Question

To account for atomic mass of nitrogen as 14.00674.what sholud be the ratio of 15N and 14N atoms in natural nitrogen

Answer 1

n-15:n-14=0.0036:1

The relative atomic mass is a weighted average of the individual atomic masses.

That is, we multiply each isotopic mass by its relative importance (percent or fraction of the mixture).

Let

x

l

=

the fraction of

N-15

. Then,

1

−

x

=

the fraction of

N-14

and

14.003 07

(

1

−

x

)

+

15.001

x

=

14.0067

14.003 07 - 14.003 07

x

+

15.001

x

=

14.0067

0.998

x

=

0.0036

x

=

0.0036

0.998

=

0.0036

1

−

x

=

1

−

0.0036

=

0.9964

N-15

N-14

=

x

1

−

x

=

0.0036

0.9964

=

0.0036

Answer 2

....... Hope this helps you...