Q3.38P - Chemistry: Molecular Nature Of Matter And Change

Question

Find the empirical formula of the following compounds:

(a) 0.039 mol of iron atoms combined with 0.052 mol of oxygen atoms

(b) 0.903 g of phosphorus combined with 6.99 g of bromine

(c) A hydrocarbon with 79.9 mass % carbon

Step-by-Step Solution

Answer

The empirical formula of the compounds is:

(a) ${\mathrm{Fe}}_3{\mathrm{O}}_4$

(b) ${\mathrm{PBr}}_3$

(c) ${\mathrm{CH}}_3$

Step 1: Introduction of empirical formula

The chemical formula of a compound that provides the proportions (ratios) of the elements present but not the exact numbers or arrangement of atoms is known as an empirical formula. This is the compound's lowest whole-number ratio.

Step 2: Determine the empirical formula of 0.039 mol of Fe atoms combined with 0.052 mol of O atoms

The ratio of moles of iron and oxygen atoms is:

$\mathbf{\text{n}}\left(\text{Fe}\right)\mathbf{\text{:n}}\left(\text{O}\right)\mathbf{\text{=0.039\hspace{0.17em}mol:0.052\hspace{0.17em}mol}}$

On dividing it by using 0.039 mol

Then,

$\mathbf{\text{n}}\left(\text{Fe}\right)\mathbf{\text{:n}}\left(\text{O}\right)\mathbf{\text{=1\hspace{0.17em}mol:1.333\hspace{0.17em}mol}}$

Then,

$\mathbf{\text{n}}\left(\text{Fe}\right)\mathbf{\text{:n}}\left(\text{O}\right)\mathbf{\text{=3\hspace{0.17em}mol:4\hspace{0.17em}mol}}$

Thus, the empirical formula is ${\mathbf{Fe}}_{\mathbf{3}}{\mathbf{O}}_{\mathbf{4}}$

Step 3: Determine the empirical formula of 0.903 g of phosphorus combined with 6.99 g of bromine

The given is,

$\begin{array}{l}\mathrm{m}\left(\mathrm{P}\right)=0.903\mathrm{g}\\ \mathrm{m}\left(\mathrm{Br}\right)=6.99\mathrm{g}\end{array}$

On simplify,

$\begin{array}{l}\mathrm{n}\left(\mathrm{P}\right)=\frac{\mathrm{m}}{\mathrm{M}}=\frac{0.903\mathrm{g}}{30.97\mathrm{g}/\mathrm{mol}}\\ \mathrm{n}\left(\mathrm{P}\right)=0.029\mathrm{mol}\end{array}$

Similarly,

$\begin{array}{l}\mathrm{n}\left(\mathrm{Br}\right)=\frac{\mathrm{m}}{\mathrm{M}}=\frac{6.99\mathrm{g}}{79.9\mathrm{g}/\mathrm{mol}}\\ \mathrm{n}\left(\mathrm{Br}\right)=0.087\mathrm{mol}\end{array}$

Then,

$\begin{array}{l}\mathrm{n}\left(\mathrm{P}\right):\mathrm{n}\left(\mathrm{Br}\right)=0.029:0.087\\ \mathrm{n}\left(\mathrm{P}\right):\mathrm{n}\left(\mathrm{Br}\right)=1:3\end{array}$

Thus, the empirical formula is ${\mathbf{PBr}}_{\mathbf{3}}$

Step 4: Determine the empirical formula of a hydrocarbon having 79.9 mass % carbon

The mass % of carbon atoms present in hydrocarbon is 79.9 %. 

The total mass % of hydrocarbon = 100 %

Since the chemical formula of the hydrocarbon is ${\mathbf{\text{C}}}_{\mathbf{\text{x}}}{\mathbf{\text{H}}}_{\mathbf{\text{y}}}$

Thus, the mass % of hydrogen = 100 % - 79.9 % = 20.1 %

The number of moles of C and H is:

$\begin{array}{c}\mathrm{n}\left(\mathrm{C}\right):\mathrm{n}\left(\mathrm{H}\right)=\frac{\frac{0.799}{12.01}}{\frac{1-0.799}{1}}\\ \mathrm{n}\left(\mathrm{C}\right):\mathrm{n}\left(\mathrm{H}\right)=\frac{0.067}{0.201}\\ \mathrm{n}\left(\mathrm{C}\right):\mathrm{n}\left(\mathrm{H}\right)=1:3\end{array}$

Thus, the empirical formula is ${\mathbf{CH}}_{\mathbf{3}}$