# 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?

Jun 20, 2018

$\text{N-15":"N-14} = 0.0036 : 1$

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 $x \textcolor{w h i t e}{l} =$ the fraction of $\text{N-15}$. Then,
$1 - x =$ the fraction of $\text{N-14}$

and

$\text{14.003 07} \left(1 - x\right) + 15.001 x = 14.0067$

$\text{14.003 07 - 14.003 07} x + 15.001 x = 14.0067$

$0.998 x = 0.0036$

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

$1 - x = 1 - 0.0036 = 0.9964$

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