Quiz on empirical and molecular formula

  • #1
ghostanime2001
256
0

Homework Statement


An unknown organic compound was analyzed in a carbon-hydrogen analyzer. The combustion of a sample of the compound produced 35.2 g of carbon dioxide and 18.0 g of water. The molar mass of the compound was found to be 116.28 g/mol. Determine the molecular formula of the compound.


Homework Equations


I am not going to write latex here for every step I reached to getting the answer. By the way, this was a in-class quiz and I forgot to ask the teacher few months ago. The question was worth 7 marks and I got 0/7.

Anyway, I was so confused that I assumed it was a hydrated salt question. The answer I wrote down is C2O4.H2O. Which is obviously wrong. I did this same question this morning and the answer I got was C8H2O which can't be true because this is not chemically possible.

The maximum number of hydrogens in alkanes is 2n+2 with alkenes and alkynes possessing fewer hydrogens. What am I doing wrong ? Please help me!

Thanks
 

Answers and Replies

  • #2
sjb-2812
445
5
How many moles of carbon dioxide did you produce? how many moles of water?
 
  • #3
ghostanime2001
256
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0.8 mol of CO2 and 1 mol H2O (because mass was equal to molar mass)
 
  • #4
sjb-2812
445
5
OK, so from a balanced combustion equation how many moles of carbon and of hydrogen did that some from?
 
  • #5
ghostanime2001
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there are 1 mol of carbon in carbon dioxide so 0.8 mol C, there are 2 mols of hydrogen in water, so 2 mol H.
 
  • #6
sjb-2812
445
5
So you have a ratio of C0.8H2. What is that in whole number terms?
 
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  • #7
ghostanime2001
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0.8/0.8=1 and 2/0.8=2.5 but we want in whole numbers so.. 1x2=2 for carbon and 2.5x2=5 for hydrogen so in whole number ratios C2H5 ?? but that is physically impossible because each carbon can form a maximum of 4 bonds. C2H5 is not possible because the maximum number of hydrogen atoms to 2 carbon atoms is CnH2n+2 so C2H6 and for an alkene is C2H4. What is going on... :(
 
  • #8
sjb-2812
445
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What would the molecular weight of an entity of formula C2H5?
 
  • #9
ghostanime2001
256
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29 g/mol
 
  • #10
sjb-2812
445
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How does this relate to the molecular weight you've been given?
 
  • #11
ghostanime2001
256
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molar mass/empirical formula mass = 116.28 g/mol / 29 g/mol = 4 so 4 x C2H5 = C8H20
 
  • #12
sjb-2812
445
5
Sounds good to me.
 
  • #13
ghostanime2001
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you do know that the maximum number of hydrogens in an alkane is CnH2n+2, alkene CnH2n and alkyne is CnH2n-2. C8H20 does not fit into any of the 3 categories so was this question a test so the teacher wanted to know if we knew how to solve this problem ?
 
  • #14
Yanick
398
22
The compound could also contain Oxygen atoms but because O2 is used to combust the compound you cannot assume all of the Oxygen in the water and carbon dioxide came from you compound. It is not immediately obvious to me how to figure out the Oxygen content of the compound. I have to think about it because I'm busy at the moment (just a quick coffee/PF break).
 
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  • #15
ghostanime2001
256
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I can calculate the number of oxygen atoms but when I try to find out the molecular formula from empirical formula, the number is quite off and not a nice number to work out with (not like a good number like 4).
 
  • #16
Yanick
398
22
How do you calculate the number of Oxygen atoms in your compound?
 
  • #17
ghostanime2001
256
0
I have the number of mols of oxygen atom in say CO2 or H2O so all I have to do is multiply by avagadro's number and add to get the total number of oxygen atoms. I can multiply moles of oxygen, hydrogen or carbon with molar mass of each to get mass of each one respectively. I had trouble with questions like these, questions related to specific proportions of elements in compounds until I saw the relationship with moles of each element with moles of the compound as a whole.
 
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  • #18
Yanick
398
22
Write the chemical equation describing combustion of an organic compound then tell me why that approach won't work.
 
  • #19
ghostanime2001
256
0
for complete combustion, organic compound (with C and H) + O2 --> xCO2 + yH2O. If you create a table, you can see the mol relationship of C, H, O and try to simplify it by dividing by the lowest value of moles to get whole number ratios of C, H and O. Then divide molar mass given in the question with empirical formula mass you will not get a number close to a whole number for example 3.98, instead you will get a number which you can't really round up or round down to a whole number. Instead you will get ratio's that are wonky.
 
  • #20
Yanick
398
22
What assumption, regarding the C and H atoms, do you make that you cannot make about the O atoms? In other words, based on your chemical equation, what can be said about the Carbon and Hydrogen atoms found in the CO2[\sub] and H2[\sub]O that is not valid for the Oxygen atoms? Another hint is: this assumption/statement is the entire basis of combustion analysis of organic compounds.

I mean no offense by the following statement but I feel it need be said anyway because many students, good and bad fall prey to this. You seem like you are just blindly throwing around procedures without actually thinking what you are doing.
 
  • #21
ghostanime2001
256
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As far as the question goes, I am not making any assumption. What assumption are you talking about ?
 
  • #22
Yanick
398
22
The forum policy is to not just give out answers so I suggest you read the section on combustion analysis in a Gen Chem text before proceeding further. One last hint is: ALL of the H atoms in water must come from your compound and ALL of the C atoms in carbon dioxide must come from your compound. Is this true for the O atoms in the products?
 
