A compound was found to contain nitrogen and oxygen in the ratio nitrogen 28 g and oxygen 80 g. The formula of the compound is:
(A) NO
(B) \[{N_2}{O_3}\]
(C) ${N_2}{O_5}$
(D) ${N_2}{O_4}$
Answer
620.7k+ views
Hint: First find out the number of moles of nitrogen and oxygen involved by dividing the given weight of the elements with their atomic weight respectively and then see their ratio to each other. This ratio will be the stoichiometric ratio of these elements to each other in the compound.
Complete answer:
-In any molecule the components are always present in a particular stoichiometry for that compound whatever the weight of the compound be.
-First we will find out the number of moles of nitrogen and oxygen we are talking about.
The questions says that the weight of nitrogen is 28 g and that of oxygen is 80 g. We know that the atomic weight of nitrogen is 14 g/mol and oxygen is 16 g/mol. Let us now find their moles.
For nitrogen: $moles = \dfrac{{weight}}{{at.wt.}}$
= $\dfrac{{28}}{{14}}$ = 2 moles
For oxygen: $moles = \dfrac{{80}}{{16}}$ = 5 moles
-So, now we can find out the ratio of moles of nitrogen and oxygen.
Moles of N : Moles of O
2 : 5
So, the formula would be: ${N_2}{O_5}$ .
It is known as nitrogen pentoxide or nitric anhydride and is a binary compound of nitrogen and oxygen. It can adopt two different structures depending on the physical conditions. In solid state it is found as a salt, nitronium nitrate and contains two separate ions: nitronium cation ($NO_2^ + $) and nitrate anions ($NO_3^ - $). But when it is present in gas phase and under some other conditions it exists as a covalently bonded molecule. But this gaseous ${N_2}{O_5}$ can absorb ultraviolet light and dissociate into radicals nitrogen dioxide ($N{O_2}$) and nitrogen trioxide ($N{O_3}$).
Both the nitrogen atoms here have oxidation state of (+5). It is a strong oxidising agent and on decomposition releases a highly toxic nitrogen dioxide gas.
The correct option is: (C) ${N_2}{O_5}$
Note:
Always remember that while calculating composition of such binary compounds use their atomic weight and not molecular weight while finding the moles of the constituent elements.
Complete answer:
-In any molecule the components are always present in a particular stoichiometry for that compound whatever the weight of the compound be.
-First we will find out the number of moles of nitrogen and oxygen we are talking about.
The questions says that the weight of nitrogen is 28 g and that of oxygen is 80 g. We know that the atomic weight of nitrogen is 14 g/mol and oxygen is 16 g/mol. Let us now find their moles.
For nitrogen: $moles = \dfrac{{weight}}{{at.wt.}}$
= $\dfrac{{28}}{{14}}$ = 2 moles
For oxygen: $moles = \dfrac{{80}}{{16}}$ = 5 moles
-So, now we can find out the ratio of moles of nitrogen and oxygen.
Moles of N : Moles of O
2 : 5
So, the formula would be: ${N_2}{O_5}$ .
It is known as nitrogen pentoxide or nitric anhydride and is a binary compound of nitrogen and oxygen. It can adopt two different structures depending on the physical conditions. In solid state it is found as a salt, nitronium nitrate and contains two separate ions: nitronium cation ($NO_2^ + $) and nitrate anions ($NO_3^ - $). But when it is present in gas phase and under some other conditions it exists as a covalently bonded molecule. But this gaseous ${N_2}{O_5}$ can absorb ultraviolet light and dissociate into radicals nitrogen dioxide ($N{O_2}$) and nitrogen trioxide ($N{O_3}$).
Both the nitrogen atoms here have oxidation state of (+5). It is a strong oxidising agent and on decomposition releases a highly toxic nitrogen dioxide gas.
The correct option is: (C) ${N_2}{O_5}$
Note:
Always remember that while calculating composition of such binary compounds use their atomic weight and not molecular weight while finding the moles of the constituent elements.
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