The density of a hydrocarbon at `N.T.P. 2.5` gram//litre. What is hydrocarbon?

Explanation

Density of `1` litre hydrocarbon `=2.5` gram//litre
`therefore` molecular weight of hydrocarbon `=2.5xx22.4=56`
After molecular weight. We calculate the molecular formula.
`C_(n)H_(2n+2)=` molecular weight (alkane) or `14n+2=` molecular weight
`C_(n)H_(2n)=` molecular weight (alkane) or `14n`=` molecular weight
`C_(n)H_(2n-2)C_(n)H_(2n-2)=` molecular weight (Alkyne) or `14n-2=` molecular weight with the help of above three formulae, we can identify the given hydrocarbon
`14n=56` (Alkene) `because` Hydrocarbon is `C_(4)H_(8)` (Butene).
`n=4`
Volume of hydrocarbon will be given and volume of `O_(2)` for complete combustion will also be give.
What is hydrocarbon is to be asked ?
The above question may be solved with the help of following three formulae.
Formula No. 1 `("Volumeof"H.C.)/("Volume of" O_(2))=(2)/3n+1)` (for Alkane)
Formula No. 2 `("Volumeof" H.C.)/("Volume of" O_(2))=(2)/(3n)` (for Alkene)
Formula No. 3 `("Volume of" H.C.)/("Volume of" O_(2))=(2)/(3n-1)` (for Alkyne)