91Ó°ÊÓ

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.

$\mathbf{\text{n}}\left(\text{Cl}\right)\mathbf{\text{:n}}\left(\text{O}\right)\mathbf{\text{=0.063″¾´Ç±ô:0.22″¾´Ç±ô}}$

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

Then,

data-custom-editor="chemistry" $\mathbf{\text{n}}\left(\text{Cl}\right)\mathbf{\text{:n}}\left(\text{O}\right)\mathbf{\text{=1″¾´Ç±ô:3.5″¾´Ç±ô}}$

On multiply by 2.

role="math" localid="1656610804973" $\mathbf{\text{n}}\left(\text{Cl}\right)\mathbf{\text{:n}}\left(\text{O}\right)\mathbf{\text{=2″¾´Ç±ô:7″¾´Ç±ô}}$

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

The given is,

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

On simplify,

data-custom-editor="chemistry" $\begin{array}{l}\mathbf{\text{n}}\left(\text{Si}\right)\mathbf{\text{=}}\frac{\mathbf{\text{m}}}{\mathbf{\text{M}}}\mathbf{\text{=}}\frac{\mathbf{\text{2.45 g}}}{\mathbf{\text{28.09 g/³¾´Ç±ô}}}\\ \mathbf{\text{n}}\left(\text{Si}\right)\mathbf{\text{=0.087″¾´Ç±ô}}\end{array}$

Similarly,

data-custom-editor="chemistry" $\begin{array}{l}\mathbf{\text{n}}\left(\text{Cl}\right)\mathbf{\text{=}}\frac{\mathbf{\text{m}}}{\mathbf{\text{M}}}\mathbf{\text{=}}\frac{\mathbf{\text{12.4 g}}}{\mathbf{\text{35.45 g/³¾´Ç±ô}}}\\ \mathbf{\text{n}}\left(\text{Cl}\right)\mathbf{\text{=0.35″¾´Ç±ô}}\end{array}$

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 isdata-custom-editor="chemistry" ${\mathbf{SiCl}}_{\mathbf{4}}$

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${\mathbf{CO}}_{\mathbf{2}}$