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?

Md Ariful Islam

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Md Ariful Islam

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If we assume that the (23) of the gas has been compared to the density of $H_2$,then the molar mass of the gas becomes $46"g/mol"$

The organic compound burnt was $=4.6g=(4.6g)/(46g/"mol")=0.1mol$

The amount of $CO_2$ produced at STP was $4.48L=(4.48L)/(22.4L/"mol")=0.2mol$

This amount of $CO_2$ will contain $0.2mol$ carbon.

Again the amount of $H_2O$ produced on burning

$=5.4g=(5.4g)/(18g/"mol")=0.3mol$

This $0.3molH_2O$ will contain $0.3mol$ of hydrogen.

Total mass of carbon and hydrogen in the $4.6g$ compound
$=0.2molxx12g/"mol"+0.3molxx2g/"mol"=3g$

So the remaining amount $1.6g$ should be due to oxygen. Hence the number of moles of oxygen $=(1.6g)/(32g/"mol")=0.05mol$

So 1 mol of the compound should contain $2mol$ Carbon,$3mol$ Hydrogen and $0.5mol$ 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 $C_2H_6O$

Two substances (1)Dimethylether($CH_3OCH_3$) and (2) Ethanol (CH_3CH_2OH) are possible with this formula but the substance being a gas it should be Dimethyl ether $color(red)((CH_3OCH_3))$.

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