hydrochloride sesquihydrate

hydrochloride sesquihydrate

molar mass | 127 g/mol formula | Cl_2H_8O_3 empirical formula | O_3Cl_2H_8 SMILES identifier | Cl.Cl.O.O.O InChI identifier | InChI=1/2ClH.3H2O/h2*1H;3*1H2 InChI key | DBODNPMIIRVQGW-UHFFFAOYSA-N

Structure diagram

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

Find the elemental composition for hydrochloride sesquihydrate 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: Cl_2H_8O_3 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 O (oxygen) | 3 Cl (chlorine) | 2 H (hydrogen) | 8 N_atoms = 3 + 2 + 8 = 13 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) | 3 | 3/13 Cl (chlorine) | 2 | 2/13 H (hydrogen) | 8 | 8/13 Check: 3/13 + 2/13 + 8/13 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent O (oxygen) | 3 | 3/13 × 100% = 23.1% Cl (chlorine) | 2 | 2/13 × 100% = 15.4% H (hydrogen) | 8 | 8/13 × 100% = 61.5% 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 O (oxygen) | 3 | 23.1% | 15.999 Cl (chlorine) | 2 | 15.4% | 35.45 H (hydrogen) | 8 | 61.5% | 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 O (oxygen) | 3 | 23.1% | 15.999 | 3 × 15.999 = 47.997 Cl (chlorine) | 2 | 15.4% | 35.45 | 2 × 35.45 = 70.90 H (hydrogen) | 8 | 61.5% | 1.008 | 8 × 1.008 = 8.064 m = 47.997 u + 70.90 u + 8.064 u = 126.961 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) | 3 | 23.1% | 47.997/126.961 Cl (chlorine) | 2 | 15.4% | 70.90/126.961 H (hydrogen) | 8 | 61.5% | 8.064/126.961 Check: 47.997/126.961 + 70.90/126.961 + 8.064/126.961 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent O (oxygen) | 3 | 23.1% | 47.997/126.961 × 100% = 37.80% Cl (chlorine) | 2 | 15.4% | 70.90/126.961 × 100% = 55.84% H (hydrogen) | 8 | 61.5% | 8.064/126.961 × 100% = 6.352%

The first step in finding the oxidation states (or oxidation numbers) in hydrochloride sesquihydrate is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In hydrochloride sesquihydrate 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: Now summarize the results: Answer: | | oxidation state | element | count -2 | O (oxygen) | 3 -1 | Cl (chlorine) | 2 +1 | H (hydrogen) | 8

hybridization | element | count sp^3 | Cl (chlorine) | 2 | O (oxygen) | 3

Orbital hybridization Structure diagram

vertex count | 13 edge count | 8 Schultz index | Wiener index | Hosoya index | Balaban index |