To account for the atomic mass of nitrogen as $14.0067$ what should be the ratio of ${}^{15}\mathrm{N}$ and ${}^{14}\mathrm{N}$ atoms in natural nitrogen?

Step 1: Given

The atomic mass of Nitrogen $\left(\mathrm{N}\right)$ $=14.0067$

Molar Mass of ${}^{15}\mathrm{N}$ is $15.001$

Molar Mass of ${}^{14}\mathrm{N}$ is $14.003$

Let the fraction of ${}^{15}\mathrm{N}$ $=\mathrm{x}$

Then the fraction of ${}^{14}\mathrm{N}$ $=1-\mathrm{x}$

Step 2: Calculation of total atomic mass of Nitrogen

Now the total atomic mass of Nitrogen $=$$Fractionof{}^{15}N\times MolarMassof{}^{15}N+Fractionof{}^{14}N\times MolarMassof{}^{14}N$

$14.0067=x\times 15.001+\left(1-x\right)\times 14.003$

Solving the above equation for $x$,

$14.003–14.003\times x+15.001x=14.0067$

$0.998\times x=0.0036$

$x=0.0036$

Therefore $\left(1-x\right)=0.9964$

Step 3: Calculation of ratio of ${}^{15}\mathrm{N}$ and ${}^{14}\mathrm{N}$ atoms

The ratio of ${}^{15}\mathrm{N}$ and ${}^{14}\mathrm{N}$ atoms will be written as,

$\frac{\left(x\right)}{\left(1-x\right)}=\frac{0.0036}{0.9964}$

$\frac{{}^{15}\mathrm{N}}{{}^{14}\mathrm{N}}=0.0036$