# Find the formula of a hydrocarbon using combustion data

• Chemistry
• lioric
In summary, the hydrocarbon has a mass of 13.5g/L and the water has a mass of 1.50g/L. The density of the hydrocarbon is 13.5g/L, which is greater than the density of the water (1.50g/L). Using the relative density per He, the hydrocarbon is 13.5 times more dense than the water. Using 1 mole of HC and 1 mole of He, the volumes of the hydrocarbon and water are the same. The molar mass of the hydrocarbon is greater than the molar mass of the water.
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?

Use it to calculate molar mass.

Borek said:
Use it to calculate molar mass.
I know that but what do I use as volume?

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.

Borek said:
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

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.

Merlin3189 said:
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

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?

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

chemisttree
Merlin3189 said:
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?

So what does the RD tell us about the masses of 1 mole of each?
Their volumes are sames at the same conditions

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

chemisttree
Merlin3189 said:
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?

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

lioric
Merlin3189 said:
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

lioric said:
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.

chemisttree said:
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

## 1. How do you determine the formula of a hydrocarbon using combustion data?

The formula of a hydrocarbon can be determined by analyzing the products of its combustion. The ratio of carbon to hydrogen in the products can be used to calculate the empirical formula of the hydrocarbon.

## 2. What is combustion data and how is it collected?

Combustion data refers to the information obtained from the process of burning a substance. This data can be collected by measuring the amount of heat released, the products formed, and the reactants consumed during the combustion reaction.

## 3. Can the formula of a hydrocarbon be determined using only the mass of the products?

No, the formula of a hydrocarbon cannot be determined using only the mass of the products. The mass of the reactants and the products must be measured in order to calculate the ratio of carbon to hydrogen and determine the empirical formula.

## 4. How accurate are the results when determining the formula of a hydrocarbon using combustion data?

The accuracy of the results depends on the precision of the measurements taken during the combustion reaction. It is important to use precise and accurate instruments to ensure reliable results.

## 5. Are there any limitations to using combustion data to determine the formula of a hydrocarbon?

Yes, there are limitations to using combustion data. This method assumes that all of the carbon and hydrogen in the hydrocarbon are converted to carbon dioxide and water, respectively. However, incomplete combustion or impurities in the sample can affect the accuracy of the results.

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