How to find the empirical formula for an nitrogen oxygen compound from the given mass of nitrogen and the product?


> 
> $5.00~\mathrm{g}$ of nitrogen is completely converted into an oxide of nitrogen. The mass of the oxide formed is $19.3~\mathrm{g}$.  
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>  The empirical formula of the oxide would be?
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My working:


$$\ce{N2 + O\_2 -> N$\_x$O$\_y$}$$


* $m(\ce{N2}) = 5~\mathrm{g}$
* $n(\ce{N2}) = 5/28~\mathrm{mol}$
* $n(\ce{N$\_x$O$\_y$}) = 19.3~\mathrm{g}$


How do I approach this? 




$\ce{x/2 N2 +y/2 O2 \rightarrow N$\_x$O$\_y$}$


We have two unknowns $x$ and $y$. So, to determine them we need two independent equations:


1 - The conservation of nitrogen mass: $5=14x$. So, $x= 0.35714$.


2- According to the stoichiometry of the above balanced chemical equation: The number of moles of $\ce{ N$\_x$O$\_y$}$ equals $n= \frac{5}{28}\times\frac{2}{x}$.


On the other hand, the number of moles of $\ce{ N$\_x$O$\_y$}$ equals $n= \frac{19.3}{14x+16y}=\frac{10}{28x} $.
By substituting $x= 0.35714 $ into the last equation, we find $y=0.89375$. 
We find $\ce{ N\_{0.35714}O\_{0.89375}}$. We divide by $ 0.35714$, We obtain $\ce{ NO\_{2.5}}$.i.e. the formula of nitrogen oxide is $\ce{ N\_2O\_5}$. 


 2

If the question specifies that only one NxOy is formed (so not also NpOq in some ratio) , you need to calculate how many atoms of Oxygen you will have added. Then, you can deduce the number of Oxygen atoms per Nitrogen atom and find x and y. If you find decimal values, adjust the stoichiometry accordingly.


 1

$$\ce{N2 +O2 -> N\_{$x$}O\_{$y$}}$$
Now using POAC (Principle of Atomic conservation),
$$\frac5{28}\cdot2=\frac{19.3}{14x+16y}\cdot x\implies \frac{y}{x}\approx2.5025\sim2.5\implies y'=5,x'=2\implies \ce{N2O5}$$


 0

Subtract the mass of $\ce{N2}$ from the mass of NXOY to get 14.3g. 
This means there is 14.3g is from the $\ce{O2}$ so you can find the moles of $\ce{O2}$ and use the coefficients to solve for the ratio between $\ce{N2}$ and $\ce{O2}$ which will give you X and Y.


 0