name of bis(trifluoromethane)sulfonimide lithium salt

bis(trifluoromethane)sulfonimide lithium salt

molar mass | 287.1 g/mol formula | C_2F_6LiNO_4S_2 empirical formula | Li_N_S_2O_4C_2F_6 SMILES identifier | C(F)(F)(F)S(=O)(=O)[N-]S(=O)(=O)C(F)(F)F.[Li+] InChI identifier | InChI=1/C2F6NO4S2.Li/c3-1(4, 5)14(10, 11)9-15(12, 13)2(6, 7)8;/q-1;+1 InChI key | QSZMZKBZAYQGRS-UHFFFAOYSA-N

vertex count | 16 edge count | 14 Schultz index | 1228 Wiener index | 340 Hosoya index | 273 Balaban index | 4.894

longest chain length | 7 atoms longest straight chain length | 7 atoms longest aliphatic chain length | 0 atoms aromatic atom count | 0 atoms H-bond acceptor count | 5 atoms H-bond donor count | 0 atoms

Find the elemental composition for bis(trifluoromethane)sulfonimide lithium 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_2F_6LiNO_4S_2 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 Li (lithium) | 1 N (nitrogen) | 1 S (sulfur) | 2 O (oxygen) | 4 C (carbon) | 2 F (fluorine) | 6 N_atoms = 1 + 1 + 2 + 4 + 2 + 6 = 16 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 Li (lithium) | 1 | 1/16 N (nitrogen) | 1 | 1/16 S (sulfur) | 2 | 2/16 O (oxygen) | 4 | 4/16 C (carbon) | 2 | 2/16 F (fluorine) | 6 | 6/16 Check: 1/16 + 1/16 + 2/16 + 4/16 + 2/16 + 6/16 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent Li (lithium) | 1 | 1/16 × 100% = 6.25% N (nitrogen) | 1 | 1/16 × 100% = 6.25% S (sulfur) | 2 | 2/16 × 100% = 12.5% O (oxygen) | 4 | 4/16 × 100% = 25.0% C (carbon) | 2 | 2/16 × 100% = 12.5% F (fluorine) | 6 | 6/16 × 100% = 37.5% 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 Li (lithium) | 1 | 6.25% | 6.94 N (nitrogen) | 1 | 6.25% | 14.007 S (sulfur) | 2 | 12.5% | 32.06 O (oxygen) | 4 | 25.0% | 15.999 C (carbon) | 2 | 12.5% | 12.011 F (fluorine) | 6 | 37.5% | 18.998403163 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 Li (lithium) | 1 | 6.25% | 6.94 | 1 × 6.94 = 6.94 N (nitrogen) | 1 | 6.25% | 14.007 | 1 × 14.007 = 14.007 S (sulfur) | 2 | 12.5% | 32.06 | 2 × 32.06 = 64.12 O (oxygen) | 4 | 25.0% | 15.999 | 4 × 15.999 = 63.996 C (carbon) | 2 | 12.5% | 12.011 | 2 × 12.011 = 24.022 F (fluorine) | 6 | 37.5% | 18.998403163 | 6 × 18.998403163 = 113.990418978 m = 6.94 u + 14.007 u + 64.12 u + 63.996 u + 24.022 u + 113.990418978 u = 287.075418978 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 Li (lithium) | 1 | 6.25% | 6.94/287.075418978 N (nitrogen) | 1 | 6.25% | 14.007/287.075418978 S (sulfur) | 2 | 12.5% | 64.12/287.075418978 O (oxygen) | 4 | 25.0% | 63.996/287.075418978 C (carbon) | 2 | 12.5% | 24.022/287.075418978 F (fluorine) | 6 | 37.5% | 113.990418978/287.075418978 Check: 6.94/287.075418978 + 14.007/287.075418978 + 64.12/287.075418978 + 63.996/287.075418978 + 24.022/287.075418978 + 113.990418978/287.075418978 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent Li (lithium) | 1 | 6.25% | 6.94/287.075418978 × 100% = 2.42% N (nitrogen) | 1 | 6.25% | 14.007/287.075418978 × 100% = 4.879% S (sulfur) | 2 | 12.5% | 64.12/287.075418978 × 100% = 22.34% O (oxygen) | 4 | 25.0% | 63.996/287.075418978 × 100% = 22.29% C (carbon) | 2 | 12.5% | 24.022/287.075418978 × 100% = 8.368% F (fluorine) | 6 | 37.5% | 113.990418978/287.075418978 × 100% = 39.71%

The first step in finding the oxidation states (or oxidation numbers) in bis(trifluoromethane)sulfonimide lithium salt is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: There are 6 carbon-fluorine bonds, 2 carbon-sulfur bonds, 2 nitrogen-sulfur bonds, and 4 oxygen-sulfur bonds in bis(trifluoromethane)sulfonimide lithium salt. 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-sulfur bonds: element | electronegativity (Pauling scale) | C | 2.55 | S | 2.58 | | | Since sulfur is more electronegative than carbon, the electrons in these bonds will go to sulfur: Next look at the nitrogen-sulfur bonds: element | electronegativity (Pauling scale) | N | 3.04 | S | 2.58 | | | Since nitrogen is more electronegative than sulfur, the electrons in these bonds will go to nitrogen: 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: Now summarize the results: Answer: | | oxidation state | element | count -3 | N (nitrogen) | 1 -2 | O (oxygen) | 4 -1 | F (fluorine) | 6 +1 | Li (lithium) | 1 +4 | C (carbon) | 2 | S (sulfur) | 2

hybridization | element | count sp^2 | O (oxygen) | 4 sp^3 | C (carbon) | 2 | F (fluorine) | 6 | N (nitrogen) | 1 | S (sulfur) | 2

Orbital hybridization Structure diagram

vertex count | 16 edge count | 14 Schultz index | 1228 Wiener index | 340 Hosoya index | 273 Balaban index | 4.894