Chemistry question

fuinneamh

When 40.5mg(7.5X10^-4 moles) of an unsaturated hydrocarbon were burned completely in excess oxygen, 67.3cm^3 of Carbon Dioxide, measured at STP were produced. Find the molecular formula of the hydrocarbon.


-i'm able to prove that its C4 but can't prove its H8, answer has to be H8 as we don't study alkynes above C2 but can't prove it!!! anyone got any ideas.

Lucas10101

You won't get a question like that in the exam.....don't bother learning it IMHO.

fuinneamh

Yeah but i would like to know how to do it anyway, the exam isn't everything in my book.

rjt

fuinneamh wrote:

When 40.5mg(7.5X10^-4 moles) of an unsaturated hydrocarbon were burned completely in excess oxygen, 67.3cm^3 of Carbon Dioxide, measured at STP were produced. Find the molecular formula of the hydrocarbon.


-i'm able to prove that its C4 but can't prove its H8, answer has to be H8 as we don't study alkynes above C2 but can't prove it!!! anyone got any ideas.

I think it should be C4H6? Here's my solution:

We have 67.3cm^3 of CO2, which is (67.3)/(22400) = 3x10^(-3) moles

So, our ratio of hydrocarbon:CO2 in moles is:
7.5x10^-4:3x10^(-3)
Which is equal to 1:4
So we have 1 hydrocarbon for every CO2, so the hydrocarbon must have 4 carbons.

Now, let x=the relative molecular mass of 1 mole of the hydrocarbon.
So (40.5x10^-3)/x = (No. of moles)
But we know that the number of moles is 7.5X10^-4
So:
(40.5x10^-3)/x = 7.5X10^-4
Therefore: x = (40.5x10^-3)/(7.5X10^-4) = 54

so (number of carbons)x12 + (number of hydrogens)x1 = 54
=> 4x12 + (number of hydrogens) = 54
=> number of hydrogens = 6

So the hydrocrbon is C4H6 (which is butyne)

fuinneamh

thank you, seems so obvious when you point it out, but isn't that always the way.