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

Question

Deandre Cervantes

Chemistry

8 January, 12:06

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

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Answers (1)

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Blaise · Answer 1

14.434 r. a. m.

Explanation:

The atomic mass of an element is a weighted average of its isotopes in which the sum of the abundance of each isotope is equal to 1 or 100%.

∵ 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.