To account for atomic mass of nitrogen is as 14.0067 , what should be the ratio of N-15 and N-14 atoms in natural nitrogen (atomic mass of N 14 is 14.00307u and N 15 is 15.001?

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Answer 1 · 2018-06-20T00:24:26

#"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 color(white)(l)=# the fraction of #"N-15"#. Then,
#1-x =# the fraction of #"N-14"#

and

#"14.003 07"(1-x) + 15.001x = 14.0067#

#"14.003 07 - 14.003 07"x + 15.001x = 14.0067#

#0.998x = 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#