Chemistry Examples

$C_{6} H_{12} O_{6}$

Step 1

Since $H$ is in column $1$ of the periodic table, it will share $1$ electrons and use an oxidation state of $1$ .

$H = 12 (1) = 12$

Step 2

Since $O$ is in column $6$ of the periodic table, it will share $6$ electrons and use an oxidation state of $6$ .

$O = 6 (- 2) = - 12$

Step 3

To find the oxidation state of $C$ , set up an equation of each oxidation state found earlier and set it equal to $0$ .

$12 - 12 + 6 x = 0$

Step 4

Solve the equation for $x$ .

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Step 4.1

Simplify $12 + - 12 + 6 x$ .

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Step 4.1.1

Add $12$ and $- 12$ .

$0 + 6 x = 0$

Step 4.1.2

Add $0$ and $6 x$ .

$6 x = 0$

Step 4.2

Divide each term in $6 x = 0$ by $6$ and simplify.

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Step 4.2.1

Divide each term in $6 x = 0$ by $6$ .

$\frac{6 x}{6} = \frac{0}{6}$

Step 4.2.2

Simplify the left side.

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Step 4.2.2.1

Cancel the common factor of $6$ .

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Step 4.2.2.1.1

Cancel the common factor.

$\frac{6 x}{6} = \frac{0}{6}$

Step 4.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{0}{6}$

Step 4.2.3

Simplify the right side.

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Step 4.2.3.1

Divide $0$ by $6$ .

$x = 0$

Step 5

$C$ is using an oxidation state of $0$ .

$0$

Step 6

This is the full list of oxidation states for this molecule.

$H = 12, O = - 12, C = 0$