Input interpretation
dichromate
Basic properties
molar mass | 216 g/mol formula | (Cr_2O_7)^2- empirical formula | Cr_2O_7 SMILES identifier | [Cr](=O)(=O)([O-])O[Cr](=O)(=O)[O-] InChI identifier | InChI=1/2Cr.7O/q;;;;;;;2*-1 InChI key | SOCTUWSJJQCPFX-UHFFFAOYSA-N
Structure diagram
vertex count | 9 edge count | 8 Schultz index | 322 Wiener index | 88 Hosoya index | 24 Balaban index | 3.746
Quantitative molecular descriptors
longest chain length | 5 atoms longest straight chain length | 5 atoms longest aliphatic chain length | 0 atoms aromatic atom count | 0 atoms H-bond acceptor count | 7 atoms H-bond donor count | 0 atoms
Elemental composition
Find the elemental composition for dichromate 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: (Cr_2O_7)^2- Use the chemical formula, (Cr_2O_7)^2-, to count the number of atoms, N_i, for each element and find the total number of atoms, N_atoms: | number of atoms Cr (chromium) | 2 O (oxygen) | 7 N_atoms = 2 + 7 = 9 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 Cr (chromium) | 2 | 2/9 O (oxygen) | 7 | 7/9 Check: 2/9 + 7/9 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent Cr (chromium) | 2 | 2/9 × 100% = 22.2% O (oxygen) | 7 | 7/9 × 100% = 77.8% 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 Cr (chromium) | 2 | 22.2% | 51.9961 O (oxygen) | 7 | 77.8% | 15.999 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 Cr (chromium) | 2 | 22.2% | 51.9961 | 2 × 51.9961 = 103.9922 O (oxygen) | 7 | 77.8% | 15.999 | 7 × 15.999 = 111.993 m = 103.9922 u + 111.993 u = 215.9852 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 Cr (chromium) | 2 | 22.2% | 103.9922/215.9852 O (oxygen) | 7 | 77.8% | 111.993/215.9852 Check: 103.9922/215.9852 + 111.993/215.9852 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent Cr (chromium) | 2 | 22.2% | 103.9922/215.9852 × 100% = 48.15% O (oxygen) | 7 | 77.8% | 111.993/215.9852 × 100% = 51.85%
Elemental oxidation states
The first step in finding the oxidation states (or oxidation numbers) in dichromate is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: There are 8 chromium-oxygen bonds in dichromate. For each of these bonds, assign the bonding electrons to the most electronegative element. First examine the chromium-oxygen bonds: element | electronegativity (Pauling scale) | Cr | 1.66 | O | 3.44 | | | Since oxygen is more electronegative than chromium, 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 chromium accordingly: Now summarize the results: Answer: | | oxidation state | element | count -2 | O (oxygen) | 7 +6 | Cr (chromium) | 2
Topological indices
vertex count | 9 edge count | 8 Schultz index | 322 Wiener index | 88 Hosoya index | 24 Balaban index | 3.746