3) Give the relative abundance of the two isotopes of nitrogen when N-14 hos o 14.0031 amu and N-15 has a mass of 15.0001 amu

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Chemistry

Skyler Parker

16 November, 16:04

3) Give the relative abundance of the two isotopes of nitrogen when N-14 hos o 14.0031 amu and N-15 has a mass of 15.0001 amu

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Selene · Answer 1

The relative abundance of N-14 = 99.64 %.

The relative abundance of N-15 = 0.361 %.

Explanation:

The average atomic mass of an element can be calculated as the summation of the atomic mass units of each isotope multiplied by its (abundance % / 100).

∴ The average atomic mass of nitrogen (N) = [ (atomic mass of N-14) (Abundance / 100) + (atomic mass of N-15) (Abundance / 100).

The average atomic mass of nitrogen (N) = [ (atomic mass of N-14) (x) ] + (atomic mass of N-15) (1-x) ].

The average atomic mass of nitrogen (N) = 14.0067 amu.

Atomic mass of N-14 = 14.0031 amu.

x is (the abundance of N-14 / 100).

Atomic mass of N-15 = 15.0001 amu.

(1-x) is (the abundance of N-15 / 100).

The average atomic mass of nitrogen (N) = [ (atomic mass of N-14) (x) ] + (atomic mass of N-15) (1-x) ].

14.0067 amu = [ (14.0031 amu) (x) ] + [ (15.0001 amu) (1-x) ]

14.0067 = 14.0031 x + 15.0001 - 15.0001 x.

(14.0067) - (15.0001) = (14.0031 x) - (15.0001 x).

- 0.9934 = - 0.997 x.

x = ( - 0.9934) / ( - 0.997) = 0.9964.

x is the abundance of N-14 / 100 = 0.9964. The relative abundance of N-14 = 99.64 %. The abundance of N-15 / 100 = (1-x) = 1 - 0.9963 = 3.61 x 10⁻³. The relative abundance of N-15 = 0.361 %.