If nitrogen (n) has 2 naturally occurring isotopes, nitrogen-14 (78.3%) and nitrogen-16 (21.7%), what is its average r. a.m.?

14.434 r.a.m.

Explanation:

∵ The atomic mass of N = ∑(atomic mass of each isotope)(its abundance)

∴ The atomic mass of N = (atomic mass of N-14)(abundance of N-14) + (atomic mass of N-16)(abundance of N-16)

atomic mass of N-14 = 14.0 r.a.m, abundance of N-14 = percent of N-14/100 = 78.3/100 = 0.783.

atomic mass of N-16 = 16.0 r.a.m, abundance of N-16 = percent of N-16/100 = 21.7/100 = 0.217.

∴ The atomic mass of N = (atomic mass of N-14)(abundance of N-14) + (atomic mass of N-16)(abundance of N-16) = (14.0 r.a.m)(0.783) + (16.0 r.a.m)(0.217) = 14.434 r.a.m.

1 is the answer to this problem