Find the Oxidation Numbers C_2H_7O_6

$C_{2} H_{7} O_{6}$

Step 1

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

$H = 7 (1) = 7$

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 $0$ .

$7 - 12 + 2 x = 0$

Step 4

Solve the equation for $x$ .

Add $7$ and $- 12$ .

$- 5 + 2 x = 0$

Add $5$ to both sides of the equation.

$2 x = 5$

Divide each term in $2 x = 5$ by $2$ and simplify.

Divide each term in $2 x = 5$ by $2$ .

$\frac{2 x}{2} = \frac{5}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{5}{2}$

Divide $x$ by $1$ .

$x = \frac{5}{2}$

Step 5

$C$ is using an oxidation state of $\frac{5}{2}$ .

$\frac{5}{2}$

Step 6

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

$H = 7, O = - 12, C = \frac{5}{2}$