Q3.36P - Chemistry: Molecular Nature Of Matter And Change

Question

What is the molecular formula for each of the following compounds?

a) Empirical formula $\mathbf{CH}\mathbf{(}{\mathbf{M}}_{\mathbf{w}}\mathbf{=}\mathbf{78}\mathbf{.11}\mathbf{g}\mathbf{/}\mathbf{mol}\mathbf{)}$

b) Empirical formula ${\mathbf{C}}_{\mathbf{3}}{\mathbf{H}}_{\mathbf{6}}{\mathbf{O}}_{\mathbf{2}}\mathbf{(}{\mathbf{M}}_{\mathbf{w}}\mathbf{=}\mathbf{74}\mathbf{.08}\mathbf{g}\mathbf{/}\mathbf{mol}\mathbf{)}$

c) Empirical formula $\mathbf{HgCl}\mathbf{(}{\mathbf{M}}_{\mathbf{w}}\mathbf{=}\mathbf{472}\mathbf{.10}\mathbf{g}\mathbf{/}\mathbf{mol}\mathbf{)}$

d) Empirical formula${\mathbf{C}}_{\mathbf{7}}{\mathbf{H}}_{\mathbf{4}}{\mathbf{O}}_{\mathbf{2}}\mathbf{(}{\mathbf{M}}_{\mathbf{w}}\mathbf{=}\mathbf{240}\mathbf{.20}\mathbf{g}\mathbf{/}\mathbf{mol}\mathbf{)}$

Step-by-Step Solution

Answer

1. The molecular formula of $\mathrm{CH}$ is ${\mathbf{C}}_{\mathbf{6}}{\mathbf{H}}_{\mathbf{6}}$.
2. The molecular formula of ${\mathrm{C}}_3{\mathrm{H}}_6{\mathrm{O}}_2$ is  ${\mathbf{C}}_{\mathbf{3}}{\mathbf{H}}_{\mathbf{6}}{\mathbf{O}}_{\mathbf{2}}$
3. The molecular formula of $\mathrm{HgCl}$ is ${\mathbf{Hg}}_{\mathbf{2}}{\mathbf{Cl}}_{\mathbf{2}}$
4. The molecular formula of ${\mathrm{C}}_7{\mathrm{H}}_4{\mathrm{O}}_2$ is ${\mathbf{C}}_{\mathbf{14}}{\mathbf{H}}_{\mathbf{8}}{\mathbf{O}}_{\mathbf{4}}$

Step 1: Defining Empirical formula and Molecular formula

The 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.

Step 2: To calculate the Molecular formula

The whole number multiple is given as,

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

The empirical formula (EF) mass is given as,

${\mathrm{M}}_{\mathrm{w}}$of data-custom-editor="chemistry" $\mathrm{CH}$,

$\begin{array}{c}={\mathrm{M}}_{\mathrm{w}}\left(\mathrm{C}\right)+{\mathrm{M}}_{\mathrm{w}}\left(\mathrm{H}\right))\\ =12.01\mathrm{g}/\mathrm{mol}+1.01\mathrm{g}/\mathrm{mol})\\ =13.02\mathrm{g}/\mathrm{mol}\end{array}$

The whole-number multiple for compound data-custom-editor="chemistry" $\mathrm{CH}$,

Whole-number multiple$=\frac{78.11(\mathrm{g}/\mathrm{mol})}{13.02(\mathrm{g}/\mathrm{mol})}\approx 6$

Multiplying subscripts of data-custom-editor="chemistry" ${\mathrm{C}}_1{\mathrm{H}}_1$ by 6, the molecular formula is given by

data-custom-editor="chemistry" $={\mathrm{C}}_{(1\times 6)}\times {\mathrm{H}}_{(1\times 6)}$

Molecular formuladata-custom-editor="chemistry" $={\mathrm{C}}_6{\mathrm{H}}_6$

Step 3: To calculate the Molecular formula

The whole number multiple is given as,

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

The empirical formula (EF) mass is given as,

data-custom-editor="chemistry" ${\mathrm{M}}_{\mathrm{w}}$ of data-custom-editor="chemistry" ${\mathrm{C}}_3{\mathrm{H}}_6{\mathrm{O}}_2$

