Combustion Reactions: C5H12 + O2 - Steps & Oxidation States

Click For Summary

Discussion Overview

The discussion revolves around the combustion reaction of pentane (C5H12) with oxygen (O2), focusing on the steps to derive the products and the oxidation states of the elements involved. Participants explore concepts related to combustion reactions, oxidation states, and the molecular structure of hydrocarbons.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant requests clarification on the product of the combustion reaction and the oxidation states of carbon and hydrogen, expressing confusion about the ratios in which these elements bond.
  • Another participant identifies C5H12 as pentane, a hydrocarbon, and suggests that the original poster needs to provide more work to receive further help.
  • A participant emphasizes that hydrocarbons like pentane consist of carbon chains with hydrogen atoms filling valence spots, highlighting the covalent nature of these bonds.
  • There is a discussion about the oxidation states of carbon, with one participant stating that carbon can have oxidation states of -2, +2, or +4.
  • A participant presents a series of equations involving oxidation states but expresses confusion about the formation of water (H2O) and the charges of oxygen and hydrogen in the reaction.
  • Another participant corrects the oxidation number of oxygen in O2, stating it is 0, not -2, and argues that oxidation numbers are merely an accounting tool rather than a reflection of reality.

Areas of Agreement / Disagreement

Participants express differing views on the interpretation of oxidation states and the nature of combustion reactions. There is no consensus on the best approach to understanding the ratios of hydrogen and carbon or the application of oxidation states in this context.

Contextual Notes

Some participants note the limitations of relying solely on oxidation states for understanding chemical reactions, suggesting that further exploration of molecular behavior may be necessary.

Who May Find This Useful

This discussion may be useful for individuals interested in combustion reactions, hydrocarbon chemistry, and the application of oxidation states in chemical equations.

vanceEE
Messages
109
Reaction score
2
$$C_{5}H_{12} + O_{2} → ... $$



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.
 
Physics news on Phys.org
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.
 
  • Like
Likes   Reactions: 1 person

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 28 ·
Replies
28
Views
6K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
2
Views
2K
Replies
16
Views
4K
  • · Replies 10 ·
Replies
10
Views
18K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
2
Views
3K
Replies
1
Views
4K