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Balancing reaction equations

  1. Aug 20, 2011 #1
    I'm in my 3rd year of a chemistry course so its fairly ridiculous that I can't balance equations. I can balance simple equations in my head in seconds but what I need is an algorithm that I can use to balance any equation I will ever encounter. Can any of you teach me a method for balancing equations that I will be able to apply to all reaction equations that I ever come across in chemistry?
     
  2. jcsd
  3. Aug 20, 2011 #2
    There is no one size fits all approach. You just need to do many problems to gain experience. Sometimes a bit of trial and error is involved. Try this out.

    Butane undergoes complete combustion to yield carbon dioxide and water:

    C4H10 + O2 => H20 + CO2

    Can you balance this?
     
  4. Aug 20, 2011 #3
  5. Aug 21, 2011 #4

    Borek

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  6. Aug 23, 2011 #5
    No. In my summer exams I had a Hess' Law question that I couldn't even start because I couldn't balance the equation for the combustion of acetaldehyde.

    EDIT: I looked it up and I see now that when I have trouble balancing 1 element like that I just put a fraction beside it then multiply everything by whats under the fraction. Knowing that should make things a whole lot easier for me from now on. I woulda had no trouble with that Hess' Law question if I just knew how to balance that equation lol.
     
    Last edited: Aug 23, 2011
  7. Aug 23, 2011 #6
    Ok the only way you will learn to balance equations is to practice by doing problems. There is no silver bullet, there is no algorithm, you just have to practice over and over. If you can't balance equations, you won't get far in chemistry (I think you are aware of this). Look in your general chemistry textbook and work problems. Look on google for practice problems. Do whatever you have to do to learn this.

    And by the way, expect a little trial and error with these problems.

    Ok lets look at the combustion of butane:

    C4H10 + O2 => H2O + CO2

    ON THE LEFT SIDE of the equation we have:
    4 atoms of CARBON
    10 atoms of HYDROGEN
    2 atoms of OXYGEN

    ON THE RIGHT SIDE of the equation we have:
    2 atoms of HYDROGEN
    1 atom of CARBON
    3 atoms of OXYGEN

    I'll start by balancing CARBON on BOTH sides of the equation:
    C4H10 + O2 => H2O + 4CO2

    **so now we have 4 CARBON atoms on both sides, but have 2 atoms of OXYGEN on the LEFT and 9 OXYGEN on the RIGHT

    Balancing the OXYGEN on BOTH sides:
    C4H10 + (9/2)O2 => H2O + 4CO2

    **so now on the LEFT we have 9 OXYGEN (because 9/2 X 2 = 9) and on the RIGHT we have 9 OXYGEN (1 + 8 = 9)

    now we have to balance the HYDROGEN on both side:

    C4H10 + (9/2)O2 =>

    5H2O + 4CO2

    **so now we have 10 HYDROGEN on each side. But we have a problem, on the RIGHT we have 13 OXYGEN on the RIGHT side and on the LEFT we have 9 OXYGEN.

    this next step is really important: DON'T FREAK OUT!!!!!!!!!!!!!!!!!!!!!

    Adding 13 OXYGEN to the LEFT:
    C4H10 + (13/2)O2 =>

    5H2O + 4CO2

    So now we look at our equation:

    LEFT SIDE: 4 CARBON, 10 HYDROGEN, 13 OXYGEN
    RIGHT SIDE: 4 CARBON, 10 HYDROGEN (5 x 2 = 10), 13 OXYGEN (5 + ((4 x 2)) = 13)

    So the law of conservation of mass is satisfied.


    If you have any chemistry questions please ask me. You can send me private messages. I am trying to do good things with my life and this chemistry degree I spend so much money on so please ask for help, it makes me feel good to share information.
     
  8. Aug 24, 2011 #7

    Borek

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    Not true. Algorithm exists and was even already mentioned in this thread. How do you think equations are balanced by programs like Equation Balancing and Stoichiometry calculator? Algorithmically, using algebraic method. Whether this method is suited for manual balancing is a completely different problem.

    Agreed. For a training in balancing equations go to

    http://www.chemistry-quizzes.info/

    You will need to register (for free), then there are several hundred practice problems, grouped into easy, hard, insane and redox. After logging you just select a section from the main page - and try.
     
  9. Aug 24, 2011 #8
    Thanks c03rcion. What I was struggling with was balancing the oxygens at the end, I didn't think of using a fraction to balance it then multiplying every coefficient by the denominator of the fraction. I've been practicing a fair bit and now I have no trouble with these combustion equations. I'm gonna look for some worst case scenario equations and see if I can balance them with this method. That algebraic method Borek mentioned does look like its a silver bullet algorithm but its pretty complicated, I haven't figured out how to use it yet. I send you a PM c03rcion. Right now I badly need some assistance with physical chemistry related fields.
     
    Last edited: Aug 24, 2011
  10. Aug 24, 2011 #9

    Borek

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    Please note: by forum rules PMs should be not used for asking for help.

    Apart from the fact that someone may not want to get such PMs (not the case here), it is better to make your questions visible to everyone, this way you may get an answer much faster, as you don't rely on this one, particular person. Besides, in the case someone trying to help is wrong, if others watch the thread they have a chance to correct the mistake.
     
  11. Aug 24, 2011 #10
    Yeah I'll post the questions in the forums anyway, that way I'll get explanations from more than 1 person. Borek thats a brilliant site you started there. I was thinking about making a site like this for a long time but didn't have the time to get it started. If you need any help I'm fairly experienced with PHP, SQL and javascript and have put a lot of thought into how to build online chemistry learning tools. For example I can contribute new questions and write the scripts for dealing with them. I can't post on the forum for some reason though. I don't see any reply or new thread buttons.
     
  12. Aug 24, 2011 #11

    Borek

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    I see you tried reaction balancing. Have you checked if you can post AFTER trying to answer the questions?

    Edit: OK, I see you already posted. Forum was extensively spammed by bots, now - to be able to post - you need to try to answer at least one question. It works :smile:
     
    Last edited: Aug 24, 2011
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