Question

An atom has the following three isotopes: 24 amu (percent abundance = 78.99%)25 amu (percent abundance =10.00\%) , and 26 amu (percent bundance=11.01\%) . Calculate the average atomic mass. What is the identity of this atom based on your calculation?

Answer 1

We multiply each mass with its percentage.
24*(78.99/100) = 18.96
25*(10/100)= 2.5
26*(11.01/100)= 2.86
Then add these together.
18.96+2.5+2.86= 24.32amu
An element that has an average mass of nearly 24.32amu would be Mg (magnesium)

Answer 2

1. The average atomic mass of the element is 24.32 amu

2. The identity of the atom is Magnesium (Mg)

Let the 1st isotope be A

Let the 2nd isotope be B

Let the 3rd isotope be C

From the question given above, the following data were obtained:

For Isotope A

Mass of A = 24 amu

Abundance of A (A%) = 78.99%

For Isotope B

Mass of B = 25 amu

Abundance of B (B%) = 10%

For Isotope C

Mass of C = 26 amu

Abundance of C (C%) = 11.01%

Average atomic mass =?

The average atomic mass of the atom can be obtained as follow:

Atomic mass = [(Mass of A × A%)/100] + [(Mass of B × B%)/100] + [(Mass of C × C%)/100]

= [(24 × 78.99)/100] + [(25 × 10)/100] + [(26 × 11.01)/100]

= 18.9576 + 2.5 + 2.8626

= 24.32 amu

Therefore, the atomic mass of the atom is 24.32 amu.

From the periodic table, the element having atomic mass of 24.32 amu is Magnesium (Mg).

Therefore, the identity of the atom is Magnesium (Mg)

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