Q3.39P - Chemistry: Molecular Nature Of Matter And Change

Question

An oxide of nitrogen contains 30.45 mass %. 

(a) What is the empirical formula of the oxide? 

(b) If the molar mass is $\mathbf{90}\mathbf{\pm }\mathbf{5}\mathbf{g}\mathbf{/}\mathbf{mol}\mathbf{,}$ what is the molecular formula?

Step-by-Step Solution

Answer

1. The empirical formula is ${\mathbf{NO}}_{\mathbf{2}}$
2. The molecular formula is ${\mathbf{N}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{4}}$

Step 1: Determine the concept of the whole-number multiple

Empirical formula is given by the simplest number ratio of the number of different atoms present in the compound. 

The molecular formula gives the actual number of different atoms present in the same molecule. 

The whole number multiple given as the ratio of molar mass to the empirical formula mass of each element deduces the molecular formula.

The whole number multiple is given as,

$\mathrm{whole}-\mathrm{number}\mathrm{multiple}=\frac{\mathrm{molar}\mathrm{mass}(\mathrm{g}/\mathrm{mol})}{\mathrm{empirical}\mathrm{formula}\mathrm{mass}(\mathrm{g}/\mathrm{mol})}$

Step 2: Determine the empirical formula of the oxide

The mass fraction of the N atom is:

$\mathbf{\text{N=0.3045}}$

The mass fraction of the O atom is calculated as:

$\mathbf{\text{O=1-0.3045=0.6955}}$

Then, the ratio of the number of moles of N and O atoms is:

$\begin{array}{l}\mathrm{n}\left(\mathrm{N}\right):\mathrm{n}\left(\mathrm{O}\right)=\frac{\frac{0.3045}{14.01}}{\frac{0.6955}{16}}\\ \mathrm{n}\left(\mathrm{N}\right):\mathrm{n}\left(\mathrm{O}\right)=0.0217:0.0435\\ \mathrm{n}\left(\mathrm{N}\right):\mathrm{n}\left(\mathrm{O}\right)=1:2\end{array}$

Thus, the empirical formula is ${\mathbf{NO}}_{\mathbf{2}}$

Step 3: Determine the molecular formula of the compound

The empirical formula is$\mathbf{ }{\mathbf{NO}}_{\mathbf{2}}$

The empirical formula mass is calculated as:

$\begin{array}{c}{\mathrm{M}}_{\mathrm{w}}\mathrm{of}\mathrm{empirical}\mathrm{formula}={\mathrm{M}}_{\mathrm{w}}\mathrm{of}\mathrm{N}+2\times \left({\mathrm{M}}_{\mathrm{w}}\mathrm{of}\mathrm{O}\right)\\ =14.01\mathrm{g}/\mathrm{mol}+2\times \left(16.00\mathrm{g}/\mathrm{mol}\right)\\ =46.01\mathrm{g}/\mathrm{mol}\end{array}$

The molecular formula mass = 90 g/mol

The whole-number ratio is:

$\mathrm{whole}-\mathrm{number}\mathrm{multiple}=\frac{90(\mathrm{g}/\mathrm{mol})}{46.01(\mathrm{g}/\mathrm{mol})}\approx 2$

Multiplying subscripts of ${\mathbf{NO}}_{\mathbf{2}}$ by 2, the molecular formula is given by

$\mathbf{=}{\mathbf{N}}_{\mathbf{(}\mathbf{1}\mathbf{\times }\mathbf{2}\mathbf{)}}\mathbf{\times }{\mathbf{O}}_{\mathbf{(}\mathbf{2}\mathbf{\times }\mathbf{2}\mathbf{)}}$

Thus, the molecular formula is ${\mathbf{N}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{4}}$