Combustion Reactions: C5H12 + O2 - Steps & Oxidation States

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SUMMARY

The discussion centers on the combustion reaction of pentane (C5H12) with oxygen (O2) and the associated oxidation states of carbon and hydrogen. Participants clarify that pentane is a hydrocarbon with a specific structure where carbon atoms are bonded to hydrogen atoms to satisfy valence requirements. The oxidation state of carbon in combustion reactions is typically +4, leading to the formation of carbon dioxide (CO2) and water (H2O) as products. Understanding these concepts is crucial for accurately predicting the products of combustion reactions.

PREREQUISITES
  • Understanding of hydrocarbon structures, specifically pentane (C5H12).
  • Knowledge of oxidation states and their application in chemical reactions.
  • Familiarity with combustion reactions and their typical products (CO2 and H2O).
  • Basic grasp of covalent bonding and valence electron configurations.
NEXT STEPS
  • Research the combustion reaction of other hydrocarbons, such as hexane (C6H14).
  • Learn about the concept of oxidation states in organic compounds.
  • Study the principles of balancing combustion reactions.
  • Explore the role of covalent bonds in hydrocarbon structures and their implications in combustion.
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Chemistry students, educators, and anyone interested in understanding combustion reactions and the behavior of hydrocarbons in chemical processes.

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$$C_{5}H_{12} + O_{2} → ... $$



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

http://en.wikipedia.org/wiki/Pentane

I've given you a hint: per the forum rules, you've got to do more work than this to get further help.
 
SteamKing said:
I've given you a hint: per the forum rules, you've got to do more work than this to get further help.

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.
 
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.
 
SteamKing said:
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.
SteamKing said:
Carbon doesn't have a multiplicity of valence states.
Carbon can have an oxidation state of -2,+2, or +4.
 
Last edited:
$$ C_{5}H_{12} + O_{2} → $$
$$ C_{5}^{5x+}H_{12}^{12-} + O_{2}^{4-} $$
$$ C_{5}^{5x+}H_{12}^{12-} + O^{2-} O^{2-}$$
$$ C_{5}^{5x+}O^{2-} + H_{12}^{12-} O^{2-}$$

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:
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.
 
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