name of xanthommatin

xanthommatin

formula | C_20H_13N_3O_8 name | xanthommatin mass fractions | C (carbon) 57.4% | H (hydrogen) 1.93% | N (nitrogen) 10% | O (oxygen) 30.6%

Draw the Lewis structure of xanthommatin. Start by drawing the overall structure of the molecule, ignoring potential double and triple bonds: Count the total valence electrons of the carbon (n_C, val = 4), hydrogen (n_H, val = 1), nitrogen (n_N, val = 5), and oxygen (n_O, val = 6) atoms, including the net charge: 20 n_C, val + 8 n_H, val + 3 n_N, val + 8 n_O, val - n_charge = 153 Calculate the number of electrons needed to completely fill the valence shells for carbon (n_C, full = 8), hydrogen (n_H, full = 2), nitrogen (n_N, full = 8), and oxygen (n_O, full = 8): 20 n_C, full + 8 n_H, full + 3 n_N, full + 8 n_O, full = 264 Subtracting these two numbers shows that 264 - 153 = 111 bonding electrons are expected (in such cases with an odd number, round down to 110 bonding electrons). Each bond has two electrons, so in addition to the 42 bonds already present in the diagram add 12 bonds. To minimize formal charge carbon wants 4 bonds, oxygen wants 2 bonds, and nitrogen wants 3 bonds. Identify the atoms that want additional bonds and the number of electrons remaining on each atom. The net charge has been given to the most electronegative atom, oxygen, in 2 places: Fill in the 12 bonds by pairing electrons between adjacent highlighted atoms, noting the formal charges of the atoms. Double bonding carbon to the other highlighted oxygen atoms would result in an equivalent molecule: Answer: | |

molar mass | 418.3 g/mol

CAS number | 521-58-4 PubChem CID number | 25203027 SMILES identifier | C([O-])(=O)C([N+])CC(=O)C1(C=CC=C4(C=1N=C3(C(=CC(=O)C2(=NC(C([O-])=O)=CC(=C23)[O-]))O4)))