2,3-dimethylphenylhydrazine hydrochloride hydrate

2, 3-dimethylphenylhydrazine hydrochloride hydrate

molar mass | 190.7 g/mol formula | C_8H_15ClN_2O empirical formula | O_Cl_C_8N_2H_15 SMILES identifier | CC1=C(C)C(=CC=C1)NN.Cl.O InChI identifier | InChI=1/C8H12N2.ClH.H2O/c1-6-4-3-5-8(10-9)7(6)2;;/h3-5, 10H, 9H2, 1-2H3;1H;1H2 InChI key | CWLHVQVBNFKKSG-UHFFFAOYSA-N

Structure diagram

longest chain length | 6 atoms longest straight chain length | 2 atoms longest aliphatic chain length | 0 atoms aromatic atom count | 6 atoms H-bond acceptor count | 3 atoms H-bond donor count | 3 atoms

Find the elemental composition for 2, 3-dimethylphenylhydrazine hydrochloride hydrate 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_8H_15ClN_2O Use the chemical formula, C_8H_15ClN_2O, to count the number of atoms, N_i, for each element and find the total number of atoms, N_atoms: | number of atoms O (oxygen) | 1 Cl (chlorine) | 1 C (carbon) | 8 N (nitrogen) | 2 H (hydrogen) | 15 N_atoms = 1 + 1 + 8 + 2 + 15 = 27 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 O (oxygen) | 1 | 1/27 Cl (chlorine) | 1 | 1/27 C (carbon) | 8 | 8/27 N (nitrogen) | 2 | 2/27 H (hydrogen) | 15 | 15/27 Check: 1/27 + 1/27 + 8/27 + 2/27 + 15/27 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent O (oxygen) | 1 | 1/27 × 100% = 3.70% Cl (chlorine) | 1 | 1/27 × 100% = 3.70% C (carbon) | 8 | 8/27 × 100% = 29.6% N (nitrogen) | 2 | 2/27 × 100% = 7.41% H (hydrogen) | 15 | 15/27 × 100% = 55.6% 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 O (oxygen) | 1 | 3.70% | 15.999 Cl (chlorine) | 1 | 3.70% | 35.45 C (carbon) | 8 | 29.6% | 12.011 N (nitrogen) | 2 | 7.41% | 14.007 H (hydrogen) | 15 | 55.6% | 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 O (oxygen) | 1 | 3.70% | 15.999 | 1 × 15.999 = 15.999 Cl (chlorine) | 1 | 3.70% | 35.45 | 1 × 35.45 = 35.45 C (carbon) | 8 | 29.6% | 12.011 | 8 × 12.011 = 96.088 N (nitrogen) | 2 | 7.41% | 14.007 | 2 × 14.007 = 28.014 H (hydrogen) | 15 | 55.6% | 1.008 | 15 × 1.008 = 15.120 m = 15.999 u + 35.45 u + 96.088 u + 28.014 u + 15.120 u = 190.671 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 O (oxygen) | 1 | 3.70% | 15.999/190.671 Cl (chlorine) | 1 | 3.70% | 35.45/190.671 C (carbon) | 8 | 29.6% | 96.088/190.671 N (nitrogen) | 2 | 7.41% | 28.014/190.671 H (hydrogen) | 15 | 55.6% | 15.120/190.671 Check: 15.999/190.671 + 35.45/190.671 + 96.088/190.671 + 28.014/190.671 + 15.120/190.671 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent O (oxygen) | 1 | 3.70% | 15.999/190.671 × 100% = 8.391% Cl (chlorine) | 1 | 3.70% | 35.45/190.671 × 100% = 18.59% C (carbon) | 8 | 29.6% | 96.088/190.671 × 100% = 50.39% N (nitrogen) | 2 | 7.41% | 28.014/190.671 × 100% = 14.69% H (hydrogen) | 15 | 55.6% | 15.120/190.671 × 100% = 7.930%

The first step in finding the oxidation states (or oxidation numbers) in 2, 3-dimethylphenylhydrazine hydrochloride hydrate is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In 2, 3-dimethylphenylhydrazine hydrochloride hydrate 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 1 carbon-nitrogen bond, 8 carbon-carbon bonds, and 1 nitrogen-nitrogen bond. For each of these bonds, assign the bonding electrons to the most electronegative element. First examine the carbon-nitrogen bond: element | electronegativity (Pauling scale) | C | 2.55 | N | 3.04 | | | Since nitrogen is more electronegative than carbon, the electrons in this bond will go to nitrogen. Decrease the oxidation number for nitrogen (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: Next look at the nitrogen-nitrogen bond: element | electronegativity (Pauling scale) | N | 3.04 | N | 3.04 | | | 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) | 2 -2 | N (nitrogen) | 2 | O (oxygen) | 1 -1 | C (carbon) | 3 | Cl (chlorine) | 1 0 | C (carbon) | 2 +1 | C (carbon) | 1 | H (hydrogen) | 15

hybridization | element | count sp^2 | C (carbon) | 6 | N (nitrogen) | 1 sp^3 | C (carbon) | 2 | Cl (chlorine) | 1 | N (nitrogen) | 1 | O (oxygen) | 1

Orbital hybridization Structure diagram

vertex count | 27 edge count | 25 Schultz index | Wiener index | Hosoya index | Balaban index |