praseodymium nitrate pentahydrate nickel

praseodymium nitrate pentahydrate nickel

molar mass | 599.7 g/mol formula | H_10N_5NiO_20Pr empirical formula | Ni_O_20N_5Pr_H_10 SMILES identifier | [N+](=O)([O-])[O-].[N+](=O)([O-])[O-].[N+](=O)([O-])[O-].[N+](=O)([O-])[O-].[N+](=O)([O-])[O-].[Ni+2].O.O.O.O.O.[Pr+3] InChI identifier | InChI=1/5NO3.Ni.5H2O.Pr/c5*2-1(3)4;;;;;;;/h;;;;;;5*1H2;/q5*-1;+2;;;;;;+3 InChI key | NKPGKBYNCMKBGP-UHFFFAOYSA-N

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 | 5 atoms H-bond donor count | 5 atoms

Find the elemental composition for praseodymium nitrate pentahydrate nickel 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: H_10N_5NiO_20Pr Use the chemical formula, H_10N_5NiO_20Pr, to count the number of atoms, N_i, for each element and find the total number of atoms, N_atoms: | number of atoms Ni (nickel) | 1 O (oxygen) | 20 N (nitrogen) | 5 Pr (praseodymium) | 1 H (hydrogen) | 10 N_atoms = 1 + 20 + 5 + 1 + 10 = 37 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 Ni (nickel) | 1 | 1/37 O (oxygen) | 20 | 20/37 N (nitrogen) | 5 | 5/37 Pr (praseodymium) | 1 | 1/37 H (hydrogen) | 10 | 10/37 Check: 1/37 + 20/37 + 5/37 + 1/37 + 10/37 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent Ni (nickel) | 1 | 1/37 × 100% = 2.70% O (oxygen) | 20 | 20/37 × 100% = 54.1% N (nitrogen) | 5 | 5/37 × 100% = 13.5% Pr (praseodymium) | 1 | 1/37 × 100% = 2.70% H (hydrogen) | 10 | 10/37 × 100% = 27.0% 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 Ni (nickel) | 1 | 2.70% | 58.6934 O (oxygen) | 20 | 54.1% | 15.999 N (nitrogen) | 5 | 13.5% | 14.007 Pr (praseodymium) | 1 | 2.70% | 140.90766 H (hydrogen) | 10 | 27.0% | 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 Ni (nickel) | 1 | 2.70% | 58.6934 | 1 × 58.6934 = 58.6934 O (oxygen) | 20 | 54.1% | 15.999 | 20 × 15.999 = 319.980 N (nitrogen) | 5 | 13.5% | 14.007 | 5 × 14.007 = 70.035 Pr (praseodymium) | 1 | 2.70% | 140.90766 | 1 × 140.90766 = 140.90766 H (hydrogen) | 10 | 27.0% | 1.008 | 10 × 1.008 = 10.080 m = 58.6934 u + 319.980 u + 70.035 u + 140.90766 u + 10.080 u = 599.69606 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 Ni (nickel) | 1 | 2.70% | 58.6934/599.69606 O (oxygen) | 20 | 54.1% | 319.980/599.69606 N (nitrogen) | 5 | 13.5% | 70.035/599.69606 Pr (praseodymium) | 1 | 2.70% | 140.90766/599.69606 H (hydrogen) | 10 | 27.0% | 10.080/599.69606 Check: 58.6934/599.69606 + 319.980/599.69606 + 70.035/599.69606 + 140.90766/599.69606 + 10.080/599.69606 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent Ni (nickel) | 1 | 2.70% | 58.6934/599.69606 × 100% = 9.787% O (oxygen) | 20 | 54.1% | 319.980/599.69606 × 100% = 53.36% N (nitrogen) | 5 | 13.5% | 70.035/599.69606 × 100% = 11.68% Pr (praseodymium) | 1 | 2.70% | 140.90766/599.69606 × 100% = 23.50% H (hydrogen) | 10 | 27.0% | 10.080/599.69606 × 100% = 1.681%

The first step in finding the oxidation states (or oxidation numbers) in praseodymium nitrate pentahydrate nickel is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In praseodymium nitrate pentahydrate nickel 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 15 nitrogen-oxygen bonds. For each of these bonds, assign the bonding electrons to the most electronegative element. First examine the nitrogen-oxygen bonds: element | electronegativity (Pauling scale) | N | 3.04 | O | 3.44 | | | Since oxygen is more electronegative than nitrogen, 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 nitrogen accordingly: Now summarize the results: Answer: | | oxidation state | element | count -2 | O (oxygen) | 20 +1 | H (hydrogen) | 10 +2 | Ni (nickel) | 1 +3 | Pr (praseodymium) | 1 +5 | N (nitrogen) | 5

vertex count | 37 edge count | 25 Schultz index | Wiener index | Hosoya index | Balaban index |