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Indium has only two naturally occurring isotopes. the mass of indium-113 is 112.9041 amu and the mass of indium-115 is 114.9039 amu . part a use the atomic mass of indium to calculate the relative abundance of indium-113.

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meerkat18
The average atomic mass written in the periodic table is determined from the relative abundances of the element's isotopes. The equation would be:

Average Atomic Mass = ∑(Relative Abundance×Mass)

Let be the relative abundance of Indium-113. Because there are only 2 isotopes, their relative abundances should equal to 1, such that the relative abundance for Indium-15 is (1-x). The atomic mass of indium is 114.818 amu.

x(112.9041) + (1 - x)(114.9039) = 114.818
Solving for x,
x = 0.043 or 4.3%
Eduard22sly

The relative abundance of indium–113 is 4.28%

Isotopy is a phenomenon where by two or more atoms have the same atomic number but different mass number due to the difference in the neutron number of the atoms involved.

We can obtain the abundance of indium–113 as illustrated below:

Let indium–113 be isotope A

Let indium–115 be isotope B

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

isotope A (indium–113):

Mass of A = 112.9041 amu

Abundance of A (A%) =?

Isotope B (indium–115):

Mass of B = 114.9039 amu

Abundance of B (B%) = 100 – A%

Atomic mass of indium = 114.818 amu

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

114.818 = [(112.9041 × A%)/100] + [(114.9039 × (100 – A%)/100]

114.818 = 1.129041A% + 114.9039 – 1.149039A%

Collect like terms

114.818 – 114.9039 = 1.129041A%  – 1.149039A%

–0.0856 = –0.019998A%

Divide both side by –0.019998

A% = –0.0856 / –0.019998

A% = 4.28%

Therefore, the relative abundance of indium–113 is 4.28%

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