Input interpretation
neodymium(III) bromide hydrate | molar mass
Result
Find the molar mass, M, for neodymium(III) bromide hydrate: 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: NdBr_3·xH_2O Use the chemical formula to count the number of atoms, N_i, for each element: | N_i Br (bromine) | 3 H (hydrogen) | 2 Nd (neodymium) | 1 O (oxygen) | 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 H (hydrogen) | 2 | 1.008 Nd (neodymium) | 1 | 144.242 O (oxygen) | 1 | 15.999 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 H (hydrogen) | 2 | 1.008 | 2 × 1.008 = 2.016 Nd (neodymium) | 1 | 144.242 | 1 × 144.242 = 144.242 O (oxygen) | 1 | 15.999 | 1 × 15.999 = 15.999 M = 239.712 g/mol + 2.016 g/mol + 144.242 g/mol + 15.999 g/mol = 401.969 g/mol
Unit conversion
0.40197 kg/mol (kilograms per mole)
Comparisons
≈ 0.56 × molar mass of fullerene ( ≈ 721 g/mol )
≈ 2.1 × molar mass of caffeine ( ≈ 194 g/mol )
≈ 6.9 × molar mass of sodium chloride ( ≈ 58 g/mol )
Corresponding quantities
Mass of a molecule m from m = M/N_A: | 6.7×10^-22 grams | 6.7×10^-25 kg (kilograms) | 402 u (unified atomic mass units) | 402 Da (daltons)
Relative molecular mass M_r from M_r = M_u/M: | 402