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

The relative abundance of N-14 = 99.64 %.

The relative abundance of N-15 = 0.361 %.

Explanation:

∴ 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) = 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).

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.

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