Question

The element iridium exists in nature as two isotopes: 191Ir has a mass of 190.9606 u, and 193Ir has a mass of 192.9629 u. The average atomic mass of iridium is 192.22 u. Calculate the relative abundance (as percentages) of the two iridium isotopes.

Answer 1

To calculate the relative abundance of the isotopes of iridium, we set up equations based on their isotopic masses and the average atomic mass. We solve this system to find the relative abundances in decimal form, which can then be converted to percentages.

Step-by-step explanation:

To calculate the relative abundance of the two isotopes of iridium, we can set up a system of equations using the given atomic masses and the average atomic mass of iridium. Let x represent the abundance of 191Ir and y represent the abundance of 193Ir. We know that x + y = 1 because the total abundance is 100%, and x and y will be in decimal form (which can later be converted to percentages). To get the weighted average, we use the formula average atomic mass = (mass of isotope 1 × relative abundance of isotope 1) + (mass of isotope 2 × relative abundance of isotope 2). For iridium, this is 192.22 u = (190.9606 u × x) + (192.9629 u × y).

With the two equations:

x + y = 1

192.22 = 190.9606x + 192.9629y

We solve this system to find the values of x and y. After doing the calculations (which involve multiplying the second equation by 100 to avoid dealing with decimals and subtracting the first equation from the second), we can find the relative abundances of 191Ir and 193Ir.

Answer 2

Answer: The percentage abundance of
`_(77)^(191)\textrm{Ir}` and
`_(77)^(193)\textrm{Ir}` isotopes are 37.10% and 62.90% respectively.

Step-by-step 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* \text{(Fractional abundance})_i` .....(1)

Let the fractional abundance of
`_(77)^(191)\textrm{Ir}` isotope be 'x'. So, fractional abundance of
`_(77)^(193)\textrm{Ir}` isotope will be '1 - x'

• For
  `_(77)^(191)\textrm{Ir}` isotope:

Mass of
`_(77)^(191)\textrm{Ir}` isotope = 190.9606 amu

Fractional abundance of
`_(77)^(191)\textrm{Ir}` isotope = x

• For
  `_(77)^(193)\textrm{Ir}` isotope:

Mass of
`_(77)^(193)\textrm{Ir}` isotope = 192.9629 amu

Fractional abundance of
`_(77)^(193)\textrm{Ir}` isotope = 1 - x

Average atomic mass of iridium = 192.22 amu

Putting values in equation 1, we get:

`192.22=[(190.9606* x)+(192.9629* (1-x))]\\\\x=0.3710`

Percentage abundance of
`_(77)^(191)\textrm{Ir}` isotope =
`0.3710* 100=37.10\%`

Percentage abundance of
`_(77)^(193)\textrm{Ir}` isotope =
`(1-0.3710)=0.6290* 100=62.90\%`

Hence, the percentage abundance of
`_(77)^(191)\textrm{Ir}` and
`_(77)^(193)\textrm{Ir}` isotopes are 37.10% and 62.90% respectively.