Question

Determine the molecular formula of a hydrocarbon if the combustion of 5.3 mg obtained 16.6 mg `"CO"_2`. The density of the gas at standard conditions is `"2.504 g/dm"^3`. What is the molar mass of the gas and its molecular and empirical formula?

Answer 1

Molecular formula is `"C"_4"H"_8`
Molar mass is `56.0896`
Empirical formula is `"CH"_2`

Explanation:

Original question stated
student mentioned that the question used 6.48 mg of carbon dioxide, as opposed to 8.48 mL.

Solution with revised inputs from student.
All things same except
"if the combustion of 5.3 mg obtained 8.48 mg CO2"
Both values of carbon dioxide gave erroneous results for the working out empirical formula of the hydrocarbon.
=.=.=.=.=.=.=.=.=.=.=.=.=.=.=.=.=.=.=.=.=.=.=.=.=.=.=.
from the original question this value for calculation purpose modified original value of `16.6" mg"` has been taken.
It has been shown that a value taken as above gives a possible solution.
If we assume that in the original question there was a typo and "obtained 16.48mg of carbon dioxide" is the correct text, it will also lead to same empirical formula.
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Given the sample is a hydrocarbon. As such its constituent elements are Carbon `"C"` and Hydrogen `"H"`

We know that 1 GMW of any gas occupies `22.4dm^3` or litres of volume under standard conditions.
Therefore calculating mass of `22.4 dm^3` of hydrocarbon from the given density gives it molar mass.

Molar mass of hydrocarbon`=2.504×22.4=56.0896`

We also know that 1 GMW of `"CO"_2` has mass `=12+2xx24=` `44gm` which is equivalent to `12gm` of carbon.
Now `44gm` of carbon dioxide is equivalent to carbon`=12gm`
`:.16.6mg` of carbon dioxide is equivalent to carbon`=12/44xx16.6`
`=4.5mg`
Percent of carbon in sample`=(4.5)/5.3xx100=84.9`
`=>`Percent hydrogen in the sample`=15.1`

To find out the empirical formula of hydrocarbon we divide the above percentages by respective Average atomic masses.
`"Carbon":"Hydrogen"=84.9/12.011:(15.1)/1.0079`
`=>"Carbon":"Hydrogen"=7.069:14.981`
`=>"Carbon":"Hydrogen"~~1:2`
We obtain Empirical formula of Hydrocarbon as `"CH"_2`

Let molecular formula be `("CH"_2)_n`
As we already have Molar mass`=56.0896`
Equating the two we have
`(12.011+2xx1.0079)xxn=56.0896`
`=>14.0268n=56.0896`
`=>n=4`, Nearest whole number
Hence molecular formula of the hydrocarbon is `"C"_4"H"_8`

Answer 2

Molar mass of the gas `=56g/"mol"`

Molecular formula of the gas `=C_4H_8`

Empirical formula of the gas `=CH_2`

Explanation:

Given that the density of hydrocarbon at STP is `2.504g/"dm"^3`

So
`"molar mass of the HC"`

`="molar volume at STP"xx"density at STP"`

`=22.4"dm"^3/(mol)xx2.504g/"dm"^3~~56g/(mol)`

Let the molecular formula of HC be `C_xH_y`

And the balanced equation of the combustion reaction of the HC in oxygen is

`C_xH_y+(x+y/4)O_2(g)->xCO_2(g)+y/2H_2O(l)`

So by this equation the stochiometric ratio of HC and the `CO_2` produced on combustion is
`=(1mol)/(x mol)=(56g)/(44xg)=56/(44x)`
`"Where " 44g/"mol"" is the molar mass of "CO_2`

But by the given data this ratio is

`(5.3mg)/(16.6mg)=5.3/16.6`

So equating these two we get

`56/(44x)=5.3/16.6`

`=>x=(56xx16.6)/(44xx5.3)~~4`

Now calculated value of molar mass of the HC (using atomic mass of carbon as `12g/(mol)` and atomic mass of hydrogen as `1g/(mol)` ) will be `(12x+y)g/(mol)`

So `12x+y=56`

Inserting the value of `x=4` we get

`12xx4+y=56`

`=>y=8`

Hence the molecular formula of HC is `C_4H_8`

And obviously the empirical formula of HC is `CH_2`