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Non-integer amounts of atoms

  1. Mar 14, 2010 #1
    Hello.

    I have a non-stoichiometric reaction that balances out to:

    C12H23.4 + 12.265(O2) ---> 11.7(H2O) + 6.415(CO2) + 5.585(C)

    The problem started with the compound C12H23.4 which is the published formula for a high-grade kerosene of the form CnH1.95n. So, in the beginning, there exists a non-integer value of hydrogen atoms (which seems crazy), and this propagates through the entire equation to create even MORE "nonsensical" fractional particles.

    What is going on here? How is one to interpret this result?
     
  2. jcsd
  3. Mar 14, 2010 #2
    It seems like the only "solution" is to multiply all by 1000 and divide by 5 to get:
    200(C12H23.4) + 2453(O2) ---> 2340(H2O) + 1283(CO2) + 1117(C)

    But that still leaves the question of the 23.4(H)!? How can you have a .4(H) atom clinging to a carbon atom? It can't really exist to do ANY combining.
     
  4. Mar 14, 2010 #3
    It's an average composition. You're dealing with a mixture of several different substances, and some have more hydrogen than others.
     
  5. Mar 14, 2010 #4
    That makes sense. If so, when dealing with an actual molecule and a real reaction, do you simply round to the nearest integer, always round up...or always round down, once you know what (n) is, and therefore, 1.95(n)?
     
  6. May 10, 2010 #5
    The interpretation of

    H2 + 1/2O2 => H2O

    isn't "one molecule of hydrogen reacts with half molecule of oxygen to form one molecule of water"

    The correct way to read is : " one mole of hydrogen molecules react with half a mole of oxygen molecules to form one mole of water molecules". The same reasoning goes to your equation. So, there are no fractional atoms involved.

    It's a common mistake, though.
     
  7. May 11, 2010 #6

    Borek

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    Yes and no. When there are fractional ceofficients used situation gets out of control as there are no fractional atoms/molecules (no doubt about it), but what you wrote suggests that reaction equations are written only in terms of moles - but I don't think there is anything wrong in interpreting them in terms of atoms/molecules, when all coefficients are integer.

    (BTW: use sub/sup to format formulas).

    No fractional atoms, but this is a different problem. Reaction you wrote can be easily converted to integer coefficients, reaction posted by treddie doesn't have and will never have integer coefficients, as it is only an approximation of the observed case.
     
  8. May 11, 2010 #7

    Borek

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    You don't do neither.

    Imagine you have mixture of ethane C2H6 and ethene C2H4. You burn them in oxygen:

    2C2H6 + 7O2 -> 4CO2 + 6H2O
    C2H4 + 3O2 -> 2CO2 + 2H2O

    Now, if you know your mixture of them is 1/3 of ethane (by mass) and 2/3 of ethene, and you know you have a 1 kg of mixture, you can either split the mass into two substances and calculate results of each reaction separately, then add them, or you can use a trick - add these reactions (in correct proportions!) getting something "reaction equation like" - and use this "reaction equation" for further calculations.

    In this particular case masses are 1:2, so moles are 1:2.14. You multiply the first reaction by 1, second by 2.14 and add them. To make things more complicated, first reaction has to be divided by 2 first, so that we start with coefficient of 1 for ethane:

    C2H6 + 3.5O2 -> 2CO2 + 3H2O

    and second reaction becomes

    2.14C2H4 + 6.42O2 -> 4.28CO2 + 4.28H2O

    Sum is:

    2.14C2H4 + C2H6 + 9.92O2 -> 6.28CO2 + 7.28H2O

    Now, we can go even further and combine both ethane and ethene:

    2.14C2H4 + C2H6 = C4.28H8.56 + C2H6 = C6.28H14.56

    and finally we have "reaction equation"

    C6.28H14.56 + 9.92O2 -> 6.28CO2 + 7.28H2O

    that describes observed stoichiometry of reaction of 1:2 mass by mass mixture of ethane and ethene burnt completely in oxygen. This is just a mathematical trick that simplifies further calculations, and all coefficients and formulas should be not treated the same way they are treated in normal circumstances (there is no compound with formula C6.28H14.56 - this formula nicely describes composition of the mix used) - but they can be used in normal stoichiometric calculations.

