trifluoride name of iodine

trifluoride iodine

molar mass | 183.9 g/mol formula | F_3I empirical formula | I_F_3 SMILES identifier | [F-].[F-].[F-].[I+3] InChI identifier | InChI=1/3FH.I/h3*1H;/q;;;+3/p-3/f3F.I/h3*1h;/q3*-1;m InChI key | OKILTMQCKKIQBX-UHFFFAOYSA-K

Structure diagram

melting point | -156.3 °C boiling point | 18.6 °C critical temperature | 469.7 K critical pressure | 4.672 MPa critical volume | 195.5 cm^3/mol molar heat of vaporization | 22.8 kJ/mol molar heat of fusion | 6.036 kJ/mol molar enthalpy | -666.4 kJ/mol molar free energy | -682 kJ/mol (computed using the Joback method)

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 trifluoride iodine 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: F_3I Use the chemical formula, F_3I, to count the number of atoms, N_i, for each element and find the total number of atoms, N_atoms: | number of atoms I (iodine) | 1 F (fluorine) | 3 N_atoms = 1 + 3 = 4 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 I (iodine) | 1 | 1/4 F (fluorine) | 3 | 3/4 Check: 1/4 + 3/4 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent I (iodine) | 1 | 1/4 × 100% = 25.0% F (fluorine) | 3 | 3/4 × 100% = 75.0% 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 I (iodine) | 1 | 25.0% | 126.90447 F (fluorine) | 3 | 75.0% | 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 I (iodine) | 1 | 25.0% | 126.90447 | 1 × 126.90447 = 126.90447 F (fluorine) | 3 | 75.0% | 18.998403163 | 3 × 18.998403163 = 56.995209489 m = 126.90447 u + 56.995209489 u = 183.899679489 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 I (iodine) | 1 | 25.0% | 126.90447/183.899679489 F (fluorine) | 3 | 75.0% | 56.995209489/183.899679489 Check: 126.90447/183.899679489 + 56.995209489/183.899679489 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent I (iodine) | 1 | 25.0% | 126.90447/183.899679489 × 100% = 69.01% F (fluorine) | 3 | 75.0% | 56.995209489/183.899679489 × 100% = 30.99%

The first step in finding the oxidation states (or oxidation numbers) in trifluoride iodine 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) | 3 +3 | I (iodine) | 1

vertex count | 4 edge count | 0 Schultz index | Wiener index | Hosoya index | Balaban index |