Question

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?

Answer 1

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"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))`.