# 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 $19.3~\mathrm{g}$.
The empirical formula of the oxide would be?

My working:

$$\ce{N2 + O_2 -> N_xO_y}$$

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

How do I approach this?

• I don't see an explicit question, except for "How do I approach this?" I guess you want to determine x and y, right? – Philipp Nov 7 '14 at 11:26
• yes that is right. – confused Nov 7 '14 at 20:33
• Is it just me? I would have subtracted 5 g from 19.3 g to get 14.3 g of oxygen used. Find the number of atoms of oxygen and divide by the number of nitrogen atoms to get the ratio. – LDC3 Nov 8 '14 at 5:08

$\ce{x/2 N2 +y/2 O2 \rightarrow N$_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 $\ce{ N_2O_5}$.

• why did you write the equations with x/2 and y/2 I love the algebraic approach though - using 14x + 16y=19.3 – confused Nov 7 '14 at 20:31
• We have to balance the chemical equation. – Yomen Atassi Nov 8 '14 at 5:59

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.

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

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.