The abundance of three isotopes of oxygen are as follows :
% of `O^(16) = 90%`
% of `O^(17) + %` of `O^(18) = 10 %`
Assume at. Mass same as mass no. find `%` of `O^(17)`, if the isotopic mass is 16.12

Calculate the mass % of oxygen in Fe_(3)O_(4)

In chemical scale, the relative mass of the isotopic mixture of oxygen atoms (O^(16),O^(17),O^(18)) is assumed to be equal to

The abundance of three isotopes of oxygen (atomic mass 16.12) contains 8, 9, 10 neutrons respectively. One of the heaviest isotopes has 2% abundance. The other two are :

The percentage of oxygen in CH_(2) O is

The percentage of oxygen in CH_(2) O is

In chemical scale of relative mass of the isotopic mixture of oxygen atom ( O^16 , O^17 ,O^18 ) assumed to be equal to:

The relative abundance of three isotopes of carbon C^(12), C^(13) and C^14 are 98.892%, 1.108% and 2 xx 10^(-10) % respectively. If the relative atomic masses of these isotopes are 12.00, 13.00335 and 14.00317 amu respectively, find the average atomic mass of carbon.