91Ó°ÊÓ

Consider that the total mass of the compoundis 100g.

Given that the mass of nitrogen is25.9g.

So, the mass of oxygen will be $\text{=100-25.4=}\text{74.1g}$.

Moles of nitrogen $\text{=}\frac{\text{³¾²¹²õ²õ´Ç´Ú n¾±³Ù´Ç²µ±ð²Ô}}{\text{³¾´Ç±ð±ô³¦³Ü±ô²¹°ù″¾²¹²õ²õ o´Ú n¾±³Ù´Ç²µ±ð²Ô}}$

Moles of oxygen $\text{=}\frac{\text{³¾²¹²õ²õ o´Ú o³æ²â²µ±ð²Ô}}{\text{moelcular ³¾²¹²õ²õ o´Ú o³æ²â²µ±ð²Ô}}$

The ratio of nitrogen and oxygen is the empirical formula of the compound, i.e.,${\mathbf{N}}_1{\mathbf{O}}_{\mathbf{2}.\mathbf{5}}$.

Now, the empirical mass of nitrogen oxide .

$\begin{array}{c}\text{Empiricalmass=³¾´Ç±ô±ð³¦³Ü±ô²¹°ù″¾²¹²õ²õ of nitrogen + 2.5× molecular ³¾²¹²õ²õ o´Ú o³æ²â²µ±ð²Ô}\\ \text{=14²µ/³¾´Ç±ô + 2×16²µ/³¾´Ç±ô}\\ \text{=54g/mol}\end{array}$

$\begin{array}{c}\text{Empiricalformulaunits=}\frac{\text{³¾´Ç±ô±ð³¦³Ü±ô²¹°ù″¾²¹²õ²õ}}{\text{±ð³¾±è¾±°ù¾±³¦²¹±ô â¶Ä‹m²¹²õ²õ}}\\ \text{=}\frac{\text{108}}{\text{54}}\\ \text{=​â¶Ä‰2²µ/³¾´Ç±ô}\end{array}$

Therefore, the molecular formula of compound is ${\text{N}}_{\text{2}}{\text{O}}_{\text{5}}$.

The structure in which a line shows the bonding between the atoms and a dot indicates lone pair of electrons is known as the Lewis dot structure.