Input interpretation
lead(II) acetate trihydrate | molar mass
Result
Find the molar mass, M, for lead(II) acetate trihydrate: 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(CH_3CO_2)_2·3H_2O Use the chemical formula to count the number of atoms, N_i, for each element: | N_i C (carbon) | 4 H (hydrogen) | 12 O (oxygen) | 7 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) C (carbon) | 4 | 12.011 H (hydrogen) | 12 | 1.008 O (oxygen) | 7 | 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) C (carbon) | 4 | 12.011 | 4 × 12.011 = 48.044 H (hydrogen) | 12 | 1.008 | 12 × 1.008 = 12.096 O (oxygen) | 7 | 15.999 | 7 × 15.999 = 111.993 Pb (lead) | 1 | 207.2 | 1 × 207.2 = 207.2 M = 48.044 g/mol + 12.096 g/mol + 111.993 g/mol + 207.2 g/mol = 379.3 g/mol
Unit conversion
0.3793 kg/mol (kilograms per mole)
Comparisons
≈ 0.53 × molar mass of fullerene ( ≈ 721 g/mol )
≈ 2 × molar mass of caffeine ( ≈ 194 g/mol )
≈ 6.5 × molar mass of sodium chloride ( ≈ 58 g/mol )
Corresponding quantities
Mass of a molecule m from m = M/N_A: | 6.3×10^-22 grams | 6.3×10^-25 kg (kilograms) | 379 u (unified atomic mass units) | 379 Da (daltons)
Relative molecular mass M_r from M_r = M_u/M: | 379