Input interpretation
iron(III) bromide | molar mass
Result
Find the molar mass, M, for iron(III) bromide: 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: FeBr_3 Use the chemical formula, FeBr_3, to count the number of atoms, N_i, for each element: | N_i Br (bromine) | 3 Fe (iron) | 1 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) Br (bromine) | 3 | 79.904 Fe (iron) | 1 | 55.845 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) Br (bromine) | 3 | 79.904 | 3 × 79.904 = 239.712 Fe (iron) | 1 | 55.845 | 1 × 55.845 = 55.845 M = 239.712 g/mol + 55.845 g/mol = 295.557 g/mol
Unit conversion
0.29556 kg/mol (kilograms per mole)
Comparisons
≈ 0.41 × molar mass of fullerene ( ≈ 721 g/mol )
≈ 1.5 × molar mass of caffeine ( ≈ 194 g/mol )
≈ 5.1 × molar mass of sodium chloride ( ≈ 58 g/mol )
Corresponding quantities
Mass of a molecule m from m = M/N_A: | 4.9×10^-22 grams | 4.9×10^-25 kg (kilograms) | 296 u (unified atomic mass units) | 296 Da (daltons)
Relative molecular mass M_r from M_r = M_u/M: | 296