$\begin{array}{l}=3\times {\mathrm{M}}_{\mathrm{w}}\left(\mathrm{C}\right)+6\times \left({\mathrm{M}}_{\mathrm{w}}\right(\mathrm{H}\left)\right)+2\times {\mathrm{M}}_{\mathrm{w}}\left(\mathrm{O}\right)\\ =3\times 12.01\mathrm{g}/\mathrm{mol}+6\times 1.01\mathrm{g}/\mathrm{mol}+2\times 16.00\mathrm{g}/\mathrm{mol}\\ =74.09\mathrm{g}/\mathrm{mol}\end{array}$

The whole-number multiple for compound data-custom-editor="chemistry" ${\mathrm{C}}_3{\mathrm{H}}_6{\mathrm{O}}_2$

Whole-number multiple$=\frac{74.08(\mathrm{g}/\mathrm{mol})}{74.09(\mathrm{g}/\mathrm{mol})}\approx 1$

Multiplying subscripts of data-custom-editor="chemistry" ${\mathrm{C}}_3{\mathrm{H}}_6{\mathrm{O}}_2$ by 1, the molecular formula is given by

Molecular formuladata-custom-editor="chemistry" $={\mathrm{C}}_3{\mathrm{H}}_6{\mathrm{O}}_2$

Step 4: To calculate the Molecular formula

The whole number multiple is given as,

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

The empirical formula (EF) mass is given as,

${\mathrm{M}}_{\mathrm{w}}$of data-custom-editor="chemistry" $\mathrm{HgCl}$

$\begin{array}{l}={\mathrm{M}}_{\mathrm{w}}\left(\mathrm{Hg}\right)+{\mathrm{M}}_{\mathrm{w}}\left(\mathrm{Cl}\right)\\ =200.59\mathrm{g}/\mathrm{mol}+35.45\mathrm{g}/\mathrm{mol})\\ =236.04\mathrm{g}/\mathrm{mol}\end{array}$

The whole-number multiple for compound data-custom-editor="chemistry" $\mathrm{HgCl}$

Whole-number multiple$=\frac{472.10(\mathrm{g}/\mathrm{mol})}{236.04(\mathrm{g}/\mathrm{mol})}\approx 2$

Multiplying subscripts of data-custom-editor="chemistry" ${\mathrm{Hg}}_1{\mathrm{Cl}}_1$ by 2, the molecular formula is given by

data-custom-editor="chemistry" $={\mathrm{Hg}}_{(1\times 2)}\times {\mathrm{Cl}}_{(1\times 2)}$

Molecular formuladata-custom-editor="chemistry" $={\mathrm{Hg}}_2{\mathrm{Cl}}_2$

Step 5: To calculate the Molecular formula

The whole number multiple is given as,

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

The empirical formula (EF) mass is given as,

data-custom-editor="chemistry" ${\mathrm{M}}_{\mathrm{w}}$ of data-custom-editor="chemistry" ${\mathrm{C}}_7{\mathrm{H}}_4{\mathrm{O}}_2$

$\begin{array}{l}=7\times {\mathrm{M}}_{\mathrm{w}}\left(\mathrm{C}\right)+4\times \left({\mathrm{M}}_{\mathrm{w}}\right(\mathrm{H}\left)\right)+2\times {\mathrm{M}}_{\mathrm{w}}\left(\mathrm{O}\right)\\ =7\times 12.01\mathrm{g}/\mathrm{mol}+4\times 1.01\mathrm{g}/\mathrm{mol}+2\times 16.00\mathrm{g}/\mathrm{mol}\\ =120.11\mathrm{g}/\mathrm{mol}\end{array}$

The whole-number multiple for compound data-custom-editor="chemistry" ${\mathrm{C}}_7{\mathrm{H}}_4{\mathrm{O}}_2$

Whole-number multiple$=\frac{240.20(\mathrm{g}/\mathrm{mol})}{120.11(\mathrm{g}/\mathrm{mol})}\approx 2$

Multiplying subscripts of ${\mathrm{C}}_7{\mathrm{H}}_4{\mathrm{O}}_2$ by 2, the molecular formula is given by

$={\mathrm{C}}_{(7\times 2)}\times {\mathrm{H}}_{(4\times 2)}\times {\mathrm{O}}_{(2\times 2)}$

Molecular formuladata-custom-editor="chemistry" $={\mathrm{C}}_{14}{\mathrm{H}}_8{\mathrm{O}}_4$