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Homework Help: Quiz on empirical and molecular formula

  1. Jul 23, 2013 #1
    1. The problem statement, all variables and given/known data
    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.

    2. Relevant 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!

  2. jcsd
  3. Jul 24, 2013 #2
    How many moles of carbon dioxide did you produce? how many moles of water?
  4. Jul 24, 2013 #3
    0.8 mol of CO2 and 1 mol H2O (because mass was equal to molar mass)
  5. Jul 25, 2013 #4
    OK, so from a balanced combustion equation how many moles of carbon and of hydrogen did that some from?
  6. Jul 25, 2013 #5
    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.
  7. Jul 25, 2013 #6
    So you have a ratio of C0.8H2. What is that in whole number terms?
    Last edited: Jul 25, 2013
  8. Jul 25, 2013 #7
    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... :(
  9. Jul 25, 2013 #8
    What would the molecular weight of an entity of formula C2H5?
  10. Jul 25, 2013 #9
  11. Jul 25, 2013 #10
    How does this relate to the molecular weight you've been given?
  12. Jul 25, 2013 #11
    molar mass/empirical formula mass = 116.28 g/mol / 29 g/mol = 4 so 4 x C2H5 = C8H20
  13. Jul 25, 2013 #12
    Sounds good to me.
  14. Jul 25, 2013 #13
    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 ?
  15. Jul 25, 2013 #14
    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).
    Last edited: Jul 25, 2013
  16. Jul 25, 2013 #15
    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).
  17. Jul 25, 2013 #16
    How do you calculate the number of Oxygen atoms in your compound?
  18. Jul 25, 2013 #17
    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.
    Last edited: Jul 25, 2013
  19. Jul 25, 2013 #18
    Write the chemical equation describing combustion of an organic compound then tell me why that approach won't work.
  20. Jul 25, 2013 #19
    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.
  21. Jul 25, 2013 #20
    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.
  22. Jul 25, 2013 #21
    As far as the question goes, I am not making any assumption. What assumption are you talking about ?
  23. Jul 26, 2013 #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?
  24. Jul 26, 2013 #23
    Hmm, I'm sure this said hydrocarbon before, maybe I'm misreading. But sorry for leading you down a slightly false path.
  25. Jul 26, 2013 #24
    Yanik I really have no idea what you're talking about
  26. Jul 26, 2013 #25
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