heptadecafluorooctanesulfonic acid tetraethylammonium salt

heptadecafluorooctanesulfonic acid tetraethylammonium salt

molar mass | 629.4 g/mol formula | C_16H_20F_17NO_3S empirical formula | C_16N_F_17S_O_3H_20 SMILES identifier | C(C(C(C(C(F)(F)S(=O)(=O)[O-])(F)F)(F)F)(F)F)(C(C(C(F)(F)F)(F)F)(F)F)(F)F.CC[N+](CC)(CC)CC InChI identifier | InChI=1/C8HF17O3S.C8H20N/c9-1(10, 3(13, 14)5(17, 18)7(21, 22)23)2(11, 12)4(15, 16)6(19, 20)8(24, 25)29(26, 27)28;1-5-9(6-2, 7-3)8-4/h(H, 26, 27, 28);5-8H2, 1-4H3/q;+1/p-1/fC8F17O3S.C8H20N/q-1;m InChI key | JHDXAQHGAJXNBY-UHFFFAOYSA-M

Structure diagram

longest chain length | 11 atoms longest straight chain length | 11 atoms longest aliphatic chain length | 8 atoms aromatic atom count | 0 atoms H-bond acceptor count | 3 atoms H-bond donor count | 0 atoms

Find the elemental composition for heptadecafluorooctanesulfonic acid tetraethylammonium 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_16H_20F_17NO_3S 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 C (carbon) | 16 N (nitrogen) | 1 F (fluorine) | 17 S (sulfur) | 1 O (oxygen) | 3 H (hydrogen) | 20 N_atoms = 16 + 1 + 17 + 1 + 3 + 20 = 58 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 C (carbon) | 16 | 16/58 N (nitrogen) | 1 | 1/58 F (fluorine) | 17 | 17/58 S (sulfur) | 1 | 1/58 O (oxygen) | 3 | 3/58 H (hydrogen) | 20 | 20/58 Check: 16/58 + 1/58 + 17/58 + 1/58 + 3/58 + 20/58 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent C (carbon) | 16 | 16/58 × 100% = 27.6% N (nitrogen) | 1 | 1/58 × 100% = 1.72% F (fluorine) | 17 | 17/58 × 100% = 29.3% S (sulfur) | 1 | 1/58 × 100% = 1.72% O (oxygen) | 3 | 3/58 × 100% = 5.17% H (hydrogen) | 20 | 20/58 × 100% = 34.5% 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 C (carbon) | 16 | 27.6% | 12.011 N (nitrogen) | 1 | 1.72% | 14.007 F (fluorine) | 17 | 29.3% | 18.998403163 S (sulfur) | 1 | 1.72% | 32.06 O (oxygen) | 3 | 5.17% | 15.999 H (hydrogen) | 20 | 34.5% | 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 C (carbon) | 16 | 27.6% | 12.011 | 16 × 12.011 = 192.176 N (nitrogen) | 1 | 1.72% | 14.007 | 1 × 14.007 = 14.007 F (fluorine) | 17 | 29.3% | 18.998403163 | 17 × 18.998403163 = 322.972853771 S (sulfur) | 1 | 1.72% | 32.06 | 1 × 32.06 = 32.06 O (oxygen) | 3 | 5.17% | 15.999 | 3 × 15.999 = 47.997 H (hydrogen) | 20 | 34.5% | 1.008 | 20 × 1.008 = 20.160 m = 192.176 u + 14.007 u + 322.972853771 u + 32.06 u + 47.997 u + 20.160 u = 629.372853771 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 C (carbon) | 16 | 27.6% | 192.176/629.372853771 N (nitrogen) | 1 | 1.72% | 14.007/629.372853771 F (fluorine) | 17 | 29.3% | 322.972853771/629.372853771 S (sulfur) | 1 | 1.72% | 32.06/629.372853771 O (oxygen) | 3 | 5.17% | 47.997/629.372853771 H (hydrogen) | 20 | 34.5% | 20.160/629.372853771 Check: 192.176/629.372853771 + 14.007/629.372853771 + 322.972853771/629.372853771 + 32.06/629.372853771 + 47.997/629.372853771 + 20.160/629.372853771 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent C (carbon) | 16 | 27.6% | 192.176/629.372853771 × 100% = 30.53% N (nitrogen) | 1 | 1.72% | 14.007/629.372853771 × 100% = 2.226% F (fluorine) | 17 | 29.3% | 322.972853771/629.372853771 × 100% = 51.32% S (sulfur) | 1 | 1.72% | 32.06/629.372853771 × 100% = 5.094% O (oxygen) | 3 | 5.17% | 47.997/629.372853771 × 100% = 7.626% H (hydrogen) | 20 | 34.5% | 20.160/629.372853771 × 100% = 3.203%

The first step in finding the oxidation states (or oxidation numbers) in heptadecafluorooctanesulfonic acid tetraethylammonium salt is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In heptadecafluorooctanesulfonic acid tetraethylammonium 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 17 carbon-fluorine bonds, 4 carbon-nitrogen bonds, 1 carbon-sulfur bond, 3 oxygen-sulfur bonds, and 11 carbon-carbon bonds. For each of these bonds, assign the bonding electrons to the most electronegative element. First examine the carbon-fluorine bonds: element | electronegativity (Pauling scale) | C | 2.55 | F | 3.98 | | | Since fluorine is more electronegative than carbon, the electrons in these bonds will go to fluorine. Decrease the oxidation number for fluorine 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-nitrogen bonds: element | electronegativity (Pauling scale) | C | 2.55 | N | 3.04 | | | Since nitrogen is more electronegative than carbon, the electrons in these bonds will go to nitrogen: Next look at the carbon-sulfur bond: element | electronegativity (Pauling scale) | C | 2.55 | S | 2.58 | | | Since sulfur is more electronegative than carbon, the electrons in this bond will go to sulfur: Next look at the oxygen-sulfur bonds: element | electronegativity (Pauling scale) | O | 3.44 | S | 2.58 | | | Since oxygen is more electronegative than sulfur, the electrons in these bonds will go to oxygen: 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 -3 | C (carbon) | 4 | N (nitrogen) | 1 -2 | O (oxygen) | 3 -1 | C (carbon) | 4 | F (fluorine) | 17 +1 | H (hydrogen) | 20 +2 | C (carbon) | 6 +3 | C (carbon) | 2 +4 | S (sulfur) | 1

hybridization | element | count sp^2 | O (oxygen) | 2 sp^3 | C (carbon) | 16 | F (fluorine) | 17 | N (nitrogen) | 1 | O (oxygen) | 1 | S (sulfur) | 1

Orbital hybridization Structure diagram

vertex count | 58 edge count | 56 Schultz index | Wiener index | Hosoya index | Balaban index |