Question

Chromium has an atomic mass of 51.9961 u and consists of four isotopes, 50 Cr , 52 Cr , 53 Cr , and 54 Cr . The 52 Cr isotope has a natural abundance of 83.79 % and an atomic mass of 51.9405 u. The 54 Cr isotope has a natural abundance of 2.37 % and an atomic mass of 53.9389 u. The natural abundances of the 50 Cr and 53 Cr isotopes exist in a ratio of 0.4579 : 1 , and the 50 Cr isotope has an atomic mass of 49.9460 u. Determine the atomic mass of the 53 Cr isotope. atomic mass of 53 Cr:______.

Answer 1

Answer : The atomic mass of
`_(24)^(53)\textrm{Cr}` isotope is 52.8367 amu

Explanation :

We know that:

Total percentage abundance of the isotope = 100 %

Percentage abundance of
`_(24)^(50)\text{Cr}\text{ and }_(24)^(53)\textrm{Cr}\text{ isotopes}=[100-(83.79+2.37)]=13.84\%`

We are given:

Ratio of
`_(24)^(50)\textrm{Cr}\text{ and }_(24)^(53)\textrm{Cr}` isotopes = 0.4579 : 1

Percentage abundance of
`_(24)^(50)\textrm{Cr}` isotope =
`(0.4579)/((0.4579+1))* 13.84\%=4.37\%`

Percentage abundance of
`_(24)^(53)\textrm{Cr}` isotope =
`(1)/((0.4579+1))* 13.84\%=9.49\%`

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 mass of
`_(24)^(53)\textrm{Cr}` isotope be 'x'

For
`_(24)^(50)\textrm{Cr}` isotope:

Mass of
`_(24)^(50)\textrm{Cr}` isotope = 49.9460 amu

Percentage abundance of
`_(24)^(50)\textrm{Cr}` = 4.37 %

Fractional abundance of
`_(24)^(50)\textrm{Cr}` isotope = 0.0437

For
`_(24)^(52)\textrm{Cr}` isotope:

Mass of
`_(24)^(52)\textrm{Cr}` isotope = 51.9405 amu

Percentage abundance of
`_(24)^(52)\textrm{Cr}` isotope = 83.79 %

Fractional abundance of
`_(24)^(52)\textrm{Cr}` isotope = 0.8379

or
`_(24)^(53)\textrm{Cr}` isotope:

Mass of
`_(24)^(53)\textrm{Cr}` isotope = x amu

Percentage abundance of
`_(24)^(53)\textrm{Cr}` isotope = 9.49 %

Fractional abundance of
`_(24)^(53)\textrm{Cr}` isotope = 0.0949

For
`_(24)^(54)\textrm{Cr}` isotope:

Mass of
`_(24)^(54)\textrm{Cr}` isotope = 53.9389 amu

Percentage abundance of
`_(24)^(54)\textrm{Cr}` isotope = 2.37 %

Fractional abundance of
`_(24)^(54)\textrm{Cr}` isotope = 0.0237

Average atomic mass of chromium = 51.9961 amu

Putting values in equation 1, we get:

`51.9961=[(49.9460* 0.0437)+(51.9405* 0.8379)+(x* 0.0949)+(53.9389* 0.0237)]\\\\x=52.8367amu`

Hence, the atomic mass of
`_(24)^(53)\textrm{Cr}` isotope is 52.8367 amu