How can I balance an equation with odd numbers of certain elements?

[FTCC.student]

I know this is elementary stuff, but I'm in Gen Chem 1 and this is the only thing I am having trouble with. I am fine balancing most equations, but when there are odd numbers of certain elements I get sort of confused. For example:

C₂H₆ + O₂ $\rightarrow$ CO₂ + H₂O

The Attempt at a Solution

I have tried several times, but the closest I could get was:

2C₂H₆ + 5O₂ $\rightarrow$ 2CO₂ + 6H₂O

It is the Oxygen andHydrogen being uneven that is screwing me up. Right away I realize I have 3 Oxygen on the right and 2 on the left, so the first thing I tried to do was make both sides have 6 oxygen...

C₂H₆ + 3O₂ $\rightarrow$ CO₂ + 4H₂O

But then that leaves me with 8 Hydrogen on the right, and only 6 on the left...I can't think of any ways to get that 6 to equal 8! Then I tried for 12 Oxygen on both sides instead of 6...then my problem is still Hydrogen.

So maybe y'all see what I'm confused about...like I said, this is the only thing I am having trouble with so far in Gen Chem is balancing this specific type of equation, and we have our first test on Tuesday! So hopefully someone can help :)

 

[Borek]

Let's start with 1 molecule of C₂H₆. It is obvious we will produce two CO₂ and three H₂O:

C₂H₆ + xO₂ → 2CO₂ + 3H₂O

Can you find x? It will be a fraction now - but you can multiply everything by some small integer to get rid of the denominator...

 

[FTCC.student]

It would be 3$\frac{1}{2}$, yes? So I should then multiply by a small integer to get rid of denom, multiply all by 2? Does that include the 3.5?

 

[Borek]

Yes, multiply 3.5 as well.

 

[DeeNos]

I completely understand your frustration with balancing equations. It may seem like a simple task, but it requires careful attention to detail and practice to master. In this particular equation, the key is to first balance the carbon and hydrogen atoms, and then balance the oxygen atoms.

To start, let's look at the carbon atoms. On the left side, we have 2 carbon atoms from the C2H6 molecule. On the right side, we have 1 carbon atom from the CO2 molecule. To balance these, we need to add a coefficient of 2 in front of the CO2 molecule, giving us 2CO2. Now we have 2 carbon atoms on both sides.

Next, let's look at the hydrogen atoms. On the left side, we have 6 hydrogen atoms from the C2H6 molecule. On the right side, we have 2 hydrogen atoms from the H2O molecule and 4 hydrogen atoms from the 2CO2 molecules (2 x 2 = 4). To balance these, we need to add a coefficient of 3 in front of the H2O molecule, giving us 3H2O. Now we have 6 hydrogen atoms on both sides.

Finally, let's look at the oxygen atoms. On the left side, we have 6 oxygen atoms from the O2 molecule. On the right side, we have 2 oxygen atoms from the H2O molecule and 4 oxygen atoms from the 2CO2 molecules (2 x 2 = 4). To balance these, we need to add a coefficient of 3 in front of the O2 molecule, giving us 3O2. Now we have 6 oxygen atoms on both sides.

The final balanced equation is:

C2H6 + 3O2 → 2CO2 + 3H2O

Remember, balancing equations takes practice and patience. Keep working at it and you will get the hang of it. Good luck on your test!