Question

How would you convert the empirical formula $N{{O}_{2}}$ into the molecular formula, given the molar mass of this molecule is $\text{92g/mol}$?

Answer 1

The molecular formula of the compound is equal to the n units of the empirical formulae.
$\text{Molecular}\,\text{formulae=n }\\!\\!\times\\!\\!\text{ empirical}\,\text{formulae}$
The molar mass of N is $\text{14g/mol}$ and of O is $\text{16g/mol}$

Complete step by step answer:
- So in the question, it is asked how will we convert the empirical formulae $N{{O}_{2}}$ into the molecular formulae, if we are provided with the molecular mass of the compound.
- So first let’s understand the difference between molecular formula and empirical formulae which will help us to understand the question more clearly.
- Empirical formulae is the formula which gives the simplest whole number ratio of the various atoms present in the compound whereas the molecular formula gives exactly the number of atoms and different kinds of atoms which have combined to give the compound.
For better clarity we may take an example of a brick and a wall. To build the wall we need a few numbers of bricks, so here the empirical formula resembles the brick that is required to build the whole wall,a small building block and the molecular formulae resembles the complete wall.
- Therefore we can conclude that like few units of building blocks generate the whole wall, we will get the molecular formula of a compound depending on the number of units of empirical formulae required.
Hence we can write an equation relating the empirical and molecular formulae as,
$\text{Molecular}\,\text{formulae=n }\\!\\!\times\\!\\!\text{ empirical}\,\text{formulae}$
n is the number of empirical formula units required to generate the molecular formulae.
- Now in the question the molecular mass of the compound is given,hence we will find the value of n by the equation.
${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}}$ $\text{n=}\dfrac{\text{Molecular}\,\text{mass}}{\text{Empirical}\,\text{mass}}$
So now we know the molecular mass of the compound which is $\text{92g/mol}$ and have to calculate the empirical mass for the empirical formulae $N{{O}_{2}}$.
The atomic mass of N is $\text{14g/mol}$ and the atomic mass of O is $\text{16g/mol}$.Since in the formulae there is two O atom, the atomic mass should be added twice or multiplied with 2.
$\text{Empirical}\,\text{mass}\,\text{of}\,\text{N}{{\text{O}}_{\text{2}}}\text{=14+2}\left( \text{16} \right)\text{=46g/mol}$
- Let’s substitute the obtained value for the calculation of n, in the above given equation.
$\text{n=}\dfrac{92}{46}=2$
- Hence the value of n is 2 and the value has to be multiplied with the empirical formula to obtain the molecular formulae.
$\text{Molecular}\,\text{formulae = 2 }\\!\\!\times\\!\\!\text{ N}{{\text{O}}_{\text{2}}}\text{=}{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}}$
Therefore, the correct molecular formula for the given empirical formula is ${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}}$, which is dinitrogen tetroxide

Note: Empirical formula is useful for only the primary analysis ,when there is no data available for us.Empirical formulae suggest the simplest whole ratio in which the chemical combinations of the atoms may occur.It is just a theoretical approach which helps us in data analysis and in the prediction of the compounds that has formed.