Question
Question: If bromine atom is available in the form of, say 2 isotopes \(_{35}B{r^{79}}\) (49.7%) and \(_{35}B{...
If bromine atom is available in the form of, say 2 isotopes 35Br79 (49.7%) and 35Br81 (50.3%), calculate the average atomic mass of bromine atom.
Solution
The average atomic weight of any element is the sum of the masses of the isotopes multiplies with their respective percentage abundances.
Formula used:
-Average atomic weight:
=i=1∑n(mass(i))(abundance(i)) (1)
Where, mass = atomic weight of 1 isotope
Abundance = percentage abundance in which the respective isotope is present.
Complete answer:
-Isotopes are those atoms of a chemical element which have the same atomic number, same position in the periodic table and nearly the same chemical properties but they differ in the mass number and physical properties.
The elements have the same number of electrons and protons but it is the number of neutrons that differs.
Examples: -carbon has 3 isotopes: C-12, C-13, C-14
-hydrogen has 3 isotopes: H-1, D-2, T-3
-bromine has 2 stable isotopes: Br-79, Br-81
-The question gives us the abundance of 2 isotopes of Br: 35Br79 = 49.7% and 35Br81 = 50.3%
We use equation (1) to calculate the average atomic mass of Br atom:
Average atomic weight=i=1∑n(mass(i))(abundance(i))
=\left\\{ {79 \times \frac{{49.7}}{{100}}} \right\\} + \left\\{ {81 \times \frac{{50.3}}{{100}}} \right\\}
= 39.263 + 40.743
= 80.006
So, the average atomic weight of Br is 80.006 amu.
Note: Isotopes are those which have the same atomic number but different mass number. Isobars are those atoms which have the same mass number but different atomic number. Isotones are those which have the same neutron number but different proton number. It should not be confused with each other.