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

$N-15 : N-14 = 0.0036 : 1$ $"N-15":"N-14" = 0.0036:1$

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

and

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

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

$0.998 x = 0.0036$ $0.998x = 0.0036$

$x = \frac{0.0036}{0.998} = 0.0036$ $x = 0.0036/0.998 = 0.0036$

$1 - x = 1 - 0.0036 = 0.9964$ $1-x = 1 - 0.0036 = 0.9964$

$\frac{N-15}{N-14} = \frac{x}{1 - x} = \frac{0.0036}{0.9964} = 0.0036$ $"N-15"/"N-14" = x/(1-x) = 0.0036/0.9964 = 0.0036$