Q3.37P - Chemistry: Molecular Nature Of Matter And Change

Question

Find the empirical formula of the following compounds:

(a)  0.063 mol of chlorine atoms combined with 0.22 mol of oxygen atoms

(b)  2.45 g of silicon combined with 12.4 g of chlorine

(c)  27.3 mass % carbon and 72.7 mass % oxygen

Step-by-Step Solution

Answer

The empirical formula of the compounds is:

(a) ${\mathrm{Cl}}_2{\mathrm{O}}_7$

(b) ${\mathrm{SiCl}}_4$

(c) ${\mathrm{CO}}_2$

Step 1: Introduction of the 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 0.063 mol of Cl atoms combined with 0.22 mol of O atoms

$\mathbf{\text{n}}\left(\text{Cl}\right)\mathbf{\text{:n}}\left(\text{O}\right)\mathbf{\text{=0.063\hspace{0.17em}mol:0.22\hspace{0.17em}mol}}$

On divide it by using $0.063\mathrm{mol}$

Then,

data-custom-editor="chemistry" $\mathbf{\text{n}}\left(\text{Cl}\right)\mathbf{\text{:n}}\left(\text{O}\right)\mathbf{\text{=1\hspace{0.17em}mol:3.5\hspace{0.17em}mol}}$

On multiply by 2.

$\mathbf{\text{n}}\left(\text{Cl}\right)\mathbf{\text{:n}}\left(\text{O}\right)\mathbf{\text{=2\hspace{0.17em}mol:7\hspace{0.17em}mol}}$

Thus, the empirical formula is data-custom-editor="chemistry" ${\mathbf{Cl}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{7}}$

Step 3: Determine the empirical formula of 2.45 g of silicon combined with 12.4 g of chlorine

The given is,

$\begin{array}{l}\mathbf{\text{m}}\left(\text{Si}\right)\mathbf{\text{=2.45\hspace{0.17em}g}}\\ \mathbf{\text{m}}\left(\text{Cl}\right)\mathbf{\text{=12.4\hspace{0.17em}g}}\end{array}$

On simplify,

data-custom-editor="chemistry" $\begin{array}{l}\mathbf{\text{n}}\left(\text{Si}\right)\mathbf{\text{=}}\frac{\mathbf{\text{m}}}{\mathbf{\text{M}}}\mathbf{\text{=}}\frac{\mathbf{\text{2.45\hspace{0.17em}g}}}{\mathbf{\text{28.09\hspace{0.17em}g/mol}}}\\ \mathbf{\text{n}}\left(\text{Si}\right)\mathbf{\text{=0.087\hspace{0.17em}mol}}\end{array}$

Similarly,

data-custom-editor="chemistry" $\begin{array}{l}\mathbf{\text{n}}\left(\text{Cl}\right)\mathbf{\text{=}}\frac{\mathbf{\text{m}}}{\mathbf{\text{M}}}\mathbf{\text{=}}\frac{\mathbf{\text{12.4\hspace{0.17em}g}}}{\mathbf{\text{35.45\hspace{0.17em}g/mol}}}\\ \mathbf{\text{n}}\left(\text{Cl}\right)\mathbf{\text{=0.35\hspace{0.17em}mol}}\end{array}$

Then,

data-custom-editor="chemistry" $\begin{array}{l}\mathbf{\text{n}}\left(\text{Si}\right)\mathbf{\text{:n}}\left(\text{Cl}\right)\mathbf{\text{=0.087:0.35}}\\ \mathbf{\text{n}}\left(\text{Si}\right)\mathbf{\text{:n}}\left(\text{Cl}\right)\mathbf{\text{=1:4}}\end{array}$

Thus, the empirical formula is data-custom-editor="chemistry" ${\mathbf{SiCl}}_{\mathbf{4}}$

Step 4: Determine the empirical formula of 27.3 mass % carbon and 72.7 mass % of oxygen

The mass % of carbon and oxygen is:

$\begin{array}{ccc}\mathbf{\text{n}}\left(\text{C}\right)& \mathbf{\text{=}}& \mathbf{\text{27.3\%}}\\ \mathbf{\text{n}}\left(\text{O}\right)& \mathbf{\text{=}}& \mathbf{\text{72.7\%}}\end{array}$

On simplify,

$\begin{array}{l}\mathbf{\text{n}}\left(\text{C}\right)\mathbf{\text{:n}}\left(\text{O}\right)\mathbf{\text{=}}\frac{\frac{\mathbf{\text{0.273}}}{\mathbf{\text{12.01}}}}{\frac{\mathbf{\text{0.727}}}{\mathbf{\text{16}}}}\\ \mathbf{\text{n}}\left(\text{C}\right)\mathbf{\text{:n}}\left(\text{O}\right)\mathbf{\text{=}}\frac{\mathbf{\text{0.023}}}{\mathbf{\text{0.045}}}\\ \mathbf{\text{n}}\left(\text{C}\right)\mathbf{\text{:n}}\left(\text{O}\right)\mathbf{\text{=1:2}}\end{array}$

Thus, the empirical formula is ${\mathbf{CO}}_{\mathbf{2}}$