pentafluoride name of arsenic

pentafluoride arsenic

molar mass | 169.9 g/mol formula | AsF_5 empirical formula | As_F_5 SMILES identifier | [As+5].[F-].[F-].[F-].[F-].[F-] InChI identifier | InChI=1/As.5FH/h;5*1H/q+5;;;;;/p-5/fAs.5F/h;5*1h/qm;5*-1 InChI key | VYCHFPKKGGZYCD-UHFFFAOYSA-I

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 | 0 atoms H-bond donor count | 0 atoms

Find the elemental composition for pentafluoride arsenic 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: AsF_5 Use the chemical formula, AsF_5, to count the number of atoms, N_i, for each element and find the total number of atoms, N_atoms: | number of atoms As (arsenic) | 1 F (fluorine) | 5 N_atoms = 1 + 5 = 6 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 As (arsenic) | 1 | 1/6 F (fluorine) | 5 | 5/6 Check: 1/6 + 5/6 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent As (arsenic) | 1 | 1/6 × 100% = 16.7% F (fluorine) | 5 | 5/6 × 100% = 83.3% 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 As (arsenic) | 1 | 16.7% | 74.921595 F (fluorine) | 5 | 83.3% | 18.998403163 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 As (arsenic) | 1 | 16.7% | 74.921595 | 1 × 74.921595 = 74.921595 F (fluorine) | 5 | 83.3% | 18.998403163 | 5 × 18.998403163 = 94.992015815 m = 74.921595 u + 94.992015815 u = 169.913610815 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 As (arsenic) | 1 | 16.7% | 74.921595/169.913610815 F (fluorine) | 5 | 83.3% | 94.992015815/169.913610815 Check: 74.921595/169.913610815 + 94.992015815/169.913610815 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent As (arsenic) | 1 | 16.7% | 74.921595/169.913610815 × 100% = 44.09% F (fluorine) | 5 | 83.3% | 94.992015815/169.913610815 × 100% = 55.91%

The first step in finding the oxidation states (or oxidation numbers) in pentafluoride arsenic is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: Now summarize the results: Answer: | | oxidation state | element | count -1 | F (fluorine) | 5 +5 | As (arsenic) | 1

vertex count | 6 edge count | 0 Schultz index | Wiener index | Hosoya index | Balaban index |