Theory:

Atomic masses of some elements:
 
Atomic number
Name
Symbol
Atomic mass
\(1\)
Hydrogen
\(H\)
\(1.008\)
\(2\)
Helium
\(He\)
\(4.003\)
\(3\)
Lithium
\(Li\)
\(6.941\)
\(4\)
Beryllium
\(Be\)
\(9.012\)
\(5\)
Boron
\(B\)
\(10.811\)
 
How to calculate the average atomic mass?
question.JPG
 
Let us understand this with the help of the following examples.
  
Example: 1
  
Oxygen is the most abundant element in both the Earth’s crust and the human body. In nature, it exists as a combination of three stable isotopes, as shown in the table:
 
Isotopes of oxygen:
 
Isotope
Mass
% abundance
\(_{8}\textrm{O}^{16}\)
\(15.9949\)
\(99.757\)
\(_{8}\textrm{O}^{17}\)
\(16.9991\)
\(0.038\)
\(_{8}\textrm{O}^{18}\)
\(17.9992\)
\(0.205\)
 
Atomicmassofoxygen=15.9949×0.99757+16.9991×0.00038+17.9992×0.00205=15.999amu
 
Hence, the atomic mass of oxygen is \(15.999\) \(amu\).
 
Example: 2
  
Boron naturally found as a mixture of boron-\(10\) (\(5\) protons + \(5\) neutrons) and boron-\(11\) (\(5\) protons + \(6\) neutrons) isotopes. The percentage abundance of boron-\(10\) is \(20\), and that of boron-\(11\) is \(80\). The atomic mass of boron is then determined as follows:
 
Atomicmassofboron=10×20100+11×80100=10×0.2+11×0.8=2+8.8=10.8amu
 
Hence, the atomic mass of boron is \(10.8\) \(amu\).