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[Main]

A 20mol sample of a hydrocarbon C(x)H(y) Yields, after a complete combustion with excess 02 gas, 0.80 mol of CO2 and 1.0 mol of H2O. The Molecular formula for CxHy is ...
a C4H10
b C4H8
c C4H5
d C8H16
e C8H10

I think its a.. but I am not to sure..

[Brain]

well lets see if I remember how to do this stuff.....at the end you 0.8mol of CO2, so there is 0.8 mole of carbon in the out gas, so there must 0.8 mole going in as well.
and for the H, we got 2 mole going out, so we need 2 mole going in.

can't balance moles, gotta balance mass!, that's why I was confused.

12.01g/mole*0.8mole=9.61g C (out=in)
1.01g/mole * 2mole = 2.02g H (out=in)

hmmmm....now what do I do? :p uhhhh......I can't really remember much of this :(

[Nava2]

Get the number of moles of Carbon and Hydrogen, use these numbers as the x,y. Then, divide all by the smallest number and multiply till you have an empirical formula.

Take the Mm of the Empirical formula, and divide it into 20 mol. Multiply the x, y values by the factor you recieve and there is your molecular formula.

:)

[Brain]

if you know it's an alkane then A is the only choice :p but it might not be.

[Main]

A,C,E are alkane
B,D are alkene

[Brain]

I think that only A is alkane, the alkane is all single bonded right? so there has to be twice the H + 2 (2 on the ends)
and the only one that satisfies that equation is A

[tarajunky]

That must be a 0.20 mol sample, (or a 20 mmol sample) or it doesn't make sense.

Since you have excess O2 and there aren't any O's in your possible answers, just ignore them.

So you've got 0.8 mol C and 2*1.0 mol H. C8H20. That's just relative amounts, so C2H5 or C4H10 would also give the same outcome.

A -> C4H10

[3Garrett3]

That must be a 0.20 mol sample, (or a 20 mmol sample) or it doesn't make sense.

Since you have excess O2 and there aren't any O's in your possible answers, just ignore them.

So you've got 0.8 mol C and 2*1.0 mol H. C8H20. That's just relative amounts, so C2H5 or C4H10 would also give the same outcome.

A -> C4H10

The 20 mol sample kinda messed with my head.

I ended up getting a 1:1 ratio, but changing the numbers around might mess the answer XD

[Main]

lol WOW!
blame university of toronto and the Canadian institute of chemistry.
They = noob.
http://www.chem.utoronto.ca/IChO.Ontario/en/problemsets.html

The problem is from the first problem set.

TY all!
We > University of toronto + Canadian Institute of chemistry

[Nava2]

lol WOW!
blame university of toronto and the Canadian institute of chemistry.
They = noob.
http://www.chem.utoronto.ca/IChO.Ontario/en/problemsets.html

The problem is from the first problem set.

TY all!
We > University of toronto + Canadian Institute of chemistry

U of T blows.

UWO UWO UWO UWO UWO