Respuesta :
Answer : The atomic mass of [tex]_{24}^{53}\textrm{Cr}[/tex] isotope is 52.8367 amu
Explanation :
We know that:
Total percentage abundance of the isotope = 100 %
Percentage abundance of [tex]_{24}^{50}\text{Cr}\text{ and }_{24}^{53}\textrm{Cr}\text{ isotopes}=[100-(83.79+2.37)]=13.84\%[/tex]
We are given:
Ratio of [tex]_{24}^{50}\textrm{Cr}\text{ and }_{24}^{53}\textrm{Cr}[/tex] isotopes = 0.4579 : 1
Percentage abundance of [tex]_{24}^{50}\textrm{Cr}[/tex] isotope = [tex]\frac{0.4579}{(0.4579+1)}\times 13.84\%=4.37\%[/tex]
Percentage abundance of [tex]_{24}^{53}\textrm{Cr}[/tex] isotope = [tex]\frac{1}{(0.4579+1)}\times 13.84\%=9.49\%[/tex]
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:[tex]\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i[/tex] .....(1)
Let the mass of [tex]_{24}^{53}\textrm{Cr}[/tex] isotope be 'x'
For [tex]_{24}^{50}\textrm{Cr}[/tex] isotope:
Mass of [tex]_{24}^{50}\textrm{Cr}[/tex] isotope = 49.9460 amu
Percentage abundance of [tex]_{24}^{50}\textrm{Cr}[/tex] = 4.37 %
Fractional abundance of [tex]_{24}^{50}\textrm{Cr}[/tex] isotope = 0.0437
For [tex]_{24}^{52}\textrm{Cr}[/tex] isotope:
Mass of [tex]_{24}^{52}\textrm{Cr}[/tex] isotope = 51.9405 amu
Percentage abundance of [tex]_{24}^{52}\textrm{Cr}[/tex] isotope = 83.79 %
Fractional abundance of [tex]_{24}^{52}\textrm{Cr}[/tex] isotope = 0.8379
or [tex]_{24}^{53}\textrm{Cr}[/tex] isotope:
Mass of [tex]_{24}^{53}\textrm{Cr}[/tex] isotope = x amu
Percentage abundance of [tex]_{24}^{53}\textrm{Cr}[/tex] isotope = 9.49 %
Fractional abundance of [tex]_{24}^{53}\textrm{Cr}[/tex] isotope = 0.0949
For [tex]_{24}^{54}\textrm{Cr}[/tex] isotope:
Mass of [tex]_{24}^{54}\textrm{Cr}[/tex] isotope = 53.9389 amu
Percentage abundance of [tex]_{24}^{54}\textrm{Cr}[/tex] isotope = 2.37 %
Fractional abundance of [tex]_{24}^{54}\textrm{Cr}[/tex] isotope = 0.0237
Average atomic mass of chromium = 51.9961 amu
Putting values in equation 1, we get:
[tex]51.9961=[(49.9460\times 0.0437)+(51.9405\times 0.8379)+(x\times 0.0949)+(53.9389\times 0.0237)]\\\\x=52.8367amu[/tex]
Hence, the atomic mass of [tex]_{24}^{53}\textrm{Cr}[/tex] isotope is 52.8367 amu