# A compound with molecular formula CH_4O burns in air to form carbon dioxide and water. How do you write a balanced equation for this reaction?

Jan 1, 2017

Warning! Long answer! The balanced equation is
$\text{2CH"_4"O" + "3O"_2 → "2CO"_2 + "4H"_2"O}$

#### Explanation:

You follow a systematic procedure to balance the equation.

$\text{CH"_4"O" + "O"_2 → "CO"_2 + "H"_2"O}$

A method that often works is to balance everything other than $\text{O}$ and $\text{H}$ first, then balance $\text{O}$, and finally balance $\text{H}$.

Another useful procedure is to start with what looks like the most complicated formula.

The most complicated formula looks like $\text{CH"_4"O}$. We put a $1$ in front of it to remind ourselves that the coefficient is now fixed.

$\textcolor{red}{1} \text{CH"_4"O" + "O"_2 → "CO"_2 + "H"_2"O}$

Balance $\text{C}$:

We have fixed 1 $\text{C}$ atom on the left-hand side, so we need 1 $\text{C}$ atom on the right-hand side. We put a $1$ in front of the ${\text{CO}}_{2}$.

$\textcolor{red}{1} \text{CH"_4"O" + "O"_2 → color(blue)(1)"CO"_2 + "H"_2"O}$

Balance $\text{O}$:

We can't balance $\text{O}$ yet, because we have two formulas that contain $\text{O}$ and lack coefficients. So we balance $\text{H}$ instead.

Balance $\text{H}$:

We have fixed 4 $\text{H}$ atoms on the left-hand side, so we need 4 $\text{H}$ atoms on the right-hand side. We put an $2$ in front of the $\text{H"_2"O}$.

$\textcolor{red}{1} \text{CH"_4"O" + "O"_2 → color(blue)(1)"CO"_2 + color(green)(2)"H"_2"O}$

Now we balance $\text{O}$:

We have fixed 4 $\text{O}$ atoms on the right-hand side: 2 from the ${\text{CO}}_{2}$ and 2 from the $\text{H"_2"O}$.

We have fixed 1 $\text{O}$ atom on the left-hand side, so we need 3 more from the ${\text{O}}_{2}$.

Oops. We would need a fraction of an ${\text{O}}_{2}$ molecule.

We start over, multiplying every coefficient by 2.

$\textcolor{red}{2} \text{CH"_4"O" + "O"_2 → color(blue)(2)"CO"_2 + color(green)(4)"H"_2"O}$

Now we have fixed 8 $\text{O}$ atoms on the right and 2 on the left. We need 6 more $\text{O}$ atoms on the left. We put a 3 in front of the ${\text{O}}_{2}$.

$\textcolor{red}{2} \text{CH"_4"O" + color(brown)(3)"O"_2 → color(blue)(2)"CO"_2 + color(green)(4)"H"_2"O}$

Every formula now has a fixed coefficient. We should have a balanced equation.

Let’s check:

$m a t h b f \left(\textcolor{w h i t e}{m} \text{Element"color(white)(m)"Left-hand side"color(white)(m)"Right-hand side}\right)$
$\textcolor{w h i t e}{m m l l} \text{C} \textcolor{w h i t e}{m m m m m m l} 2 \textcolor{w h i t e}{m m m m m m m m l l} 2$
$\textcolor{w h i t e}{m m l l} \text{H} \textcolor{w h i t e}{m m m m m m l} 8 \textcolor{w h i t e}{m m m m m m m m l l} 8$
$\textcolor{w h i t e}{m m l l} \text{O} \textcolor{w h i t e}{m m m m m m l} 8 \textcolor{w h i t e}{m m m m m m m m l l} 8$

All atoms balance. The balanced equation is

$\text{2CH"_4"O" + "3O"_2 → "2CO"_2 + "4H"_2"O}$