Input interpretation
barium bromate monohydrate | molar mass
Result
Find the molar mass, M, for barium bromate monohydrate: 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: Ba(BrO_3)_2·H_2O Use the chemical formula to count the number of atoms, N_i, for each element: | N_i Ba (barium) | 1 Br (bromine) | 2 H (hydrogen) | 2 O (oxygen) | 7 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) Ba (barium) | 1 | 137.327 Br (bromine) | 2 | 79.904 H (hydrogen) | 2 | 1.008 O (oxygen) | 7 | 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) Ba (barium) | 1 | 137.327 | 1 × 137.327 = 137.327 Br (bromine) | 2 | 79.904 | 2 × 79.904 = 159.808 H (hydrogen) | 2 | 1.008 | 2 × 1.008 = 2.016 O (oxygen) | 7 | 15.999 | 7 × 15.999 = 111.993 M = 137.327 g/mol + 159.808 g/mol + 2.016 g/mol + 111.993 g/mol = 411.144 g/mol
Unit conversion
0.41114 kg/mol (kilograms per mole)
Comparisons
≈ 0.57 × molar mass of fullerene ( ≈ 721 g/mol )
≈ 2.1 × molar mass of caffeine ( ≈ 194 g/mol )
≈ 7 × molar mass of sodium chloride ( ≈ 58 g/mol )
Corresponding quantities
Mass of a molecule m from m = M/N_A: | 6.8×10^-22 grams | 6.8×10^-25 kg (kilograms) | 411 u (unified atomic mass units) | 411 Da (daltons)
Relative molecular mass M_r from M_r = M_u/M: | 411