1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Combustion Reactions

  1. Jan 31, 2014 #1
    $$C_{5}H_{12} + O_{2} → ... $$

    What would be the product? Please show in steps how you're getting your answer, and what oxidation state you're using for C, I don't understand why Carbon and Hydrogen are bonded in this ratio, please explain. Also, please provide any advice/strategies on figuring out my product from using oxidation states e.g. in Combustion Reactions.
  2. jcsd
  3. Jan 31, 2014 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    C5H12 is pentane, a hydrocarbon, specifically an alkane. You know, similar to the stuff that gasoline and diesel fuel are composed of.


    I've given you a hint: per the forum rules, you've got to do more work than this to get further help.
  4. Jan 31, 2014 #3
    I do not need help on the specific equation, simply combustion equations in general. Please solve a similar combustion equation in steps so that I can see how they are done. I just don't understand why Hydrogen and Carbon are in the ratio they are in and the oxidation state of Carbon in such an equation, because of the ratios I am encountering, I am struggling to derive the oxidation state to form the product. I do not believe that there is anything in the forum rules that prevents you from explaining a concept of ratio and bonds.
  5. Jan 31, 2014 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Frankly, I don't understand your confusion. Carbon doesn't have a multiplicity of valence states, like a transition metal. Pentane and similar hydrocarbons are composed of a chain of carbon atoms with hydrogen atoms attached to fill in the empty valence spots in the outer electron shell. These are covalent, not ionic, bonds.

    Pentane structure:

    CH3 - CH2 - CH2 - CH2 - CH3 (all carbon atoms have 4 electrons in the outer shell)

    If you will read the attached article on pentane, you will find information about the combustion of this substance.

    As someone who has posted at PF before, you are aware of the rules. We are not allowed to show you step by step solutions to problems, especially if you show no work attempt at providing you own solution.

    I don't know why you are hung up on these mysterious 'ratios', but, in general, combustion of hydrocarbons results in two very common substances. These substances flow in and out of your body with every breath.
  6. Jan 31, 2014 #5
    Carbon can have an oxidation state of -2,+2, or +4.
    Last edited: Jan 31, 2014
  7. Jan 31, 2014 #6
    $$ C_{5}H_{12} + O_{2} → $$
    $$ C_{5}^{5x+}H_{12}^{12-} + O_{2}^{4-} $$
    $$ C_{5}^{5x+}H_{12}^{12-} + O^{2-} O^{2-}$$
    $$ C_{5}^{5x+}O^{2-} + H_{12}^{12-} O^{2-}$$

    Can someone please explain how the H2O forms here. This is a known reaction, I am not seeking a solution to the problem. I simply want to know how this happens; my oxygens and hydrogens have negative charges. Can someone please explain this concept and point out my error? I'm assuming that I'm dealing with a Carbon that has an oxidation state of +4, simply because it will be the only state that will form the CO2 in this case, but is there a way that I can figure out this element's charge without this assumption? Thank you.
    Last edited: Jan 31, 2014
  8. Jan 31, 2014 #7


    User Avatar

    Staff: Mentor

    Oxygen in O2 has oxidation number of 0, not -2. If it were -2, oxygen molecule would have a -4 charge, it doesn't.

    Applying oxidation numbers won't get you far - that's not how things happen. Oxidation numbers are just an artificial accounting device, they don't exist in reality.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted