Question

A compound contains only carbon, hydrogen, and oxygen. Combustion of 10.68 mg of thecompound yields 16.01 mg CO2 and 4.37 mg H20. The molar mass of the compound is 176.1g/mol. What are the empirical and molecular formulas of the compound?

Short answer

The molecular and empirical formulas are$C_3{H}_4{O}_3\&C_6{H}_8{O}_6$

Step-by-step solution

Reaction of compound combustion

The reaction of compound combustion can be,

$a{C}_x{H}_y{O}_z+b{O}_2\to cC{O}_2+d{H}_{2}O$

To calculate the mass of H and C can be denoted as,

$\begin{array}{rcl}\mathrm{m}\left(\mathrm{C}\right)& =& 16.01\mathrm{g}{\mathrm{CO}}_2·\frac{12.011\mathrm{g}\mathrm{C}}{44.009\mathrm{g}{\mathrm{CO}}_2}\\ & =& 4.36\mathrm{mg}\mathrm{C}\\ \mathrm{m}\left(\mathrm{H}\right)& =& 4.37\mathrm{g}{\mathrm{H}}_2\mathrm{O}·\frac{2.016\mathrm{g}\mathrm{H}}{18.015\mathrm{g}{\mathrm{H}}_2\mathrm{O}}\\ & =& 0.489\mathrm{mg}\mathrm{H}\end{array}$

Calculating C and H percent

On calculating C and H percent in acid,

$\begin{array}{rcl}\mathrm{Percent}\mathrm{C}& =& \frac{\mathrm{m}\left(\mathrm{C}\right)}{\mathrm{m}\left(\mathrm{acid}\right)}·100\mathrm{Percent}\\ & =& \frac{4.369\mathrm{mg}}{10.68\mathrm{mg}}·100\mathrm{Percent}\\ & =& 40.908\mathrm{Percent}\\ \mathrm{Percent}\mathrm{H}& =& \frac{0.489\mathrm{g}}{10.68\mathrm{g}}·100\\ & =& 4.579\mathrm{Percent}\\ \mathrm{Percent}\mathrm{O}& =& 100\mathrm{Percent}-40.908\mathrm{Percent}-4.579\mathrm{Percent}\\ & =& 54.513\mathrm{Percent}\end{array}$

Calculating empirical formula

On finding empirical formula,

$\begin{array}{rcl}\mathrm{n}\left(\mathrm{C}\right)& =& \frac{40.908\mathrm{g}}{12.001{\displaystyle \raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.}}\\ & =& 3.4\mathrm{mol}\\ \mathrm{n}\left(\mathrm{H}\right)& =& \frac{4.579\mathrm{g}}{1.008{\displaystyle \raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.}}\\ & =& 4.543\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.\\ \mathrm{n}\left(\mathrm{O}\right)& =& \frac{54.513\mathrm{g}}{15.999{\displaystyle \raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.}}\\ & =& 3.4\mathrm{mol}\\ \mathrm{n}\left(\mathrm{C}\right):\mathrm{n}\left(\mathrm{H}\right):\mathrm{n}\left(\mathrm{O}\right)& =& 3:4:3\end{array}$

The empirical formula is$C_3{H}_4{O}_3$.

Calculating molecular formula

On calculating molecular formula,

$\begin{array}{rcl}\mathrm{Molecular}\mathrm{formula}& =& {\left(\mathrm{Theempirical}\mathrm{formula}\right)}_{\mathrm{x}}\\ \mathrm{n}& =& \frac{\mathrm{m}}{\mathrm{M}}\\ \mathrm{M}& =& \frac{41.5\mathrm{g}}{0.25\mathrm{mol}}\\ \mathrm{M}& =& 166\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.\\ & & \end{array}$

The mass of empirical formula is $3.12+4.1+3.16=88\mathrm{g}$. Then,

$\begin{array}{rcl}\mathrm{x}& =& \frac{176.1\mathrm{g}}{88\mathrm{g}}\\ \mathrm{x}& =& 2\end{array}$

Molecular formula is$C_6{H}_8{O}_6$.