Formula of hydrocarbon from combustion reaction

How can you have an initial volume of 10cm^3 and have it contract by a volume greater than 10cm^3? I guess you mean expands?

How can you have an initial volume of 10cm^3 and have it contract by a volume greater than 10cm^3? I guess you mean expands?

The equation is:

CxHy + (x + y/4)O2 -> (y/2)H20 + xC02

Important facts are:

(1)The volume of the water vapour is negligible at room temperature.
(2)The alkali absorbs all the CO2
(3) As these are all gases at the same temperature by Avogadro's law the number of moles of each would be proportional to their volumes.

Using (2) you could easily deduced the volume of CO2 produced and the value of x.

To find the volume of oxygen used simply use the facts to build an equation and then its simple algebra.

Change in volume = volume of gas produced - volume of gas used

Then solve for y.

First of all, thanks a lot for your quality response!

Now ...
By what you said, then volume of CO2 is 40 cm^3. So how to relate this to moles? Shall I divide it by 24?
Secondly, I still didn't arrive to the method of finding y :/ Could you please help me a bit further.

You remember this question came in our exam right? I don't know how we were supposed to know that the volume of H2O (g) is negligible? and how will it help us in solving the question.

By what I said, you can conclude that am still lost in this question

Yes - a similar question popped up in our exam. Fortunately for me I read through the revision guide just before the exam and they worked out a similar question.

I previously posted this:





In fact you don't need to convert between volumes and moles, my bad! We could instead just use ratios!


CxHy (10) + (x + y/4)O2 (v) -> (y/2)H20 (0) + xC02 (40)


The bold are the volumes.

Use ratios:

C02:CxHy = x = 40/10 = 4

To find the volume of Oxygen used, use this:

Change in volume = volume of gas produced - volume of gas used

Then again use ratios to find the value of (x + y/4) and then solve for y.

I'm sorry for taking you down the wrong route with my first post

Yes - a similar question popped up in our exam. Fortunately for me I read through the revision guide just before the exam and they worked out a similar question.

I previously posted this:





In fact you don't need to convert between volumes and moles, my bad! We could instead just use ratios!


CxHy (10) + (x + y/4)O2 (v) -> (y/2)H20 (0) + xC02 (40)


The bold are the volumes.

Use ratios:

C02:CxHy = x = 40/10 = 4

To find the volume of Oxygen used, use this:

Change in volume = volume of gas produced - volume of gas used

Then again use ratios to find the value of (x + y/4) and then solve for y.

I'm sorry for taking you down the wrong route with my first post

I found this somewhere and it contains a very similar situation, just difference in numbers.
Check:

The only bit i didn't follow is that how did he get a total volume of 110 cm^3 ? What's the total volume in our example?

Thank you once again. I really started getting the point behind it.

Now i want to understand this approach to the problem:



Unfortunately i lost the 3 marks at the exam. I was going to just write it as C4H4 without any working. Now the depressing thing is that they awarded full marks without any working, dammit :/

This question has one piece of additional data that our question lacked - the initial volume of oxygen. With this data the total initial and final volume could be given a value but in our question we should give a variable to the initial amount of oxygen if we want an expression for total initial and final volume.

I don't think this piece of data is necessary for the calculations. What is important is the change in the volume of oxygen (the volume of oxygen used) and this could be calculated without the value for the initial volume or oxygen as we did above

Perfect!

Now to tie things up ...


If the volume contracted by 20 cm^3 and became 40 cm^3 and there was 10 cm^3 of the hydrocarbon, then there must be a total of 60 cm^3 in the beginning of which 10 cm^3 is hydrocarbon, so O2 = 50 cm^3.

CxHy : O2 = 10 : 50 = 1:5

Therefore
x + y/4 = 5
from above calculation, x is 4
so 4 + y/4 = 5
y/4 =1
therefore y=4

Hence hydrocarbon is C4H4.



Correct method? Probably not, coz of excess oxygen?

You forgot the CO2 produced!

The total volume change is -20.

Total volume change = Volume of gas produced - volume of gas used

-20 = Volume of CO2 produced - (volume of CxHy used + volume or O2 used)

Solve for volume of oxygen use and continue

You forgot the CO2 produced!

The total volume change is -20.

Total volume change = Volume of gas produced - volume of gas used

-20 = Volume of CO2 produced - (volume of CxHy used + volume or O2 used)

Solve for volume of oxygen use and continue

10 cm3 of a gaseous hydrocarbon, CxHy, was mixed with excess oxygen and ignited. The total gas volume was measured at room temperature and pressure before and after combustion, and it was found that it had contracted by 20 cm3. On shaking the remaining gases with excess potassium hydroxide solution, the total gas volume contracted by a further 40 cm3.
What is the molecular formula of the hydrocarbon CxHy.


Really can't figure out how to go about this question.