Input interpretation
bis(trifluoromethane)sulfonimide lithium salt | molar mass
Result
Find the molar mass, M, for bis(trifluoromethane)sulfonimide lithium salt: 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_2F_6LiNO_4S_2 Use the chemical formula, C_2F_6LiNO_4S_2, to count the number of atoms, N_i, for each element: | N_i Li (lithium) | 1 N (nitrogen) | 1 S (sulfur) | 2 O (oxygen) | 4 C (carbon) | 2 F (fluorine) | 6 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) Li (lithium) | 1 | 6.94 N (nitrogen) | 1 | 14.007 S (sulfur) | 2 | 32.06 O (oxygen) | 4 | 15.999 C (carbon) | 2 | 12.011 F (fluorine) | 6 | 18.998403163 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) Li (lithium) | 1 | 6.94 | 1 × 6.94 = 6.94 N (nitrogen) | 1 | 14.007 | 1 × 14.007 = 14.007 S (sulfur) | 2 | 32.06 | 2 × 32.06 = 64.12 O (oxygen) | 4 | 15.999 | 4 × 15.999 = 63.996 C (carbon) | 2 | 12.011 | 2 × 12.011 = 24.022 F (fluorine) | 6 | 18.998403163 | 6 × 18.998403163 = 113.990418978 M = 6.94 g/mol + 14.007 g/mol + 64.12 g/mol + 63.996 g/mol + 24.022 g/mol + 113.990418978 g/mol = 287.08 g/mol
Unit conversion
0.2871 kg/mol (kilograms per mole)
Comparisons
≈ 0.4 × molar mass of fullerene ( ≈ 721 g/mol )
≈ 1.5 × molar mass of caffeine ( ≈ 194 g/mol )
≈ 4.9 × molar mass of sodium chloride ( ≈ 58 g/mol )
Corresponding quantities
Mass of a molecule m from m = M/N_A: | 4.8×10^-22 grams | 4.8×10^-25 kg (kilograms) | 287 u (unified atomic mass units) | 287 Da (daltons)
Relative molecular mass M_r from M_r = M_u/M: | 287