    "Reaction equation" you have listed is even more complicated, as it was most likely not calculated from the basic stoichiometry as I did above, but it was determined experimentally - amount of soot is not something that can be easily predicted.
     
    Last edited by a moderator: Aug 13, 2013
  9. May 11, 2010 #8
    @Borek: I agree with you when you say there's no problem interpreting a equation in term of molecules instead of moles when all coefficients are integers. However, all my chemistry teachers have told me that the correct way to read it is using moles.

    My intention with my post was to clarify that a fractional integer doesn't mean fractional molecules. I find using moles a extremely useful way to get around this problem, but each one should follow its tastes for rigor.
     
  10. May 11, 2010 #9
    Then, in the final analysis, what would you do in the real world with
    C6.28H14.56 + 9.92O2 -> 6.28CO2 + 7.28H2O?

    It seems that at some point, this has to be "reduced" to reflect reality by making some changes to the equation. I understand the facility of working with a "perfect" mathematical formula, but it ultimately fails to explain reality unless some changes are made to it. Is the answer to split it all back up somehow, to get back to the original two equations, or would that be impossible by ending up with an infinite amount of possible solutions? Much like 3 + 4 always = 7, but 7 can be broken up into an infinite amount of two values that sum to 7.

    Then again, in most real-world situations, I would suppose the amount of particles actually involved in a reaction would probably never be known to such accuracy that it even matters...it would all reduce to a statistical average. Unless you were dealing with some study of quantum amounts, then it would definitely matter.
     
  11. May 11, 2010 #10
    That C6.28H14.56 is non-standard and I suspect represents an averge which I suppose stoichmetrically would agree with empirical results. However, I've never seen a molecular formula written with non-integer numbers. And the non-integer coefficients in a balanced equation represents moles like [itex]6.28 CO_2[/itex] represents 6.28 X 44 grams of [itex]CO_2[/itex]. It's an empirical expression. The 6.28 does not represent molecules.
     
    Last edited: May 11, 2010
  12. May 11, 2010 #11

    alxm

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    You could probably get an okay approximation of the heat of combustion from that, if you interpolated the hydrocarbon value between, say, C6H12 and C7H16.
     
    Last edited: May 11, 2010
  13. May 11, 2010 #12

    Borek

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    Ah, the joy of answering without reading the thread. Two posts earlier I have shown how and when you can get this equation.

    --
    methods
     
  14. May 11, 2010 #13

    Borek

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    It can be easily used to calculate mass ratio of hydrocarbon mix to oxygen, so you can use it to predict amount of oxygen necessary to burn given mass of of the mixture. It will also let you calculate mass of products. That's what the stoichiometry is about, isn't it?

    Actually it is quite the opposite - this is the "reduced" equation that reflects the reality. Ideal case was two separate equations, but they were difficult to use.

    No. Many mixtures you will deal with have relatively well known composition. Equation you have started this thread with is just an example of such situation:

    Kerosene is produced to meet some parameters, very likely composition is one of them (perhaps not directly, but for sure heat of combustion is one of the parameters, that sets limits to possible composition variation). So you can be sure if you are working with kerosene that meets the standard, it has composition that follows the formula, and it burns according to the reaction given.
     
  15. May 11, 2010 #14
    So then, going back over this thread, would you say that the fraction in C12H23.4 (and similar compounds) results because not all of the Kerosene molecules in the mixture would have the same exact amount of H atoms attached. But that the average comes out to 23.4?
     
  16. May 12, 2010 #15

    Borek

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    Exactly, kerosene is a mixture.

    That was the first answer to your question that you got in this thread - and it was spot on.
     
    Last edited by a moderator: Aug 13, 2013
  17. May 12, 2010 #16
    Well, it took me awhile, but it finally sunk in! :smile:
    Thanks everybody for all of your generous help. Much appreciated!
     
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