Input interpretation
tetraamminepalladium(II) acetate | molar mass
Result
Find the molar mass, M, for tetraamminepalladium(II) acetate: 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: Pd(NH_3)_4(CH_3CO_2)_2 Use the chemical formula, Pd(NH_3)_4(CH_3CO_2)_2, to count the number of atoms, N_i, for each element: | N_i C (carbon) | 2 H (hydrogen) | 11 N (nitrogen) | 1 O (oxygen) | 4 Pd (palladium) | 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) | 2 | 12.011 H (hydrogen) | 11 | 1.008 N (nitrogen) | 1 | 14.007 O (oxygen) | 4 | 15.999 Pd (palladium) | 1 | 106.42 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) | 2 | 12.011 | 2 × 12.011 = 24.022 H (hydrogen) | 11 | 1.008 | 11 × 1.008 = 11.088 N (nitrogen) | 1 | 14.007 | 1 × 14.007 = 14.007 O (oxygen) | 4 | 15.999 | 4 × 15.999 = 63.996 Pd (palladium) | 1 | 106.42 | 1 × 106.42 = 106.42 M = 24.022 g/mol + 11.088 g/mol + 14.007 g/mol + 63.996 g/mol + 106.42 g/mol = 219.53 g/mol
Unit conversion
0.21953 kg/mol (kilograms per mole)
Comparisons
≈ 0.3 × molar mass of fullerene ( ≈ 721 g/mol )
≈ 1.1 × molar mass of caffeine ( ≈ 194 g/mol )
≈ 3.8 × molar mass of sodium chloride ( ≈ 58 g/mol )
Corresponding quantities
Mass of a molecule m from m = M/N_A: | 3.6×10^-22 grams | 3.6×10^-25 kg (kilograms) | 220 u (unified atomic mass units) | 220 Da (daltons)
Relative molecular mass M_r from M_r = M_u/M: | 220