Chemistry Examples

K_{2} C O_{3}

$K_{2} C O_{3}$

Step 1

Since

K

$K$ is in column

1

$1$ of the periodic table, it will share

1

$1$ electrons and use an oxidation state of

1

$1$ .

K = 2 (1) = 2

$K = 2 (1) = 2$

Step 2

Since

O

$O$ is in column

6

$6$ of the periodic table, it will share

6

$6$ electrons and use an oxidation state of

6

$6$ .

O = 3 (- 2) = - 6

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

Step 3

To find the oxidation state of

C

$C$ , set up an equation of each oxidation state found earlier and set it equal to

0

$0$ .

2 - 6 + 1 x = 0

$2 - 6 + 1 x = 0$

Step 4

Solve the equation for

x

$x$ .

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Step 4.1

Simplify

2 + - 6 + 1 x

$2 + - 6 + 1 x$ .

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Step 4.1.1

Multiply

x

$x$ by

1

$1$ .

2 + - 6 + x = 0

$2 + - 6 + x = 0$

Step 4.1.2

Add

2

$2$ and

- 6

$- 6$ .

- 4 + x = 0

$- 4 + x = 0$

- 4 + x = 0

$- 4 + x = 0$

Step 4.2

Add

4

$4$ to both sides of the equation.

x = 4

$x = 4$

x = 4

$x = 4$

Step 5

C

$C$ is using an oxidation state of

4

$4$ .

4

$4$

Step 6

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

K = 2, O = - 6, C = 4

$K = 2, O = - 6, C = 4$

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