8-naphthalic anhydride potassium salt

8-naphthalic anhydride potassium salt

molar mass | 365.4 g/mol formula | C_22H_14KO_3 empirical formula | K_C_22O_3H_14 SMILES identifier | C1=CC=C2C(=C1)C=CC=C2C(=O)OC(=O)C3=CC=CC4=CC=CC=C43.[K] InChI identifier | InChI=1/C22H14O3.K/c23-21(19-13-5-9-15-7-1-3-11-17(15)19)25-22(24)20-14-6-10-16-8-2-4-12-18(16)20;/h1-14H; InChI key | IQAXXKVJLFMHJI-UHFFFAOYSA-N

vertex count | 26 edge count | 28 Schultz index | 6758 Wiener index | 1508 Hosoya index | 242292 Balaban index | 1.365

longest chain length | 13 atoms longest straight chain length | 5 atoms longest aliphatic chain length | 0 atoms aromatic atom count | 20 atoms H-bond acceptor count | 3 atoms H-bond donor count | 0 atoms

Find the elemental composition for 8-naphthalic anhydride potassium salt in terms of the atom and mass percents: atom percent = N_i/N_atoms × 100% mass percent = (N_im_i)/m × 100% Plan: • Write the chemical formula and gather atomic masses from the periodic table. • Determine values for N_i, m_i, N_atoms and m using these items. • Finally, compute the percents and check the results. Write the chemical formula: C_22H_14KO_3 Use the chemical formula to count the number of atoms, N_i, for each element and find the total number of atoms, N_atoms, per molecule: | number of atoms K (potassium) | 1 C (carbon) | 22 O (oxygen) | 3 H (hydrogen) | 14 N_atoms = 1 + 22 + 3 + 14 = 40 Divide each N_i by N_atoms to calculate atom fractions. Then use the property that atom fractions must sum to one to check the work: | number of atoms | atom fraction K (potassium) | 1 | 1/40 C (carbon) | 22 | 22/40 O (oxygen) | 3 | 3/40 H (hydrogen) | 14 | 14/40 Check: 1/40 + 22/40 + 3/40 + 14/40 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent K (potassium) | 1 | 1/40 × 100% = 2.50% C (carbon) | 22 | 22/40 × 100% = 55.0% O (oxygen) | 3 | 3/40 × 100% = 7.50% H (hydrogen) | 14 | 14/40 × 100% = 35.0% Look up the atomic mass, m_i, in unified atomic mass units, u, for each element in the periodic table: | number of atoms | atom percent | atomic mass/u K (potassium) | 1 | 2.50% | 39.0983 C (carbon) | 22 | 55.0% | 12.011 O (oxygen) | 3 | 7.50% | 15.999 H (hydrogen) | 14 | 35.0% | 1.008 Multiply N_i by m_i to compute the mass for each element. Then sum those values to compute the molecular mass, m: | number of atoms | atom percent | atomic mass/u | mass/u K (potassium) | 1 | 2.50% | 39.0983 | 1 × 39.0983 = 39.0983 C (carbon) | 22 | 55.0% | 12.011 | 22 × 12.011 = 264.242 O (oxygen) | 3 | 7.50% | 15.999 | 3 × 15.999 = 47.997 H (hydrogen) | 14 | 35.0% | 1.008 | 14 × 1.008 = 14.112 m = 39.0983 u + 264.242 u + 47.997 u + 14.112 u = 365.4493 u Divide the mass for each element by m to calculate mass fractions. Then use the property that mass fractions must sum to one to check the work: | number of atoms | atom percent | mass fraction K (potassium) | 1 | 2.50% | 39.0983/365.4493 C (carbon) | 22 | 55.0% | 264.242/365.4493 O (oxygen) | 3 | 7.50% | 47.997/365.4493 H (hydrogen) | 14 | 35.0% | 14.112/365.4493 Check: 39.0983/365.4493 + 264.242/365.4493 + 47.997/365.4493 + 14.112/365.4493 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent K (potassium) | 1 | 2.50% | 39.0983/365.4493 × 100% = 10.70% C (carbon) | 22 | 55.0% | 264.242/365.4493 × 100% = 72.31% O (oxygen) | 3 | 7.50% | 47.997/365.4493 × 100% = 13.13% H (hydrogen) | 14 | 35.0% | 14.112/365.4493 × 100% = 3.862%

The first step in finding the oxidation states (or oxidation numbers) in 8-naphthalic anhydride potassium salt is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In 8-naphthalic anhydride potassium salt hydrogen is not bonded to a metal with lower electronegativity, so it will have an oxidation state of +1. Any element bonded to hydrogen gains the bonding electrons, decreasing their oxidation state by 1 for every bond: With hydrogen out of the way, look at the remaining bonds. There are 4 carbon-oxygen bonds, and 24 carbon-carbon bonds. For each of these bonds, assign the bonding electrons to the most electronegative element. First examine the carbon-oxygen bonds: element | electronegativity (Pauling scale) | C | 2.55 | O | 3.44 | | | Since oxygen is more electronegative than carbon, the electrons in these bonds will go to oxygen. Decrease the oxidation number for oxygen in every highlighted bond (by 1 for single bonds, 2 for double bonds, and 3 for triple bonds), and increase the oxidation number for carbon accordingly: Next look at the carbon-carbon bonds: element | electronegativity (Pauling scale) | C | 2.55 | C | 2.55 | | | Since these elements are the same the bonding electrons are shared equally, and there is no change to the oxidation states: Now summarize the results: Answer: | | oxidation state | element | count -2 | O (oxygen) | 3 -1 | C (carbon) | 14 0 | C (carbon) | 6 | K (potassium) | 1 +1 | H (hydrogen) | 14 +3 | C (carbon) | 2

vertex count | 40 edge count | 42 Schultz index | 18728 Wiener index | 4492 Hosoya index | 6.576×10^7 Balaban index | 1.638