Input interpretation
lead(II) nitrate | molar mass
Result
Find the molar mass, M, for lead(II) nitrate: 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(NO_3)_2 Use the chemical formula to count the number of atoms, N_i, for each element: | N_i N (nitrogen) | 2 O (oxygen) | 6 Pb (lead) | 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) N (nitrogen) | 2 | 14.007 O (oxygen) | 6 | 15.999 Pb (lead) | 1 | 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) N (nitrogen) | 2 | 14.007 | 2 × 14.007 = 28.014 O (oxygen) | 6 | 15.999 | 6 × 15.999 = 95.994 Pb (lead) | 1 | 207.2 | 1 × 207.2 = 207.2 M = 28.014 g/mol + 95.994 g/mol + 207.2 g/mol = 331.2 g/mol
Unit conversion
0.3312 kg/mol (kilograms per mole)
Comparisons
≈ 0.46 × molar mass of fullerene ( ≈ 721 g/mol )
≈ 1.7 × molar mass of caffeine ( ≈ 194 g/mol )
≈ 5.7 × molar mass of sodium chloride ( ≈ 58 g/mol )
Corresponding quantities
Mass of a molecule m from m = M/N_A: | 5.5×10^-22 grams | 5.5×10^-25 kg (kilograms) | 331 u (unified atomic mass units) | 331 Da (daltons)
Relative molecular mass M_r from M_r = M_u/M: | 331