Question

Naturally occuring boron is 80.20% boron-11 (atomic mass 11.01 amu) and 19.80% of some other isotope. What must the atomic mass of this second isotope be in order to account for the 10.81 amu average atomic mass of boron?.

Answer 1

You will have to do some math. 

80.2 × 11.01 + 19.8 × x = 100 * 10.81. 

x is the mass in amu of the second isotope. 

Solve for x. 

19.8 × x = 100 * 10.81 - 80.2 × 11.01 
19.8 × x = 1081 - 883.00 = 198.00 
x = 1964.00 / 19.8 = 10.00 

The mass of the other isotope is 10.00 amu.