trimethylsilyl amide name of lithium

trimethylsilyl amide lithium

molar mass | 95.14 g/mol formula | C_3H_10LiNSi empirical formula | Li_C_3Si_N_H_10 SMILES identifier | C[Si](C)(C)[NH-].[Li+] InChI identifier | InChI=1/C3H10NSi.Li/c1-5(2, 3)4;/h4H, 1-3H3;/q-1;+1 InChI key | HMRCZKQIOFZACX-UHFFFAOYSA-N

vertex count | 6 edge count | 5 Schultz index | 64 Wiener index | 16 Hosoya index | 5 Balaban index | 3.024

longest chain length | 3 atoms longest straight chain length | 3 atoms longest aliphatic chain length | 0 atoms aromatic atom count | 0 atoms H-bond acceptor count | 1 atom H-bond donor count | 0 atoms

Find the elemental composition for trimethylsilyl amide lithium 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_3H_10LiNSi Use the chemical formula, C_3H_10LiNSi, to count the number of atoms, N_i, for each element and find the total number of atoms, N_atoms: | number of atoms Li (lithium) | 1 C (carbon) | 3 Si (silicon) | 1 N (nitrogen) | 1 H (hydrogen) | 10 N_atoms = 1 + 3 + 1 + 1 + 10 = 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 C (carbon) | 3 | 3/16 Si (silicon) | 1 | 1/16 N (nitrogen) | 1 | 1/16 H (hydrogen) | 10 | 10/16 Check: 1/16 + 3/16 + 1/16 + 1/16 + 10/16 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent Li (lithium) | 1 | 1/16 × 100% = 6.25% C (carbon) | 3 | 3/16 × 100% = 18.8% Si (silicon) | 1 | 1/16 × 100% = 6.25% N (nitrogen) | 1 | 1/16 × 100% = 6.25% H (hydrogen) | 10 | 10/16 × 100% = 62.5% 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 C (carbon) | 3 | 18.8% | 12.011 Si (silicon) | 1 | 6.25% | 28.085 N (nitrogen) | 1 | 6.25% | 14.007 H (hydrogen) | 10 | 62.5% | 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 Li (lithium) | 1 | 6.25% | 6.94 | 1 × 6.94 = 6.94 C (carbon) | 3 | 18.8% | 12.011 | 3 × 12.011 = 36.033 Si (silicon) | 1 | 6.25% | 28.085 | 1 × 28.085 = 28.085 N (nitrogen) | 1 | 6.25% | 14.007 | 1 × 14.007 = 14.007 H (hydrogen) | 10 | 62.5% | 1.008 | 10 × 1.008 = 10.080 m = 6.94 u + 36.033 u + 28.085 u + 14.007 u + 10.080 u = 95.145 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% | 6.94/95.145 C (carbon) | 3 | 18.8% | 36.033/95.145 Si (silicon) | 1 | 6.25% | 28.085/95.145 N (nitrogen) | 1 | 6.25% | 14.007/95.145 H (hydrogen) | 10 | 62.5% | 10.080/95.145 Check: 6.94/95.145 + 36.033/95.145 + 28.085/95.145 + 14.007/95.145 + 10.080/95.145 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent Li (lithium) | 1 | 6.25% | 6.94/95.145 × 100% = 7.29% C (carbon) | 3 | 18.8% | 36.033/95.145 × 100% = 37.87% Si (silicon) | 1 | 6.25% | 28.085/95.145 × 100% = 29.52% N (nitrogen) | 1 | 6.25% | 14.007/95.145 × 100% = 14.72% H (hydrogen) | 10 | 62.5% | 10.080/95.145 × 100% = 10.59%

The first step in finding the oxidation states (or oxidation numbers) in trimethylsilyl amide lithium is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In trimethylsilyl amide lithium 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 3 carbon-silicon bonds, and 1 nitrogen-silicon bond. For each of these bonds, assign the bonding electrons to the most electronegative element. First examine the carbon-silicon bonds: element | electronegativity (Pauling scale) | C | 2.55 | Si | 1.90 | | | Since carbon is more electronegative than silicon, the electrons in these bonds will go to carbon. Decrease the oxidation number for carbon in every highlighted bond (by 1 for single bonds, 2 for double bonds, and 3 for triple bonds), and increase the oxidation number for silicon accordingly: Next look at the nitrogen-silicon bond: element | electronegativity (Pauling scale) | N | 3.04 | Si | 1.90 | | | Since nitrogen is more electronegative than silicon, the electrons in this bond will go to nitrogen: Now summarize the results: Answer: | | oxidation state | element | count -4 | C (carbon) | 3 -3 | N (nitrogen) | 1 +1 | H (hydrogen) | 10 | Li (lithium) | 1 +4 | Si (silicon) | 1

hybridization | element | count sp^3 | C (carbon) | 3 | N (nitrogen) | 1 | Si (silicon) | 1

Orbital hybridization Structure diagram

vertex count | 16 edge count | 14 Schultz index | 1060 Wiener index | 298 Hosoya index | 288 Balaban index | 5.776