Input interpretation
iodide 2 methylphenylzinc iodide
Basic properties
![molar mass | 410.3 g/mol formula | C_7H_7I_2Zn empirical formula | I_2C_7Zn_H_7 SMILES identifier | CC1=CC=CC=C1[Zn+2].[I-].[I-] InChI identifier | InChI=1/C7H7.2HI.Zn/c1-7-5-3-2-4-6-7;;;/h2-5H, 1H3;2*1H;/q;;;+2/p-2/fC7H7.2I.Zn/h;2*1h;/q;2*-1;m InChI key | SCDMKQIKWBVGCH-UHFFFAOYSA-L](../image_source/f097a866c3cff1785c532f5c9d0351e5.png)
molar mass | 410.3 g/mol formula | C_7H_7I_2Zn empirical formula | I_2C_7Zn_H_7 SMILES identifier | CC1=CC=CC=C1[Zn+2].[I-].[I-] InChI identifier | InChI=1/C7H7.2HI.Zn/c1-7-5-3-2-4-6-7;;;/h2-5H, 1H3;2*1H;/q;;;+2/p-2/fC7H7.2I.Zn/h;2*1h;/q;2*-1;m InChI key | SCDMKQIKWBVGCH-UHFFFAOYSA-L
Structure diagram
vertex count | 10 edge count | 8 Schultz index | 264 Wiener index | 60 Hosoya index | 39 Balaban index | 2.279
Quantitative molecular descriptors
longest chain length | 5 atoms longest straight chain length | 0 atoms longest aliphatic chain length | 0 atoms aromatic atom count | 6 atoms H-bond acceptor count | 0 atoms H-bond donor count | 0 atoms
Elemental composition
Find the elemental composition for iodide 2 methylphenylzinc iodide 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: C_7H_7I_2Zn Use the chemical formula, C_7H_7I_2Zn, 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) | 2 C (carbon) | 7 Zn (zinc) | 1 H (hydrogen) | 7 N_atoms = 2 + 7 + 1 + 7 = 17 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) | 2 | 2/17 C (carbon) | 7 | 7/17 Zn (zinc) | 1 | 1/17 H (hydrogen) | 7 | 7/17 Check: 2/17 + 7/17 + 1/17 + 7/17 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent I (iodine) | 2 | 2/17 × 100% = 11.8% C (carbon) | 7 | 7/17 × 100% = 41.2% Zn (zinc) | 1 | 1/17 × 100% = 5.88% H (hydrogen) | 7 | 7/17 × 100% = 41.2% 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) | 2 | 11.8% | 126.90447 C (carbon) | 7 | 41.2% | 12.011 Zn (zinc) | 1 | 5.88% | 65.38 H (hydrogen) | 7 | 41.2% | 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 I (iodine) | 2 | 11.8% | 126.90447 | 2 × 126.90447 = 253.80894 C (carbon) | 7 | 41.2% | 12.011 | 7 × 12.011 = 84.077 Zn (zinc) | 1 | 5.88% | 65.38 | 1 × 65.38 = 65.38 H (hydrogen) | 7 | 41.2% | 1.008 | 7 × 1.008 = 7.056 m = 253.80894 u + 84.077 u + 65.38 u + 7.056 u = 410.32194 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) | 2 | 11.8% | 253.80894/410.32194 C (carbon) | 7 | 41.2% | 84.077/410.32194 Zn (zinc) | 1 | 5.88% | 65.38/410.32194 H (hydrogen) | 7 | 41.2% | 7.056/410.32194 Check: 253.80894/410.32194 + 84.077/410.32194 + 65.38/410.32194 + 7.056/410.32194 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent I (iodine) | 2 | 11.8% | 253.80894/410.32194 × 100% = 61.86% C (carbon) | 7 | 41.2% | 84.077/410.32194 × 100% = 20.49% Zn (zinc) | 1 | 5.88% | 65.38/410.32194 × 100% = 15.93% H (hydrogen) | 7 | 41.2% | 7.056/410.32194 × 100% = 1.720%
Topological indices
vertex count | 17 edge count | 15 Schultz index | 1239 Wiener index | 318 Hosoya index | 582 Balaban index | 2.969