Input interpretation
platinum(IV) bromide | molecular mass
Result
Find the molecular mass, m, for platinum(IV) 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: PtBr_4 Use the chemical formula, PtBr_4, to count the number of atoms, N_i, for each element: | N_i Br (bromine) | 4 Pt (platinum) | 1 Look up the atomic mass, m_i, in unified atomic mass units, u, for each element in the periodic table: | N_i | m_i/u Br (bromine) | 4 | 79.904 Pt (platinum) | 1 | 195.084 Multiply N_i by m_i to compute the mass for each element. Then sum those values to compute the molecular mass, m: Answer: | | | N_i | m_i/u | mass/u Br (bromine) | 4 | 79.904 | 4 × 79.904 = 319.616 Pt (platinum) | 1 | 195.084 | 1 × 195.084 = 195.084 m = 319.616 u + 195.084 u = 514.700 u
Unit conversions
514.7 Da (daltons)
0.5147 kDa (kilodaltons)
8.5468×10^-22 grams
8.5468×10^-25 kg (kilograms)
514.72 chemical atomic mass units (unit officially deprecated)
514.86 physical atomic mass units (unit officially deprecated)
Comparisons as mass of molecule
≈ 0.71 × molecular mass of fullerene ( ≈ 721 u )
≈ 2.7 × molecular mass of caffeine ( ≈ 194 u )
≈ 8.8 × molecular mass of sodium chloride ( ≈ 58 u )
Corresponding quantities
Relative atomic mass A_r from A_r = m_aN_A/M_u: | 515
Molar mass M from M = m_aN_A: | 515 g/mol (grams per mole)
Relative molecular mass M_r from M_r = m_mN_A/M_u: | 515
Molar mass M from M = m_mN_A: | 515 g/mol (grams per mole)