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molecular mass of lutetium(III) trifluoromethanesulfonate

Input interpretation

lutetium(III) trifluoromethanesulfonate | molecular mass
lutetium(III) trifluoromethanesulfonate | molecular mass

Result

Find the molecular 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, Lu(CF_3SO_3)_3, 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 unified atomic mass units, u, for each element in the periodic table:  | N_i | m_i/u  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 molecular mass, m: Answer: |   | | N_i | m_i/u | mass/u  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 u + 170.985628467 u + 174.9668 u + 143.991 u + 96.18 u = 622.16 u
Find the molecular 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, Lu(CF_3SO_3)_3, 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 unified atomic mass units, u, for each element in the periodic table: | N_i | m_i/u 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 molecular mass, m: Answer: | | | N_i | m_i/u | mass/u 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 u + 170.985628467 u + 174.9668 u + 143.991 u + 96.18 u = 622.16 u

Unit conversions

622.2 Da (daltons)
622.2 Da (daltons)
0.6222 kDa (kilodaltons)
0.6222 kDa (kilodaltons)
1.0331×10^-21 grams
1.0331×10^-21 grams
1.0331×10^-24 kg (kilograms)
1.0331×10^-24 kg (kilograms)
622.2 chemical atomic mass units  (unit officially deprecated)
622.2 chemical atomic mass units (unit officially deprecated)
622.4 physical atomic mass units  (unit officially deprecated)
622.4 physical atomic mass units (unit officially deprecated)

Comparisons as mass of molecule

 ≈ 0.86 × molecular mass of fullerene ( ≈ 721 u )
≈ 0.86 × molecular mass of fullerene ( ≈ 721 u )
 ≈ 3.2 × molecular mass of caffeine ( ≈ 194 u )
≈ 3.2 × molecular mass of caffeine ( ≈ 194 u )
 ≈ 11 × molecular mass of sodium chloride ( ≈ 58 u )
≈ 11 × molecular mass of sodium chloride ( ≈ 58 u )

Corresponding quantities

Relative atomic mass A_r from A_r = m_aN_A/M_u:  | 622
Relative atomic mass A_r from A_r = m_aN_A/M_u: | 622
Molar mass M from M = m_aN_A:  | 622 g/mol (grams per mole)
Molar mass M from M = m_aN_A: | 622 g/mol (grams per mole)
Relative molecular mass M_r from M_r = m_mN_A/M_u:  | 622
Relative molecular mass M_r from M_r = m_mN_A/M_u: | 622
Molar mass M from M = m_mN_A:  | 622 g/mol (grams per mole)
Molar mass M from M = m_mN_A: | 622 g/mol (grams per mole)