Question

A compound contains only carbon, hydrogen, and oxygen. Combustion of 10.68 mg of the compound yields 16.01 mg $C{O}_2$ and 4.37 mg $H_{2}O$. The molar mass of the compound is 176.1 g/mol. What are the empirical and molecular formulas of the compound?

Answer 1

$C_x{H}_y{O}_z+O_2⟶xC{O}_2+\frac{y}{z}{H}_{2}O$

moles of $C{O}_2=0.364mmol$

It contains $0.364mmol$ of $C.$

$\therefore$ Mass of $C=0.364\times 12=4.368mg$$

moles of $H_{2}O=\frac{4.37}{18}=0.243mmol$

It contains $2\times 0.243mmol$ of $H$

$\therefore$ Mass of $H=0.486mg$

Mass of $O=10.68-(4.368+0.486)=5.826mg$

$\therefore$ Moles of $O=0.364$

$\therefore$ Ratio $C:H:O=0.364:0.486:0.364=1:1.34:1=3:4:3$

$\therefore$ Empirical formula is $C_3{H}_4{O}_3$,

Empirical formula mass $=88=2\times$ Molar mass.

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