Question

Question #7fcaa

Answer 1

The first step in the process is converting to the standard units for us to use in the later calculations.

which are,

Litres for volume (L)
Kilopascals for pressure (kPa)
Kelvin for temperature (K)

So to convert from `cm^3` to Litres which is 1000 `cm^3`

so,

`20cm^3 => 0.02L` of gaseous hydrocarbon.
`60cm^3 => 0.06L` of carbon (iv) oxide (`CO_"2"`)
`40cm^3 => 0.04L` of water vapour

so since all volumes are measured at the same temperature and pressure we can assume standard temperature and pressure, STP which is 0 degrees celsius and 100Kpa.

so what has occurred is a combustion reaction following the general formula,

hydrocarbon + `O_"2" => CO_"2" +H_"2"O`

and since it is in excess oxygen we can assume it has reached completion (ie there is no more Hydrocarbons left).
There is also no other substances produced so we can assume that they are not present in our hydrocarbon and that all H, and C produced came from our hydrocarbon.

this uses the formula `n=V/22.71`

so to work out H,

`n(H_"2"O)=0.04/22.71`

`n(H_"2"O)=1.76134*10^-3 mol`

`n(H)=2*n(H_"2"O)`

`n(H)=2*1.76134*10^-3`

`n(H)=3.523*10^-3 mol`

now C,

`n(CO_"2") = 0.06/22.71`

`n(CO_"2") = 2.642*10^-3 mol`

`n(CO_"2") = n(C)`

`n(C) = 2.642*10^-3 mol`

leaving us with,

`n(H)=3.523*10^-3 mol` `" , "` `n(C) = 2.642*10^-3 mol`

`n(hydrocarbon) = 0.02/22.71`

`n(hydrocarbon) = 8.81*10^-4mol`

So now without mass, we have to use molar ratios to work out how many times of our empirical formula is our molecular formula. we will use x as this factor.

`n(C)= x*n(hydrocarbon)`

`x=(2.642*10^-3)/(8.81*10^-4)`

`x=3`

`n(H)=y*n(hydrocarbon)`

`y=(3.523*10^-3) /(8.81*10^-4)`

`y=4`

leaving us with a molecular formula of,

`C_"3"H_"4"`

which could be Propyne or Cyclopropene etc.