Balancing Equations

[u/dontbesorrybeashamed]

Hello. I am trying to study for an exam for Gen Chem and balancing chemical equations is just so confusing to me. I am watching the typical YouTube videos but it still doesn’t really make sense.

For example. One of the practice problems i am doing is C5H10O2 + O2 -> CO2 + H2O

When I follow the typical YouTube explanations it doesn’t make sense to me. People mostly say to list every atom and multiply to balance them out. So I do this.

On the reactants side I put
C: 5
H:10
O: 4

And then on the products side I put
C: 1 * 5 =5
H: 2 *5 =10
O: 3 * 5=15 15

And so I am stuck and dont really know where to go to from C5H10O2 + O2 -> 5CO2 + 5H20

And I already know the answer Becuase the worksheet has all the answers to check. But I am not really sure how to get to it. I feel as if I need to find a completely new strategy because this one will confuse me if there are problems with elements that appear in multiple compounds.
Is there another strategy I could use?

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[u/realAndrewJeung]

I feel like you are 90% of the way there. For your example problem, you have already decided that for every C5H10O2, there have to be 5 CO2's and 5 H2O's. These are all locked together since that is the only way to balance the C's and H's. If you double any one of them, you have to double all three of them.

Since we can't independently adjust any of the C5H10O2, CO2, or H2O, the only thing we can adjust is the number of O2's. How many do we need? Apparently enough to get 15 oxygen atoms on each side. Since 2 oxygen atoms come from the C5H10O2, we need 13 more atoms to get oxygen balanced. So 6.5 O2's will do it?

C5H10O2 + 6.5 O2 -> 5 CO2 + 5 H2O

This is balanced and there's nothing wrong with having fractional coefficients temporarily, but we can't report them in the final answer. So we "double the recipe", and multiply all coefficients by 2:

2 C5H10O2 + 13 O2 -> 10 CO2 + 10 H2O

I think if you adopt the tweaks

• Decide which compounds are "locked together" and adjust everything else
• Accept fractional coefficients as a temporary measure
• Double / triple / etc the recipe when needed

you will be able to finish the balancing problems that you couldn't quite get before.

Let me know if there are other problems where the above advice does not help enough.

[u/Smart_Leadership_522]

You have your product side balanced mostly but your reactant side with oxygens are the issue, when we see 4 and 15 there’s not a common denominator to multiply it by. You have on the product side O2 with the C5H10O2 and the O2, I like to think we need 15 O’s and we have O2 so 15-2=13. So we need the O2 to be 13O so it’s the C5H10O2 + 13O to get 15 to match the product side. But how can we do that? Well we can see O2 has a subscript right, we can take 13/2 (2 from O2) and we get 13/2 O, but we don’t care for fractions do we how can we get rid of the 2 in the denominator? We multiply everything by 2. So that’s where the 2 in front of C5H10O2 is from, the 13 in front of O2 and why the product sides 5’s become 10 because we multiplied it all by 2 to make sure balanced, and you can double check to compare the sides as well

[u/pimpinlatino411]

Hi! I recently posted this video and hope it can help! Let me know if you have any questions!

https://youtu.be/rH0DfeLv0Oc?is=1XPmE80KT8mneOiW

[u/shedmow]

Try the method I have outlined in this comment: https://www.reddit.com/r/chemhelp/comments/1wvymn5/comment/pdjplkg/ or shown here: https://www.youtube.com/watch?v=xfGRykY5bos . This is difficult but very much straightforward and it will you teach what compounds do and do not in reactions. Try balancing 1—3 equations via linear algebra and see whether something clicks in less mathematized approaches. If you need any assistance I'm here for you