[Big-Daddy]

10cm³ of a gaseous hydrocarbon was mixed with excess oxygen and ignited. The gas volumes were measured at room temperature and pressure before and after the conversion and it was found that the total gas volume had contracted by 20 cm³. Given that combustion was complete, what was the formula of the hydrocarbon?

The answer is C₄H₄. But how on earth was it calculated? Without knowing the volume of oxygen reacted how can this possibly be reached ...

[Corribus]

Interesting problem. I kind of got halfway but I'm out of time. Still maybe this will help.

First, find a generic balanced equation for combustion of a hydrocarbon in the form  C_xH_y + αO₂ βCO₂ + γH₂O

Where the Greek letters are coefficients in terms of x and y. Probably you have already done this.

If the conditions were the same before and after the reaction (when the volumes are measured), you can make some assumptions about the volumes of each participating reactant and product being proportional to molar stoichiometric equivalents, no? (Because at STP, the volume per mole is already known). So, if you consume 10 cm³ of hydrocarbon, then you also consume 10α cm³ of oxygen and produce 10β cm³ of carbon dioxide and 10γ cm³ of water. The total net volume is -20 cm³.

This gives you one equation and two unknowns, but the "x" terms appear to cancel out. I was able to get y = 4.  Haven't figured out how you get x yet.

Maybe you can find a flaw in that reasoning, but it seems to be on the right track. I did it rather quickly though.

[Big-Daddy]

Ah, so we assume that H2O condenses and doesn't contribute to the volume? Then I can get the right answer y=4 by your method.

I'm starting to suspect it is not possible to find x without knowing the volume of O₂ you started with ...

[Borek]

I'm starting to suspect it is not possible to find x without knowing the volume of O₂ you started with ...

Assume 100 mL.

Assume 200 mL.

Are you getting somewhere, or nowhere?

[Borek]

First, find a generic balanced equation for combustion of a hydrocarbon in the form  C_xH_y + αO₂ βCO₂ + γH₂O

More like

C_xH_y + (x+^y/₄)O₂ xCO₂ + ^γ/₂H₂O

(that is, two unknowns only).

I am in a bit of hurry so I could make some mistake, but the general idea should be clear).

[Corribus]

Well yes, that's what I got too. I was keeping it general so as not to just give him the solution.

[Big-Daddy]

I'm starting to suspect it is not possible to find x without knowing the volume of O₂ you started with ...

Assume 100 mL.

Assume 200 mL.

Are you getting somewhere, or nowhere?

Let's say V mL. Then V/10 = x + 1/4y, y we know is 4 so x = V/10 - 1.

Doesn't do any good, we can't eliminate V and there's no way of taking any limits usefully. I'd say we're getting nowhere - problem is, this V is volume of oxygen reacted, and we aren't told that value, only that excess oxygen is present.

We can say that it's unlikely x>4 when y=4, in which case we can cap off V at 50 mL. Still doesn't bring us closer to finding x though

[Corribus]

No matter what x is, there are the same moles of gas before and after the reaction (x + 2), so it appears the volume does not change as a function of x - only of y. So I agree, I'm not seeing immediately how you determine x without another piece of information. I will continue to think on it, because maybe I have missed something obvious.