Chapter 11, Problem 132IP

Anthraquinone contains only carbon, hydrogen, and oxygen. When 4.80 mg anthraquinone is burned, 14.2 mg CO₂ and 1.65 mg H₂O are produced. The freezing point of camphor is lowered by 22.3°C when 1.32 g anthraquinone is dissolved in 11.4 g camphor. Determine the empirical and molecular formulas of anthraquinone.

Interpretation Introduction

Interpretation: The empirical formula and molecular formula of Anthraquinone has to be determined.

Concept Introduction:

The ratio of the elements present in a compound and not the arrangement of the atoms are called as Empirical formula

Molecular formula is the representation of sum of number of atoms and molecules, not their arrangement in structure.

Answer to Problem 132IP

The empirical formula of Anthraquinone is ${\text{C}}_{\text{7}}{\text{H}}_{\text{4}}\text{O}$

The molecular formula of Anthraquinone is ${\text{C}}_{\text{14}}{\text{H}}_{\text{8}}{\text{O}}_{\text{2}}$

Explanation of Solution

Record the data

Mass of Carbon dioxide                = $\text{14}\text{mg}$

Mass of Water                              = $\text{1}\text{.65}\text{mg}$

Freezing point of Camphor            = $\text{22}\text{.3°C}$

Mass of Anthraquinone burned      = $\text{4}\text{.80}\text{mg}$

Mass of Anthraquinone Dissolved = $\text{1}\text{.32}\text{g}$

Mass of Camphor                         = $\text{11}\text{.4}\text{g}$

To calculate the mass percent of Carbon, Hydrogen and Oxygen

Atomic mass of Carbon             = $\text{12}\text{.01mg}$

Molar mass of Carbon dioxide   = $\text{44}\text{.01}\text{mg}$

Atomic mass of Hydrogen         = $\text{2}\text{.016}\text{mg}$

Molar mass of Water                 = $\text{18}\text{.02mg}$

Mass of Carbon = $\text{14}\text{.2mg}{\text{CO}}_{\text{2}}\text{×}\frac{\text{12}\text{.01}\text{mg}\text{C}}{\text{44}\text{.01}\text{mg}{\text{CO}}_{\text{2}}}\text{=3}\text{.88}\text{mg}$

Mass percentage of Carbon = $\frac{\text{3}\text{.88}\text{mg}}{\text{4}\text{.80}\text{mg}}\text{×100}$

                                                     = $\text{80}\text{.8\%}$

Mass of Hydrogen = $\text{1}\text{.65}\text{mg}{\text{H}}_{\text{2}}\text{O×}\frac{\text{2}\text{.016}\text{mg}\text{H}}{\text{18}\text{.02}\text{mg}{\text{H}}_{\text{2}}\text{O}}\text{=0}\text{.185}\text{mg}$

Mass percentage of Hydrogen = $\frac{\text{0}\text{.185}\text{mg}}{\text{4}\text{.80}\text{mg}}\text{×100}$

