Input interpretation
lutetium(III) trifluoromethanesulfonate | molar mass
Result
Find the molar mass, M, for lutetium(III) trifluoromethanesulfonate: 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: Lu(CF_3SO_3)_3 Use the chemical formula to count the number of atoms, N_i, for each element: | N_i C (carbon) | 3 F (fluorine) | 9 Lu (lutetium) | 1 O (oxygen) | 9 S (sulfur) | 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) C (carbon) | 3 | 12.011 F (fluorine) | 9 | 18.998403163 Lu (lutetium) | 1 | 174.9668 O (oxygen) | 9 | 15.999 S (sulfur) | 3 | 32.06 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) | 3 | 12.011 | 3 × 12.011 = 36.033 F (fluorine) | 9 | 18.998403163 | 9 × 18.998403163 = 170.985628467 Lu (lutetium) | 1 | 174.9668 | 1 × 174.9668 = 174.9668 O (oxygen) | 9 | 15.999 | 9 × 15.999 = 143.991 S (sulfur) | 3 | 32.06 | 3 × 32.06 = 96.18 M = 36.033 g/mol + 170.985628467 g/mol + 174.9668 g/mol + 143.991 g/mol + 96.18 g/mol = 622.16 g/mol
Unit conversion
0.6222 kg/mol (kilograms per mole)
Comparisons
≈ 0.86 × molar mass of fullerene ( ≈ 721 g/mol )
≈ 3.2 × molar mass of caffeine ( ≈ 194 g/mol )
≈ 11 × molar mass of sodium chloride ( ≈ 58 g/mol )
Corresponding quantities
Mass of a molecule m from m = M/N_A: | 1×10^-21 grams | 1×10^-24 kg (kilograms) | 622 u (unified atomic mass units) | 622 Da (daltons)
Relative molecular mass M_r from M_r = M_u/M: | 622