  • #23
sjb-2812
445
5
Hmm, I'm sure this said hydrocarbon before, maybe I'm misreading. But sorry for leading you down a slightly false path.
 
  • #24
ghostanime2001
256
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Yanik I really have no idea what you're talking about
 
  • #26
ghostanime2001
256
0
I understand that, I can give you the masses of carbon and hydrogen contained in the original sample. But there is no information if the compound has oxygen or not. Are you saying that there is oxygen in the sample too ? The problem statement just said organic substance, so I assume it has ONLY carbon & hydrogen.
 
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  • #27
Yanick
398
22
EDIT: Can you now tell me why we cannot determine the O content in the manner which you proposed?

The information given is that it is an organic compound. This means it can be made up of C, H, O, N etc. Combustion analysis, in a typical question like this, does not allow one to know if or how many hetero-atoms (except O) are present. Given that they tell you the products are water and carbon dioxide, you can assume that the compound has C, H and maybe O. Otherwise you'd get other combustion products of N, S etc. So we are left to decide if we have only a hydrocarbon or an O containing hydrocarbon. Seeing as the assumption that we have only a hydrocarbon, without O, gives us a molecular formula which is impossible we are left to assume that the compound should contain O atoms. Then we are left with the task of determining the relative amount of O content in the compound, but I don't think there is a way to figure that out based on the information provided. Typically the O content is calculated by using conservation of mass but that requires knowing the mass of the sample that was combusted.

I hope I didn't give you the false impression that I know how to solve the question exactly, I was only trying to get you to understand the thought process. I'm inclined now to believe that there is not enough information in the problem to get an answer. There may be some trick, or what not, which may give an answer. Guess and check might work. You know the relative amounts of C and H in your compound and you know the molar mass of the compound. It may be possible to try different molecular formulas where you keep the C and H ratio but vary the O content until you get the correct molecular weight. It is extremely annoying but maybe possible.

Are you sure that this was the question exactly as written?
 
  • #28
ghostanime2001
256
0
yes, I have typed the question exactly as it was written on my test paper. I was thinking that too, I am not given the mass of the unknown sample, so I can't find the mass of oxygen. The only way I know to find the mass of oxygen is to add the masses of oxygen in carbon dioxide and water together.

vary the O content until you get the correct molecular weight
what you mean by that ?
 
  • #29
ghostanime2001
256
0
The mole ratios I come up with if I include oxygen in addition to carbon & hydrogen is C : H : O = 1 : 2.5 : 3.25 so if I multiply each by 4 to get whole numbers, I get 4 : 10 : 13 (C4H10O13), which is just absurd!

There are more oxygen atoms than carbon and hydrogen atoms ! :(
 
  • #30
Yanick
398
22
That is because you are CANNOT assume that ALL the Oxygen atoms in your products are from the unknown compound.

Before we proceed further I need you to understand why you can't count the O atoms in the products (that is the water and carbon dioxide) as coming only from your organic compound. I'd like you to read the link I posted, or a Gen Chem textbook or any other resource you'd like and explain to me why you cannot assume the Oxygen atoms in the water and carbon dioxide are ALL from the unknown organic compound. This is the first step that you need to get straight in your head.

Before I help you any further I want you to explain to me the basic principles behind combustion analysis. You are still blindly throwing around procedures without thinking about what you are doing.
 
  • #31
ghostanime2001
256
0
ok I understand why now, because dry oxygen gas was used to burn the unknown sample in the analyzer
 
  • #32
epenguin
Homework Helper
Gold Member
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Not so obvious. I agree the C/H proportions give you C8H20 which also fits the molar mass nicely. But as the OP said there are no C8 compounds that can fit that ratio. There is too much H.

But even if you have oxygen in the compound, it doesn't help as far as I can see. If you replace any C-H with C-O-H you don't increase the H/C ratio. So i see just two ways to increase the H/C. One is by water of crystallisation. But then to be plausible you have to incorporate O into the molecule as well. Considering the molar mass I think this needs to be a C4 molecule in which case you could fit the C/H ratio even without the water of crystallisation. There is an outside possibility with a C6 molecule and water, but I can't see anything that adds up in either case.

I am taking it that any molar mass measurement that works at all will give you the MMass of compound without the water, and that we can take the accuracy of the measurements given as good.
 
  • #33
Yanick
398
22
ok I understand why now, because dry oxygen gas was used to burn the unknown sample in the analyzer

That is good that you get this because at least you got something from this situation.

Not so obvious.

I am glad that I'm not the only one who is stumped. I haven't been able to figure it out myself.

Nothing that I've tried has seemed to work. Anything beyond C6 doesn't fit the 2n+2 criteria. Trying C4H10 gets ugly numbers.

Maybe there is supposed to be some kind of other hetero-atom, but that seems like a silly possibility.
 
  • #34
ghostanime2001
256
0
looks like we are all stumped on this question. Where do we go from here ?
 
  • #35
epenguin
Homework Helper
Gold Member
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Maybe I've got it! Maybe we were meant to bring in something that was mentioned and then forgotten - nitrogen. When I include the possibility of N and use the well-known approximate whole number atomic masses there is a solution that I think is unique.

The only trouble is when I calculate it using accurate atomic masses I have disagreement with the given 116.26 by about 0.1 - a whole 1 in 1,000 discrepancy!:blushing: Perhaps the OP could try the calculation with this idea, or someone else check the figures, or maybe I've overlooked some other solution?

That incredibly accurate molar mass figure given makes the problem look to me a contrived one. Surely a figure that accurate could not be result of experimental measurement; one would only calculate that figure once one knew the answer?
 
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