Solomonium has two naturally occurring isotopes: So⁻³⁵ with a natural abundance of 75.77% and So⁻³⁷ with a natural abundance of 24.23%. What is the atomic mass of Solomonium?
The Correct Answer and Explanation is :
To calculate the atomic mass of Solomonium (So), we use the formula for the weighted average of the isotopes based on their natural abundances:
[
\text{Atomic Mass} = ( \text{mass of So}^{-35} \times \text{fractional abundance of So}^{-35} ) + ( \text{mass of So}^{-37} \times \text{fractional abundance of So}^{-37} )
]
Given Data:
- Mass of So⁻³⁵: 34.96885 amu
- Mass of So⁻³⁷: 36.96590 amu
- Natural abundance of So⁻³⁵: 75.77% (or 0.7577 in fractional form)
- Natural abundance of So⁻³⁷: 24.23% (or 0.2423 in fractional form)
Calculation:
Now we plug in the values:
[
\text{Atomic Mass} = (34.96885 \, \text{amu} \times 0.7577) + (36.96590 \, \text{amu} \times 0.2423)
]
[
\text{Atomic Mass} = (26.519074 \, \text{amu}) + (8.951170 \, \text{amu})
]
[
\text{Atomic Mass} = 35.470244 \, \text{amu}
]
Result:
The atomic mass of Solomonium is approximately 35.47 amu.
Explanation:
The atomic mass of an element is a weighted average of the masses of its isotopes, accounting for their relative abundances. In the case of Solomonium, it has two isotopes: So⁻³⁵ and So⁻³⁷.
- So⁻³⁵ is the lighter isotope, with a mass of about 34.97 amu, and it constitutes about 75.77% of naturally occurring Solomonium.
- So⁻³⁷, the heavier isotope, has a mass of approximately 36.97 amu and accounts for 24.23% of the natural abundance.
By using the natural abundances to weight the contribution of each isotope’s mass, we arrive at a more accurate representation of the element’s atomic mass. This average atomic mass is crucial in chemistry, as it influences calculations involving molar mass and stoichiometry in chemical reactions. The precision of the atomic mass is also vital for scientific applications, including isotope studies and the understanding of elemental behavior in various conditions.