Input interpretation
lead(II) borate monohydrate | molar mass
Result
Find the molar mass, M, for lead(II) borate 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: Pb(BO_2)_2·H_2O Use the chemical formula to count the number of atoms, N_i, for each element: | N_i B (boron) | 2 H (hydrogen) | 2 O (oxygen) | 7 Pb (lead) | 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) B (boron) | 2 | 10.81 H (hydrogen) | 2 | 1.008 O (oxygen) | 7 | 15.999 Pb (lead) | 3 | 207.2 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) B (boron) | 2 | 10.81 | 2 × 10.81 = 21.62 H (hydrogen) | 2 | 1.008 | 2 × 1.008 = 2.016 O (oxygen) | 7 | 15.999 | 7 × 15.999 = 111.993 Pb (lead) | 3 | 207.2 | 3 × 207.2 = 621.6 M = 21.62 g/mol + 2.016 g/mol + 111.993 g/mol + 621.6 g/mol = 757.2 g/mol
Unit conversion
0.7572 kg/mol (kilograms per mole)
Comparisons
≈ 1.1 × molar mass of fullerene ( ≈ 721 g/mol )
≈ 3.9 × molar mass of caffeine ( ≈ 194 g/mol )
≈ 13 × molar mass of sodium chloride ( ≈ 58 g/mol )
Corresponding quantities
Mass of a molecule m from m = M/N_A: | 1.3×10^-21 grams | 1.3×10^-24 kg (kilograms) | 757 u (unified atomic mass units) | 757 Da (daltons)
Relative molecular mass M_r from M_r = M_u/M: | 757