name of 2 thienylzinc bromide

2 thienylzinc bromide

molar mass | 228.4 g/mol formula | C_4H_3BrSZn empirical formula | Br_S_C_4Zn_H_3 SMILES identifier | C1=CSC(=C1)[Zn+].[Br-] InChI identifier | InChI=1/C4H3S.BrH.Zn/c1-2-4-5-3-1;;/h1-3H;1H;/q;;+1/p-1/fC4H3S.Br.Zn/h;1h;/q;-1;m InChI key | KBBMRTYVFOYNIW-UHFFFAOYSA-M

vertex count | 7 edge count | 6 Schultz index | 126 Wiener index | 26 Hosoya index | 16 Balaban index | 2.184

longest chain length | 4 atoms longest straight chain length | 0 atoms longest aliphatic chain length | 0 atoms aromatic atom count | 5 atoms H-bond acceptor count | 0 atoms H-bond donor count | 0 atoms

Find the elemental composition for 2 thienylzinc bromide 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_4H_3BrSZn Use the chemical formula, C_4H_3BrSZn, to count the number of atoms, N_i, for each element and find the total number of atoms, N_atoms: | number of atoms Br (bromine) | 1 S (sulfur) | 1 C (carbon) | 4 Zn (zinc) | 1 H (hydrogen) | 3 N_atoms = 1 + 1 + 4 + 1 + 3 = 10 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 Br (bromine) | 1 | 1/10 S (sulfur) | 1 | 1/10 C (carbon) | 4 | 4/10 Zn (zinc) | 1 | 1/10 H (hydrogen) | 3 | 3/10 Check: 1/10 + 1/10 + 4/10 + 1/10 + 3/10 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent Br (bromine) | 1 | 1/10 × 100% = 10.00% S (sulfur) | 1 | 1/10 × 100% = 10.00% C (carbon) | 4 | 4/10 × 100% = 40.0% Zn (zinc) | 1 | 1/10 × 100% = 10.00% H (hydrogen) | 3 | 3/10 × 100% = 30.0% 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 Br (bromine) | 1 | 10.00% | 79.904 S (sulfur) | 1 | 10.00% | 32.06 C (carbon) | 4 | 40.0% | 12.011 Zn (zinc) | 1 | 10.00% | 65.38 H (hydrogen) | 3 | 30.0% | 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 Br (bromine) | 1 | 10.00% | 79.904 | 1 × 79.904 = 79.904 S (sulfur) | 1 | 10.00% | 32.06 | 1 × 32.06 = 32.06 C (carbon) | 4 | 40.0% | 12.011 | 4 × 12.011 = 48.044 Zn (zinc) | 1 | 10.00% | 65.38 | 1 × 65.38 = 65.38 H (hydrogen) | 3 | 30.0% | 1.008 | 3 × 1.008 = 3.024 m = 79.904 u + 32.06 u + 48.044 u + 65.38 u + 3.024 u = 228.412 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 Br (bromine) | 1 | 10.00% | 79.904/228.412 S (sulfur) | 1 | 10.00% | 32.06/228.412 C (carbon) | 4 | 40.0% | 48.044/228.412 Zn (zinc) | 1 | 10.00% | 65.38/228.412 H (hydrogen) | 3 | 30.0% | 3.024/228.412 Check: 79.904/228.412 + 32.06/228.412 + 48.044/228.412 + 65.38/228.412 + 3.024/228.412 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent Br (bromine) | 1 | 10.00% | 79.904/228.412 × 100% = 34.98% S (sulfur) | 1 | 10.00% | 32.06/228.412 × 100% = 14.04% C (carbon) | 4 | 40.0% | 48.044/228.412 × 100% = 21.03% Zn (zinc) | 1 | 10.00% | 65.38/228.412 × 100% = 28.62% H (hydrogen) | 3 | 30.0% | 3.024/228.412 × 100% = 1.324%

The first step in finding the oxidation states (or oxidation numbers) in 2 thienylzinc bromide is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In 2 thienylzinc bromide 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 2 carbon-sulfur bonds, 1 carbon-zinc bond, and 3 carbon-carbon bonds. For each of these bonds, assign the bonding electrons to the most electronegative element. First examine 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. Decrease the oxidation number for sulfur 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-zinc bond: element | electronegativity (Pauling scale) | C | 2.55 | Zn | 1.65 | | | Since carbon is more electronegative than zinc, the electrons in this bond will go to carbon: 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 | S (sulfur) | 1 -1 | Br (bromine) | 1 | C (carbon) | 2 0 | C (carbon) | 2 +1 | H (hydrogen) | 3 +2 | Zn (zinc) | 1

vertex count | 10 edge count | 9 Schultz index | 336 Wiener index | 80 Hosoya index | 53 Balaban index | 2.544