Question

*An element has three naturally occuring isotopes with the following masses and abundances

Isotopic mass(amu):
27.977
28.976
29. 974
Fractional abundance:
0.9221
0.0470
0.0309

a. calculate the atomic weight of this element
b.What is the identity of the element?​

Answer 1

a. 28.08 amu

b. Silicon

Step-by-step explanation:

Step 1: Given data

Isotope Isotopic mass (mi) Fractional abundance (abi)

²⁸X 27.977 amu 0.9221

²⁹X 28.976 amu 0.0470

³⁰X 29. 974 amu 0.0309

Step 2: Calculate the atomic weight (aw) of this element

We will use the following expression.

aw = ∑mi × abi

aw = 27.977 amu × 0.9221 + 28.976 amu × 0.0470 + 29. 974 amu × 0.0309

aw = 28.08 amu

Looking at the periodic table, the element with atomic weight = 28.08 amu is silicon.