dichromate

dichromate

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

vertex count | 9 edge count | 8 Schultz index | 322 Wiener index | 88 Hosoya index | 24 Balaban index | 3.746

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

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 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 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%

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

vertex count | 9 edge count | 8 Schultz index | 322 Wiener index | 88 Hosoya index | 24 Balaban index | 3.746