# CHEMISTRY 101 – Drawing Lewis Structures: organic molecules, CH3COOH In this example problem, we’ll draw the Lewis structure for the organic molecule CH3COOH and use formal charge to evaluate it. Our first step is to sum the valence electrons. We have two carbon atoms with four electrons each, four hydrogen atoms with one electron each, and two oxygen atoms with six electrons each. The molecule does not have a charge, so we do not be to make any adjustments. We have 24 valence electrons. Our next step is to draw the structure with only single bonds. The chemical formulas for organic molecules can be written to give an idea of how to draw them. Our formula has a CH3, which is three hydrogen atoms bonded to a central carbon atom, and COOH, which is an oxygen atom and an OH group bonded to a carbon. By creating these seven single bonds, we’ve used up 14 electrons, so we have 10 electrons left. Our next step is to satisfy the octet rule by adding lone pairs of electrons. We’ll begin by satisfying the octet rule for peripheral atoms. All of our hydrogen atoms are already satisfied because they follow the duet rule. Oxygen is our all only other peripheral atom; it currently has one bond so it wants six more electrons. We do still have two central atoms that want additional electrons: carbon and oxygen. Since oxygen is more electronegative, we’ll satisfy it first with four lone pair electrons. We’ve now used up all 24 electrons, but our central carbon atom still wants two additional electrons. So we’ll need to create a double bond using one of the lone pairs of electrons we’ve added. We could use a lone pair of electrons from the top oxygen atom. This leaves two lone pairs of electrons on each oxygen. Or we can also use a lone pair of electrons from the oxygen on the right to create the double bond. This would leave one oxygen atom with three lone pairs of electrons and the other with just one lone pair. Let’s evaluate the structures using formal charge to determine which is better. Both of the oxygen atoms in the structure on the left have the same number of lone pairs and bonds, so they will have the same formal charge. To calculate the formal charge, oxygen normally has six valence electrons subtract the four lone pair electrons and two bonds. The formal charge for each of these oxygen atoms is zero. To calculate the formal charge of the double bonded oxygen in the structure on the right, oxygen normally has six valence electrons subtract two lone pair electrons and three bonds. This oxygen has a plus one formal charge. To calculate the formal charge of the single bonded oxygen in the structure on the right, oxygen normally has six valence electrons subtract six alone pair electrons and one bond. So this oxygen atom has a -1 formal charge. The structure on the left has smaller formal charges so it is our structure for CH3COOH.