Question

An element has two naturally occurring isotopes. Isotope 1 has a mass of 120.9038 amu and a relative abundance of 57.4%, and isotope 2 has a mass of 122.9042 amu and a relative abundance of 42.6%.A.Find the atomic mass of this elementExpress your answer using four significant figures.B. By comparison to the periodic table, identify it.Express your answer as a chemical symbol.

Answer 1

Answer:

The atomic mass of this element is 121.8 amu.

The average atomic mass calculated is of an element named antimony that is Sb.

Explanation:

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

We are given:

Mass of isotope 1 = 120.9038 amu

Percentage abundance of isotope 1 = 57.4%

Fractional abundance of isotope 1 = 0.574

Mass of isotope 2 = 122.9042 amu

Percentage abundance of isotope 2 = 42.6%.

Fractional abundance of isotope 2 = 0.426

Putting values in equation 1, we get:

$\text{Average atomic mass of Z}=\sum[(120.9038 amu\times 0.574)+( 122.9042 amu\times 0.426)]$

$\text{Average atomic mass of Z}=121.7560 amu\approx 121.8 amu$

The atomic mass of this element is 121.8 amu.

The average atomic mass calculated is of an element named antimony that is Sb.