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Question: Average atomic weight of an element M is\(51.7\). If two isotopes of M, \({M^{50}}\)and \({M^{52}}\)...

Average atomic weight of an element M is51.751.7. If two isotopes of M, M50{M^{50}}and M52{M^{52}}are present, then calculate the percentage of occurrence of M50{M^{50}}in nature.

Explanation

Solution

The average weight is defined as the sum of weights of all the isotopes multiplied by the percentage fraction of their presence in the atmosphere. Mathematically,
Average atomic weight = F1M1+F2M2+F3M3+.....+FnMn{F_1}{M_1} + {F_2}{M_2} + {F_3}{M_3} + ..... + {F_n}{M_n}
Where F1, F2, F3, Fn{F_1},{\text{ }}{F_2},{\text{ }}{F_3},{\text{ }}{F_n} correspond to percentage fraction
M1, M2, M3, Mn{M_1},{\text{ }}{{\text{M}}_2},{\text{ }}{{\text{M}}_3},{\text{ }}{{\text{M}}_n}correspond to molecular weights of isotopes
We have the average weight, thus we can find the percentage of one isotope.

Complete step by step solution:
First, let us see what isotopes are.
The isotopes are the atoms of the same element that have the same atomic number but different mass number. In nature the sum of total conc. of all the isotopes of an element is considered 100%100\% .
The average weight is the total sum of weights of all the isotopes multiplied by the percentage fraction of their presence. Mathematically, it can be written as -
Average atomic weight = F1M1+F2M2+F3M3+.....+FnMn{F_1}{M_1} + {F_2}{M_2} + {F_3}{M_3} + ..... + {F_n}{M_n}
Where F1, F2, F3, Fn{F_1},{\text{ }}{F_2},{\text{ }}{F_3},{\text{ }}{F_n} correspond to percentage fraction
M1, M2, M3, Mn{M_1},{\text{ }}{{\text{M}}_2},{\text{ }}{{\text{M}}_3},{\text{ }}{{\text{M}}_n}correspond to molecular weights of isotopes
We have average weight of M = 51.7
The two isotopes are M50{M^{50}}and M52{M^{52}}
Thus, we can write -
51.7=x×50+(1−x)×5251.7 = x \times 50 + (1 - x) \times 52
51.7=50x+52−52x51.7 = 50x + 52 - 52x
−2x=−0.3- 2x = - 0.3
x=0.32x = \frac{{0.3}}{2}
x=0.15x = 0.15
The value of x corresponds to the amount of M50{M^{50}}. So, the M50{M^{50}}is 0.150.15 in nature i.e. 15%15\% .

So, the abundance of M50{M^{50}}is 15%15\% .

Note: The weight that we generally write of the element is the average weight only which is calculated by addition of weights of all isotopes multiplied by their relative concentration w.r.t each other in the atmosphere.