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

Short answer

The empirical formula of the compounds is:

(a)${\mathrm{Fe}}_3{\mathrm{O}}_4$

(b)${\mathrm{PBr}}_3$

(c)${\mathrm{CH}}_3$

Step-by-step solution

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 anempirical formula. This is the compound's lowest whole-number ratio.

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${\mathbf{Fe}}_{\mathbf{3}}{\mathbf{O}}_{\mathbf{4}}$

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${\mathbf{PBr}}_{\mathbf{3}}$

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${\mathbf{CH}}_{\mathbf{3}}$