Chemistry Examples

$C_{3} H_{7} N_{2}$

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 $N$ is in column $5$ of the periodic table, it will share $5$ electrons and use an oxidation state of $5$ .

$N = 2 (- 3) = - 6$

Step 3

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

$7 - 6 + 3 x = 0$

Step 4

Solve the equation for $x$ .

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Step 4.1

Add $7$ and $- 6$ .

$1 + 3 x = 0$

Step 4.2

Subtract $1$ from both sides of the equation.

$3 x = - 1$

Step 4.3

Divide each term in $3 x = - 1$ by $3$ and simplify.

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Step 4.3.1

Divide each term in $3 x = - 1$ by $3$ .

$\frac{3 x}{3} = \frac{- 1}{3}$

Step 4.3.2

Simplify the left side.

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Step 4.3.2.1

Cancel the common factor of $3$ .

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Step 4.3.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{- 1}{3}$

Step 4.3.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 1}{3}$

Step 4.3.3

Simplify the right side.

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Step 4.3.3.1

Move the negative in front of the fraction.

$x = - \frac{1}{3}$

Step 5

$C$ is using an oxidation state of $- \frac{1}{3}$ .

$- \frac{1}{3}$

Step 6

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

$H = 7, N = - 6, C = - \frac{1}{3}$