Question

I need to balance this equation:
C4H10 + O2 yields CO2 + H2O
I understand how to balance equations, but I keep getting tripped up here. Can someone walk me through it (not just give me the answer)?

Answer 1

Just out of curiosity, are you looking for the purely mathematical approach to this problem? Or is this coming from a Chemistry class where you often take it by trial and error?

Answer 2

Mainly trial and error, I suppose. We just keep tweaking until the coefficients balance.

Answer 3

Got it. There is a way to do it with linear algebra methods, but I'd need to look at it again to remember how it went. >.>

Anyways, what have you tried so far?

Answer 4

I don't know what the principles are in this context; like, should I start with a certain atom, as a rule? e.g. doing oxygen last.

Answer 5

I just kept trying to balance carbon and hydrogen first (separately, obviously). But I end up with, like, 13 oxygens. How do you balance THAT? Lol.

Answer 6

I tried multiplying CO2 on the right side by a coefficient of 4 so the carbon balanced. I got stuck a few steps later.

Answer 7

So, we start:
C4 H10 + O2 yields CO2 + H2O
We balance Carbon first with a coefficient of 4 on the right/
C4 H10 + O2 yields 4CO2 + H2O

Answer 8

Then a coefficient of 5 for H2O?

Answer 9

Yes. Then we do Hydrogen next. The H2O on the left needs 5 times the number of Hydrogen (2*5 = 10) to match the H10.
C4H10 + O2 yields 4CO2 + 5H2O

Answer 10

But that makes oxygen really hard to balance. That's where I got stuck.

Answer 11

On the left, we have 2 Oxygen. On the right, we have 4*2 = 8 in CO2 and 5 in H2O for 13.
It would probably work out better to double that 5 on the right, while doubling the number of CO2 and C4H10 in the process.
2C4H10 + O2 yields 8CO2 + 10H2O
Note we now have 2*4 = 8 carbon on the left, 8 on the right.
10*2 = 20 hydrogen on the left, 2*10 = 20 hydrogen on the right.
now we have 2 oxygen on the left, and 8*2=16 on CO2 along with 10 on H2O.

Answer 12

I mashed three steps into one here, you would start with doubling the number of H2O. That makes you need twice the number of hydrogen on C4H10. Then you need twice the number of CO2 to fix the carbons.

Answer 13

C4 H10 + O2 yields CO2 + H2O <--- 4 CO2
C4 H10 + O2 yields 4CO2 + H2O <--- 5H2O
C4 H10 + O2 yields 4CO2 + 5 H2O <--- double H2O
C4 H10 + O2 yields 4CO2 + 10 H2O <--- we need twice the C4H10 now.
2 C4 H10 + O2 yields 4CO2 + 10 H2O <-- now we need twice the CO2 to balance carbons.
2 C4 H10 + O2 yields 8CO2 + 10H2O

At this point, we should now have an even number of oxygens on the right side.

Answer 14

Sorry, I got distracted. Let me read through your replies.

Answer 15

We have 26 oxygen on the right side now, correct? So we need to multiply the left side O2 atom by...13?

Answer 16

Correct. :)

Answer 17

It seems apt to include a purely Mathematical approach as well, seeing as this is actually the Math section though...
If we call the coefficients: w, x, y, and z;
w C4 H10 + x O2 yields y CO2 + z H2O
Compare the Carbons: 4w = 1y ==> y = 4w
the Hydrogen: 10w = 2z ==> z = 5w
the Oxygens: 2x = 2y + 1z ==> 2x = 2(4w) + 5w
==> x = 13/2 w
This makes the coefficients: w=w, x=13/2 w, y=4w, and z=5w.
The greatest common denominator is 2, so w=2 solves our coefficients!

Answer 18

Oh, no. I didn't mean to post in mathematics. So sorry. o.o I've never really learned my way around this site.

Answer 19

No worries
It is pretty close to a Math question as it is a Chemistry question. ;)

Answer 20

The Chemistry section is here: http://openstudy.com/study#/groups/Chemistry

Answer 21

Sorry I disappeared; I lost connection. I meant to thank you for your help!