Input interpretation
hafnium(IV) n-butoxide | molar mass
Result
Find the molar mass, M, for hafnium(IV) n-butoxide: 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: C_16H_36HfO_4 Use the chemical formula to count the number of atoms, N_i, for each element: | N_i C (carbon) | 16 H (hydrogen) | 36 Hf (hafnium) | 1 O (oxygen) | 4 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) | 16 | 12.011 H (hydrogen) | 36 | 1.008 Hf (hafnium) | 1 | 178.49 O (oxygen) | 4 | 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) C (carbon) | 16 | 12.011 | 16 × 12.011 = 192.176 H (hydrogen) | 36 | 1.008 | 36 × 1.008 = 36.288 Hf (hafnium) | 1 | 178.49 | 1 × 178.49 = 178.49 O (oxygen) | 4 | 15.999 | 4 × 15.999 = 63.996 M = 192.176 g/mol + 36.288 g/mol + 178.49 g/mol + 63.996 g/mol = 470.95 g/mol
Unit conversion
0.47095 kg/mol (kilograms per mole)
Comparisons
≈ 0.65 × molar mass of fullerene ( ≈ 721 g/mol )
≈ 2.4 × molar mass of caffeine ( ≈ 194 g/mol )
≈ 8.1 × molar mass of sodium chloride ( ≈ 58 g/mol )
Corresponding quantities
Mass of a molecule m from m = M/N_A: | 7.8×10^-22 grams | 7.8×10^-25 kg (kilograms) | 471 u (unified atomic mass units) | 471 Da (daltons)
Relative molecular mass M_r from M_r = M_u/M: | 471