# C3H8O Lewis structure

C3H8O (isopropyl alcohol) has three carbon atoms, eight hydrogen atoms, and one oxygen atom.

In C3H8O Lewis structure, there are two single bonds between the three carbon atoms. The left carbon and right carbon are attached with three hydrogen atoms, and the center carbon is attached with one oxygen atom and one hydrogen atom. And the oxygen atom (with which the other hydrogen atom is attached) has two lone pairs.

Contents

## Steps

Use these steps to correctly draw the C3H8O Lewis structure:

#1 First draw a rough sketch
#2 Mark lone pairs on the atoms
#3 Calculate and mark formal charges on the atoms, if required

Let’s discuss each step in more detail.

### #1 First draw a rough sketch

• First, determine the total number of valence electrons

In the periodic table, carbon lies in group 14, hydrogen lies in group 1, and oxygen lies in group 16.

Hence, carbon has four valence electrons, hydrogen has one valence electron, and oxygen has six valence electrons.

Since C3H8O has three carbon atoms, eight hydrogen atoms, and one oxygen atom, so…

Valence electrons of three carbon atoms = 4 × 3 = 12
Valence electrons of eight hydrogen atoms = 1 × 8 = 8
Valence electrons of one oxygen atom = 6 × 1 = 6

And the total valence electrons = 12 + 8 + 6 = 26

• Second, find the total electron pairs

We have a total of 26 valence electrons. And when we divide this value by two, we get the value of total electron pairs.

Total electron pairs = total valence electrons ÷ 2

So the total electron pairs = 26 ÷ 2 = 13

• Third, determine the central atom

Here hydrogen can not be the central atom. Because the central atom is bonded with at least two other atoms, and hydrogen has only one electron in its last shell, so it can not make more than one bond.

Now we have to choose the central atom from carbon and oxygen. Place the least electronegative atom at the center.

Since carbon is less electronegative than oxygen, assume that the central atom is carbon.

Here, there are three carbon atoms, so we can assume any one as the central atom.

Let’s assume that the central atom is center carbon.

Therefore, place carbons in the center and hydrogen and oxygen on either side.

• And finally, draw the rough sketch

### #2 Mark lone pairs on the atoms

Here, we have a total of 13 electron pairs. And eleven bonds are already marked. So we have to only mark the remaining two electron pairs as lone pairs on the sketch.

Also remember that both (carbon and oxygen) are the period 2 elements, so they can not keep more than 8 electrons in their last shell. And hydrogen is a period 1 element, so it can not keep more than 2 electrons in its last shell.

Always start to mark the lone pairs from outside atoms. Here, the outside atoms are hydrogens, left carbon, right carbon, and oxygen. But no need to mark on hydrogen, because each hydrogen has already two electrons.

So for oxygen, there are two lone pairs, and for left carbon and right carbon, there is zero lone pair because all two electron pairs are over.

Mark the lone pairs on the sketch as follows:

### #3 Calculate and mark formal charges on the atoms, if required

Use the following formula to calculate the formal charges on atoms:

Formal charge = valence electrons – nonbonding electrons – ½ bonding electrons

For left carbon and right carbon atom, formal charge = 4 – 0 – ½ (8) = 0

For each hydrogen atom, formal charge = 1 – 0 – ½ (2) = 0

For oxygen atom, formal charge = 6 – 4 – ½ (4) = 0

Here, the atoms do not have charges, so no need to mark the charges.

In the above structure, you can see that the central atom (center carbon) forms an octet. The outside atoms (left carbon, right carbon, and oxygen) also form an octet, and all hydrogens form a duet. Hence, the octet rule and duet rule are satisfied.

Therefore, this structure is the stable Lewis structure of C3H8O.

Next: CI4 Lewis structure