Question

Anthraquinone contains only carbon, hydrogen, and oxygen. When 4.80 mg of anthraquinone is burned, 14.2 mg of ${\text{CO}}_2$ and 1.65 mg of${\text{H}}_2\text{O}$ are produced. The freezing point of camphor is lowered by $\text{22.3}°\text{C}$ when 1.32 g of anthraquinone is dissolved in 11.4 g of camphor. Determine the empirical and molecular formulas of anthraquinone.

Short answer

The empirical formula and molecular formula of anthraquinone is${\text{C}}_{14}{\text{H}}_8{\text{O}}_2$ and${\text{C}}_{14}{\text{H}}_8{\text{O}}_2$ respectively.

Step-by-step solution

Step 1:

According to the question, it is given that

Mass of carbon dioxide = 14.2mg = $\text{14.2}\times {\text{10}}^{-3}\text{g}$

Mass of water = 1.65mg = $\text{1.65}\times {\text{10}}^{-3}\text{g}$

Mass of anthraquinone = 4.80 mg = $4.80\times {\text{10}}^{-3}g$

The molar mass of carbon dioxide = 44g

The molar mass of water = 18g

So, the number of moles of carbon dioxide =$\frac{\text{Mass}}{\text{Molar mass}}\text{=}\frac{\text{14.2}\times {\text{10}}^{-3}\text{g}}{44g}\text{= 3.23}\times {\text{10}}^{-4}\text{mol}$

As only one molecule of carbon is present, then the molar amount of carbon will be:

${\text{n(C) = n(CO}}_2\text{) = 3.23}\times {\text{10}}^{-4}\text{mol}$

The number of moles of water =$\frac{\text{Mass}}{\text{Molar mass}}\text{=}\frac{\text{1.65}\times {\text{10}}^{-3}\text{g}}{18g}\text{= 9.5}\times {\text{10}}^{-5}\text{mol}$

As two molecule of hydrogen is present, then the molar amount of hydrogen will be:

${\text{n(H) = n(H}}_2\text{O)}\times \text{2 = 9.15}\times {\text{10}}^{-5}\times \text{2 = 1.83}\times {\text{10}}^{-4}\text{mol}$

Step 2:

To determine the mass of hydrogen and carbon in anthraquinone, we will use mass and molar mass that is:

Mass of hydrogen = $\text{Moles}\times \text{molar mass = 1.83}\times {\text{10}}^{-4}\times \text{1 = 1.83}\times {\text{10}}^{-4}g$

Mass of carbon = $\text{Moles}\times \text{molar mass = 3.23}\times {\text{10}}^{-4}\times \text{12 = 3.87}\times {\text{10}}^{-3}g$

Now, mass of oxygen = mass of anthraquinone – (mass of hydrogen + mass of oxygen)

Mass of oxygen = $4.80\times {\text{10}}^{-3}$- ($\text{0.183}\times {\text{10}}^{-3}\text{+ 3.87}\times {\text{10}}^{-3}$) = $\text{0.75}\times {\text{10}}^{-3}g$

Thus, moles of oxygen =$\frac{\text{Mass}}{\text{Molar mass}}\text{=}\frac{\text{0.75}\times {\text{10}}^{-3}}{16}\text{= 4.66}\times {\text{10}}^{-5}\text{mol}$

Empirical formula of anthraquinone

Atom	Moles	Simple whole Ratio
Carbon	$32.3\times {10}^{-5}$	localid="1664877903774" $\text{32.3}\times {\text{10}}^{-5}$/ localid="1664877909294" $\text{4.66}\times {\text{10}}^{-5}$= 7
Hydrogen	$\text{18.3}\times {\text{10}}^{-5}$	localid="1664877914523" $\text{18.3}\times {\text{10}}^{-5}$/localid="1664877919484" $\text{4.66}\times {\text{10}}^{-5}$ = 4
Oxygen	$\text{4.66}\times {\text{10}}^{-5}$	localid="1664877926307" $\text{4.66}\times {\text{10}}^{-5}$/$\text{4.66}\times {\text{10}}^{-5}$= 1

So, the empirical formula will be localid="1664877932249" ${\text{C}}_7{\text{H}}_4\text{O}$and mass will be 104g

Molecular formula of anthraquinone

The relation between empirical formula and molecular formula is:

$\text{n =}\frac{\text{Molecular formula}}{\text{Empirical formula}}$

The value of n will be 2, because the value Van’t Hoff factor will be:

$\Delta \text{T = i}\times {\text{K}}_{\text{f}}\times \text{m}$ Here, ${\text{K}}_{\text{f}}$ will be $40°\text{C/m}$ for camphor.

$\text{22.3 = i}\times \text{40}\times \frac{1.32\times \text{1000}}{104.11\times \text{11.4}}$

$\text{i = 0.5}$

It indicates that one anthraquinone will have 2 empirical formulas as it will form dimer because only half the molecule is dissolved.

Therefore, $\text{Molecular formula = n}\times \text{Empirical formula}$

$\text{Molecular formula = 2}\times {\text{C}}_7{\text{H}}_4{\text{O =C}}_{14}{\text{H}}_8{\text{O}}_2$