To find the atomic mass of an element with multiple isotopes, you need to compute the weighted average of the masses of its isotopes, where the weights are their respective natural abundances. Let’s go through this step by step using the provided masses and abundances.
### Step-by-Step Solution:
1. Identify the masses and abundances of the isotopes:
- Isotope 1:
- Mass: 135.90714 amu
- Abundance: 0.19%
- Isotope 2:
- Mass: 137.90599 amu
- Abundance: 0.25%
- Isotope 3:
- Mass: 139.90543 amu
- Abundance: 88.43%
- Isotope 4:
- Mass: 141.90924 amu
- Abundance: 11.13%
2. Convert the percentage abundances to decimal form:
- Abundance of Isotope 1: [tex]\( \frac{0.19}{100} = 0.0019 \)[/tex]
- Abundance of Isotope 2: [tex]\( \frac{0.25}{100} = 0.0025 \)[/tex]
- Abundance of Isotope 3: [tex]\( \frac{88.43}{100} = 0.8843 \)[/tex]
- Abundance of Isotope 4: [tex]\( \frac{11.13}{100} = 0.1113 \)[/tex]
3. Calculate the weighted average for atomic mass:
The formula for the weighted average mass (atomic mass) is:
[tex]\[
\text{Atomic Mass} = (\text{Mass of Isotope 1} \times \text{Abundance of Isotope 1}) + (\text{Mass of Isotope 2} \times \text{Abundance of Isotope 2}) + (\text{Mass of Isotope 3} \times \text{Abundance of Isotope 3}) + (\text{Mass of Isotope 4} \times \text{Abundance of Isotope 4})
\][/tex]
Plugging in the values:
[tex]\[
\text{Atomic Mass} = (135.90714 \times 0.0019) + (137.90599 \times 0.0025) + (139.90543 \times 0.8843) + (141.90924 \times 0.1113)
\][/tex]
4. Summing the products:
[tex]\[
= 0.258223566 + 0.344764975 + 123.710885289 + 15.801984872
\][/tex]
5. Total Atomic Mass:
[tex]\[
\text{Atomic Mass} = 140.115858702 \, \text{amu}
\][/tex]
### Conclusion:
The atomic mass of the element, computed from its isotopes' masses and natural abundances, is approximately 140.116 amu.
To identify the element, we compare this calculated atomic mass with the atomic masses in the periodic table. An atomic mass of approximately 140.116 amu corresponds to the element Lanthanum (La).
Thus, the element with these four isotopes is Lanthanum.