                                              = $\text{3}\text{.85}\text{\%}$

Mass percentage of Oxygen = $\text{100}\text{.00-(80}\text{.8+3}\text{.85)}$

                                          = $\text{15}\text{.4\%}$

Mass percentage of Carbon            = $\text{80}\text{.8\%}$

Mass percentage of Water               = $\text{3}\text{.85}\text{\%}$

Mass percentage of Oxygen          = $\text{15}\text{.4\%}$

To calculate the empirical formula

Mass percentage of Carbon      = $\text{80}\text{.8\%}$

Mass percentage of Hydrogen   = $\text{3}\text{.85}\text{\%}$

Mass percentage of Oxygen     = $\text{15}\text{.4\%}$

Out of $\text{100}\text{.00}\text{g}$ ,

$\begin{array}{l}\text{80}\text{.8}\text{g}\text{C×}\frac{\text{1}\text{mol}}{\text{12}\text{.01}\text{g}}\text{=6}\text{.73}\text{mol}\text{C=}\frac{\text{6}\text{.73}}{\text{0}\text{.963}}\text{=6}\text{.99}\approx \text{7}\\ \\ \text{3}\text{.85}\text{g}\text{H×}\frac{\text{1}\text{mol}}{\text{1}\text{.008}\text{g}}\text{=3}\text{.82}\text{mol}\text{H=}\frac{\text{3}\text{.82}}{\text{0}\text{.963}}\text{=3}\text{.97}\approx \text{4}\\ \\ \text{15}\text{.4}\text{g}\text{O×}\frac{\text{1}\text{mol}}{\text{16}\text{.00}\text{g}}\text{=0}\text{.963}\text{mol}\text{O=}\frac{\text{0}\text{.963}}{\text{0}\text{.963}}\text{=1}\text{.00}\end{array}$ ~

Therefore, the empirical formula is ${\text{C}}_{\text{7}}{\text{H}}_{\text{4}}\text{O}$

Record the given info

Freezing point of Camphor            = $\text{22}\text{.3°C}$

Molal freezing point depression constant = $\text{40}\text{.°C/molal}$

Mass of Anthraquinone burned      = $\text{4}\text{.80}\text{mg}$

Mass of Anthraquinone Dissolved = $\text{1}\text{.32}\text{g}$

Mass of Camphor                         = $\text{11}\text{.4}\text{g}$

To calculate the mass of Anthraquinone (m),

$\begin{array}{l}{\text{ΔT}}_{\text{f}}{\text{=K}}_{\text{f}}\text{m}\\ \\ \text{m=}\frac{{\text{ΔT}}_{\text{f}}}{{\text{K}}_{\text{f}}}\\ \\ \text{=}\frac{\text{22}\text{.3°C}}{\text{40}\text{.°C/molal}}\\ \\ \text{=0}\text{.56}\text{molal}\end{array}$

Molal of Anthraquinone = $\text{0}\text{.56}\text{molal}$

Moles of Anthraquinone = $\text{0}\text{.0114}\text{kg}\text{camphor×}\frac{\text{0}\text{.56}\text{mol}\text{anthraquinone}}{\text{kg}\text{camphor}}$

                                     = $\text{6}{\text{.4×10}}^{\text{-3}}\text{mol}$

Moles of Anthraquinone = $\text{6}{\text{.4×10}}^{\text{-3}}\text{mol}$

To calculate the molar mass

Moles of Anthraquinone = $\text{6}{\text{.4×10}}^{\text{-3}}\text{mol}$

Mass of Anthraquinone Dissolved = $\text{1}\text{.32}\text{g}$

Molar mass of Anthraquinone = $\frac{\text{1}\text{.3}\text{g}}{\text{6}{\text{.4×10}}^{\text{-3}}\text{mol}}$

                                              = $\text{210}\text{g}{\text{mol}}^{\text{-1}}$

To determine the Molecular formula of Anthraquinone

Empirical formula mass = $\text{104}\text{g}{\text{mol}}^{\text{-1}}$

Molar mass of Anthraquinone = $\text{210}\text{g}{\text{mol}}^{\text{-1}}$

Molar mass of Anthraquinone is twice the empirical mass of Anthraquinone, therefore the molecular mass of Anthraquinone is ${\text{C}}_{\text{14}}{\text{H}}_{\text{8}}{\text{O}}_{\text{2}}$

Molecular Mass of Anthraquinone = ${\text{C}}_{\text{14}}{\text{H}}_{\text{8}}{\text{O}}_{\text{2}}$

Conclusion

The moles of individual elements were calculated by using the mass percentages to their molar masses. The moles of the individual elements were divided by smallest ratio of moles and approximated to determine the empirical formula. The empirical formula was found to be ${\text{C}}_{\text{7}}{\text{H}}_{\text{4}}\text{O}$.

The molecular mass of Anthraquinone is calculated by using the molar mass and empirical formula mass. The molecular mass of Anthraquinone was ${\text{C}}_{\text{14}}{\text{H}}_{\text{8}}{\text{O}}_{\text{2}}$