Input interpretation
titanium(III) chloride-aluminum chloride | molar mass
Result
Find the molar mass, M, for titanium(III) chloride-aluminum chloride: M = sum _iN_im_i Plan: • Write the chemical formula and gather atomic masses from the periodic table. • Determine values for N_i and m_i using these items. • Finally, compute the mass. Write the chemical formula: (TiCl_3)_3·AlCl_3 Use the chemical formula to count the number of atoms, N_i, for each element: | N_i Al (aluminum) | 1 Cl (chlorine) | 12 Ti (titanium) | 3 Look up the atomic mass, m_i, in g·mol^(-1) for each element in the periodic table: | N_i | m_i/g·mol^(-1) Al (aluminum) | 1 | 26.9815385 Cl (chlorine) | 12 | 35.45 Ti (titanium) | 3 | 47.867 Multiply N_i by m_i to compute the mass for each element. Then sum those values to compute the molar mass, M: Answer: | | | N_i | m_i/g·mol^(-1) | mass/g·mol^(-1) Al (aluminum) | 1 | 26.9815385 | 1 × 26.9815385 = 26.9815385 Cl (chlorine) | 12 | 35.45 | 12 × 35.45 = 425.40 Ti (titanium) | 3 | 47.867 | 3 × 47.867 = 143.601 M = 26.9815385 g/mol + 425.40 g/mol + 143.601 g/mol = 595.98 g/mol
Unit conversion
0.596 kg/mol (kilograms per mole)
Comparisons
≈ 0.83 × molar mass of fullerene ( ≈ 721 g/mol )
≈ 3.1 × molar mass of caffeine ( ≈ 194 g/mol )
≈ 10 × molar mass of sodium chloride ( ≈ 58 g/mol )
Corresponding quantities
Mass of a molecule m from m = M/N_A: | 9.9×10^-22 grams | 9.9×10^-25 kg (kilograms) | 596 u (unified atomic mass units) | 596 Da (daltons)
Relative molecular mass M_r from M_r = M_u/M: | 596