# 4.6g of an organic gaseous compound was burned. Its gas density compared to air is 23. After burning this compound, 4.48 L of CO2 (at 0 °C and 1 atm) and 5.4 g of H2O were produced. What is the name and chemical formula of this organic compound?

Jan 15, 2017

If we assume that the density (23) of the gas has been compared to the density of ${H}_{2}$,then the molar mass of the gas becomes $46 \text{g/mol}$

The organic compound burnt was $= 4.6 g = \frac{4.6 g}{46 \frac{g}{\text{mol}}} = 0.1 m o l$

The amount of $C {O}_{2}$ produced at STP was $4.48 L = \frac{4.48 L}{22.4 \frac{L}{\text{mol}}} = 0.2 m o l$

This amount of $C {O}_{2}$ will contain $0.2 m o l$ carbon.

Again the amount of ${H}_{2} O$ produced on burning

$= 5.4 g = \frac{5.4 g}{18 \frac{g}{\text{mol}}} = 0.3 m o l$

This $0.3 m o l {H}_{2} O$ will contain $0.3 m o l$ of hydrogen.

Total mass of carbon and hydrogen in the $4.6 g$ compound
$= 0.2 m o l \times 12 \frac{g}{\text{mol"+0.3molxx2g/"mol}} = 3 g$

So the remaining amount $1.6 g$ should be due to oxygen. Hence the number of moles of oxygen $= \frac{1.6 g}{32 \frac{g}{\text{mol}}} = 0.05 m o l$

So 1 mol of the compound should contain $2 m o l$ Carbon,$3 m o l$ Hydrogen and $0.5 m o l$ of Oxygen.

So 1 molecule of the compound should contain 2 molecules or 2 atoms of Carbon, 3 molecules or 6atoms of Hydrogen and 0.5molecule or 1atom of Oxygen.

Hence the molecular formula of the compound is ${C}_{2} {H}_{6} O$

Two substances (1)Dimethylether($C {H}_{3} O C {H}_{3}$) and (2) Ethanol (CH_3CH_2OH) are possible with this formula but the substance being a gas it should be Dimethyl ether $\textcolor{red}{\left(C {H}_{3} O C {H}_{3}\right)}$.