Hydrogen Isotope PLEASE HELP :)

Question

Naturally occuring H has three isotopes. H-1 is 1.001 amu, 99.82% abund. H-3 is 2.930 amu 0.070% abund. What must the atom. Mass of the third isotope be to account for the 1.008 (1.00794) amu mass of H?

Answer 1

Set up the equation:

 

1.008 = (0.9982)(1.001) + (7x10^-4)(2.930) + (0.0011)(x)

 

At this point all you have to do is perform the corresponding multiplications and solve for x. I will leave this part of the problem up to you to solve.