Question

Indium has two naturally occurring isotopes: indium-113 and indium-115. The relative atomic mass of indium is 114.82. Calculate the natural abundance of each isotope.

Answer 1

To calculate the natural abundance of each isotope, set up a system of equations using the relative atomic mass and atomic masses of the isotopes. Solve the system of equations to find the values of x and y. The natural abundance of indium-113 is approximately 4.685% and the natural abundance of indium-115 is approximately 95.315%.

Step-by-step explanation:

To calculate the natural abundance of each isotope, we can set up a system of equations using the relative atomic mass of indium and the atomic masses of the isotopes. Let's assume the natural abundances of indium-113 and indium-115 are x and y respectively. The equation for the relative atomic mass is:

114.82 = (113 × x + 115y)/100

We also know that the sum of the natural abundances is 100%: x + y = 100

Now we can solve this system of equations to find the values of x and y. Solving for y in the second equation gives us: y = 100 - x. Substituting this into the first equation and simplifying gives us: 114.82 = (113 × x + 115 × (100 - x))/100. Solving this equation, we find x ≈ 4.685 and y ≈ 95.315. Therefore, the natural abundance of indium-113 is approximately 4.685% and the natural abundance of indium-115 is approximately 95.315%.

Answer 2

114.818 amu.x(112.9041) + (1 - x)

(114.9039) = 114.818Solving for xx=0.043 or 4.3%

Step-by-step explanation:

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) = 114.818Solving for x,x = 0.043 or 4.3%