calcium sulfate dihydrate name of calcium

calcium sulfate dihydrate calcium

molar mass | 308.3 g/mol formula | Ca_2H_4O_10S_2 empirical formula | Ca_O_5S_H_2 SMILES identifier | [Ca+2].[Ca+2].O=S(=O)([O-])[O-].O=S(=O)([O-])[O-].O.O InChI identifier | InChI=1/2Ca.2H2O4S.2H2O/c;;2*1-5(2, 3)4;;/h;;2*(H2, 1, 2, 3, 4);2*1H2/q2*+2;;;;/p-4/f2Ca.2O4S.2H2O/q2m;2*-2;; InChI key | DKHYYXRUGXCJIS-UHFFFAOYSA-J

Structure diagram

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 | 10 atoms H-bond donor count | 2 atoms

Find the elemental composition for calcium sulfate dihydrate calcium 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: Ca_2H_4O_10S_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 Ca (calcium) | 2 O (oxygen) | 10 S (sulfur) | 2 H (hydrogen) | 4 N_atoms = 2 + 10 + 2 + 4 = 18 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 Ca (calcium) | 2 | 2/18 O (oxygen) | 10 | 10/18 S (sulfur) | 2 | 2/18 H (hydrogen) | 4 | 4/18 Check: 2/18 + 10/18 + 2/18 + 4/18 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent Ca (calcium) | 2 | 2/18 × 100% = 11.1% O (oxygen) | 10 | 10/18 × 100% = 55.6% S (sulfur) | 2 | 2/18 × 100% = 11.1% H (hydrogen) | 4 | 4/18 × 100% = 22.2% 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 Ca (calcium) | 2 | 11.1% | 40.078 O (oxygen) | 10 | 55.6% | 15.999 S (sulfur) | 2 | 11.1% | 32.06 H (hydrogen) | 4 | 22.2% | 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 Ca (calcium) | 2 | 11.1% | 40.078 | 2 × 40.078 = 80.156 O (oxygen) | 10 | 55.6% | 15.999 | 10 × 15.999 = 159.990 S (sulfur) | 2 | 11.1% | 32.06 | 2 × 32.06 = 64.12 H (hydrogen) | 4 | 22.2% | 1.008 | 4 × 1.008 = 4.032 m = 80.156 u + 159.990 u + 64.12 u + 4.032 u = 308.298 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 Ca (calcium) | 2 | 11.1% | 80.156/308.298 O (oxygen) | 10 | 55.6% | 159.990/308.298 S (sulfur) | 2 | 11.1% | 64.12/308.298 H (hydrogen) | 4 | 22.2% | 4.032/308.298 Check: 80.156/308.298 + 159.990/308.298 + 64.12/308.298 + 4.032/308.298 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent Ca (calcium) | 2 | 11.1% | 80.156/308.298 × 100% = 26.00% O (oxygen) | 10 | 55.6% | 159.990/308.298 × 100% = 51.89% S (sulfur) | 2 | 11.1% | 64.12/308.298 × 100% = 20.80% H (hydrogen) | 4 | 22.2% | 4.032/308.298 × 100% = 1.308%

The first step in finding the oxidation states (or oxidation numbers) in calcium sulfate dihydrate calcium is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In calcium sulfate dihydrate calcium 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 8 oxygen-sulfur bonds. For each of these bonds, assign the bonding electrons to the most electronegative element. First examine 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. Decrease the oxidation number for oxygen in every highlighted bond (by 1 for single bonds, 2 for double bonds, and 3 for triple bonds), and increase the oxidation number for sulfur accordingly: Now summarize the results: Answer: | | oxidation state | element | count -2 | O (oxygen) | 10 +1 | H (hydrogen) | 4 +2 | Ca (calcium) | 2 +6 | S (sulfur) | 2

hybridization | element | count sp^2 | O (oxygen) | 4 sp^3 | O (oxygen) | 6 | S (sulfur) | 2

Orbital hybridization Structure diagram

vertex count | 18 edge count | 12 Schultz index | Wiener index | Hosoya index | Balaban index |