Find the formula of a hydrocarbon using combustion data

[lioric]

Homework Statement

Find the formula of a hydrocarbon. Combustion of that hydrocarbon resulted
in the formation of 2 L of carbon dioxide and 1.205 g of water. The relative density
of the hydrocarbon per helium is 13.5.

Relevant Equations

number of moles of any gas =22.4L
number of moles = mass/ molar mass
density = mass / volume

number of moles of CO2 =0.089moles (using 2/22.4)
number of moles of water = 0.067 (using 1.205/18)
I know that all the carbon from the hydrocarbon is in the CO2
and all the hydrogen from the hydrocarbon is in the water and water creates x2 hydrogens

so number of moles of C : H = 0.089 : 0.067x2
0.089 : 0.134
so I get 1 : 1.50
which is like 2 : 3

now what do I do with the relative density?

 

[Borek]

Use it to calculate molar mass.

 

[lioric]

Use it to calculate molar mass.

I know that but what do I use as volume?

 

[Borek]

Doesn't matter, it will cancel out. It is ratio that matters. Do the calculations using symbols first.

You can start with 1 L if it will make it easier to grasp the concept, just don't crunch the numbers before you derive the final formula, it will actually save you time and remove potential source of errors, as most of calculations are not necessary.

 

[lioric]

Doesn't matter, it will cancel out. It is ratio that matters. Do the calculations using symbols first.

You can start with 1 L if it will make it easier to grasp the concept, just don't crunch the numbers before you derive the final formula, it will actually save you time and remove potential source of errors, as most of calculations are not necessary.

Ok. So density of the hydrocarbon is 13.5g/L
So 13.5g/L= mass/L
So that means the mass of the hydrocarbon is 13.5g/L
moles = mass/ molar mass
But what do I use for moles

 

[Merlin3189]

I'd have thought relative density wrt (=per?) He, means the hydrocarbon gas is 13.5 times more dense than He gas at the same temp & press.
He isn't 1g/litre, more like 0.178 g/l, which gives 2.40 g/l, but I think you don't really need to know that.

But I might think of Avogadro.

 

[lioric]

I'd have thought relative density wrt (=per?) He, means the hydrocarbon gas is 13.5 times more dense than He gas at the same temp & press.
He isn't 1g/litre, more like 0.178 g/l, which gives 2.40 g/l, but I think you don't really need to know that.

But I might think of Avogadro.

I think since the question says that it's relative density per helium we need to take that into account.
Thank you

But I'm still stuck with the moles that I'm going to use

 

[Merlin3189]

Sorry about the "relative density per He". I thought I'd understood that to mean "RD with respect to He", but it's a new term to me. If I've got it wrong, can you explain what it means please?

We can take RD into account without working out absolute density in g/l. That's why I said you don't need to find 2.40 g/l.

How about using 1 mole of HC and 1 mole of He?
What does Avogadro say about their volumes?

So what does the RD tell us about the masses of 1 mole of each?

 

[lioric]

Sorry about the "relative density per He". I thought I'd understood that to mean "RD with respect to He", but it's a new term to me. If I've got it wrong, can you explain what it means please?

We can take RD into account without working out absolute density in g/l. That's why I said you don't need to find 2.40 g/l.

How about using 1 mole of HC and 1 mole of He?
What does Avogadro say about their volumes?

So what does the RD tell us about the masses of 1 mole of each?

Their volumes are sames at the same conditions

 

[Merlin3189]

So 1 mole of CxHy has the same volume as 1 mole of He.
RD tells us 1 volume of CxHy has 13.5 times the mass of 1 volume of He
So 1 mole of CxHy has 13.5 times the mass of 1 mole of He.

If you know the molar mass of He, then you can work out the molar mass of CxHy.

My reference to Avogadro was to his idea that equal volumes of gas contain the same number of molecules. Since all the mass of a gas is in the molecules, not the space between, then if one is N times as dense as the other, then its molecules must be N times more massive. So you can go straight from RD to relative molecular mass, without bothering with volumes, moles

 

[lioric]

So 1 mole of CxHy has the same volume as 1 mole of He.
RD tells us 1 volume of CxHy has 13.5 times the mass of 1 volume of He
So 1 mole of CxHy has 13.5 times the mass of 1 mole of He.

If you know the molar mass of He, then you can work out the molar mass of CxHy.

I see
1 mole of He has a mass of 4.002g
1 mole of CxHy has 4.002g x 13.5= 54.027g
If the empirical formula is C2H3 =27g
CxHy = 54.027
54.027/27= 2.001
2x C2H3
C4H6

Am I ok?

 

[Merlin3189]

That's my logic.
I await a chemist to tell us what on Earth this stuff is!

 

[lioric]

That's my logic.
I await a chemist to tell us what on Earth this stuff is!

Same here. That logic makes a lot of sense

 

[chemisttree]

Same here. That logic makes a lot of sense

It should because it’s correct, whether you’re a chemist or not. I am a chemist and I’ve never seen the term “relative density” used like this before. It is a clever way to test across several concepts. And using Avogadro’s analysis is precisely how to use the relative density information.

Concepts tested:
1) Mole concept
2) convert from volume to moles
3) convert from mass to moles
4) avogadro relationship of moles to volume
5) algebraic concept of relative ratios
6) distinguishing between empirical and molecular formula
7) concept of density
8) knowledge of combustion reaction

It’s a good question.

 

[lioric]

It should because it’s correct, whether you’re a chemist or not. I am a chemist and I’ve never seen the term “relative density” used like this before. It is a clever way to test across several concepts. And using Avogadro’s analysis is precisely how to use the relative density information.

Concepts tested:
1) Mole concept
2) convert from volume to moles
3) convert from mass to moles
4) avogadro relationship of moles to volume
5) algebraic concept of relative ratios
6) distinguishing between empirical and molecular formula
7) concept of density
8) knowledge of combustion reaction

It’s a good question.

Thank you very much for your help
For your information these are from a Russian chemistry paper pre uni