Sixth Example

This sixth equation introduces another dilemma. We will see what it is in a moment. The way of approaching this equation is very much the same as the previous one.

The carbon and hydrogen on the left split up and go into the other compounds, so let's start with that.

C₄H₁₀ + O₂ CO₂ + H₂O
4 C, 10 H, 2 O	1 C, 2 H, 3 O

C₄ will have to become 4 CO₂ to give four carbons on each side. If we start with ten hydrogens in C₄H₁₀, we have to end with ten hydrogens. They come in pairs in H₂O, so it is going to be 5 H₂O.

C₄H₁₀ + O₂ 4 CO₂ + 5 H₂O
4 C, 10 H, 2 O	4 C, 10 H, 13 O

That takes care of the carbon and the hydrogen. What about the oxygen? We have 4 CO₂'s, so that is eight oxygen atoms and 5 H₂O's is five more oxygen atoms. That is a total of 13 oxygen atoms, so we need 13 oxygen atoms on the left. This is where the problem comes in. Thirteen is an odd number. The formula O₂ shows that oxygen comes in pairs. You cannot have an odd number arranged in pairs. It just doesn't work out.

There are a couple ways of getting around this dilemma, and I think this is the simplest. We need 13 oxygen atoms, that would be the same as having 6 1/2 O₂'s. We can go ahead and write that down for the moment.

C₄H₁₀ + 6˝ O₂ 4 CO₂ + 5 H₂O
4 C, 10 H, 13 O	4 C, 10 H, 13 O

However, we don't like to have fractional numbers in a balanced equation. Just think about what O₂ means. It means that oxygen comes as two atoms attached to one another. If that is the way they come, then you can't have half of two of them. You have to have two of them at a time. So now to get rid of that fraction. What we have now is one C₄H₁₀ plus 6˝ O₂ becomes 4 CO₂ plus 5 H₂O. We have the right ratio of numbers, we just don't have the right numbers.

We can clear the fraction by doubling all of the numbers going across. That gives us 2 C₄H₁₀ + 13 O₂

8 CO₂ + 10 H₂O. That, I think, is the easiest way to balance this type of equation. We can balance the equation by using a fractional number, so we do that first. Then clear the fraction.

2 C₄H₁₀ + 13 O₂ 8 CO₂ + 10 H₂O
8 C, 20 H, 26 O	8 C, 20 H, 26 O

Alternate Approach

There is another way of looking at that problem which is a little bit better conceptually, but not as easy mechanically.

That is to realize that if, indeed, 1 C₄H₁₀ is going to take 6˝ O₂'s--but we cannot have ˝ of an O₂--then we are going to have to have two C₄H₁₀'s.

C₄H₁₀ + 6˝ ? O₂ 4 CO₂ + 5 H₂O
4 C, 10 H, 13 O	4 C, 10 H, 13 O

We must have another one to take up the extra half-molecule of O₂.

2 C₄H₁₀ + ? O₂ ? CO₂ + ? H₂O
8 C, 20 H, 2 O	1 C, 2 H, 3 O

So, if we start off with a two in front of C₄H₁₀, then we have eight carbons and we need to have an eight in front of CO₂. Also, we have 20 hydrogens, so we need a ten in front of H₂O.

2 C₄H₁₀ + O₂ 8 CO₂ + 10 H₂O
8 C, 20 H, 2 O	8 C, 20 H, 26 O

Next add up the oxygens on the right. The 8 CO₂'s contain 16 oxygens, and the 10 H₂O's which contain 10 oxygen. That gives us a total of 26 oxygens on the right. They come in pairs on the left so we need 13 pairs. Note that whichever way you go about balancing the equation you get the same result.

2 C₄H₁₀ + 13 O₂ 8 CO₂ + 10 H₂O
8 C, 20 H, 26 O	8 C, 20 H, 26 O

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