \times \text{abundance}_2)
\][/tex]

4. Substitute the values into the formula and calculate:

[tex]\[
\text{atomic weight} = (138 \times 0.00089) + (139 \times 0.9990)
\][/tex]

Perform the multiplications:
[tex]\[
138 \times 0.00089 = 0.12282
\][/tex]
[tex]\[
139 \times 0.9990 = 138.861
\][/tex]

Add these values together to get the atomic weight:
[tex]\[
\text{atomic weight} = 0.12282 + 138.861 = 138.98382
\][/tex]

### Conclusion:

The atomic weight of the element is [tex]\( 138.98382 \)[/tex] amu.