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molar mass of ammonium trifluoromethanesulfonate

Input interpretation

ammonium trifluoromethanesulfonate | molar mass
ammonium trifluoromethanesulfonate | molar mass

Result

Find the molar mass, M, for ammonium 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: CF_3SO_3NH_4 Use the chemical formula to count the number of atoms, N_i, for each element:  | N_i  C (carbon) | 1  F (fluorine) | 3  H (hydrogen) | 4  N (nitrogen) | 1  O (oxygen) | 3  S (sulfur) | 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) | 1 | 12.011  F (fluorine) | 3 | 18.998403163  H (hydrogen) | 4 | 1.008  N (nitrogen) | 1 | 14.007  O (oxygen) | 3 | 15.999  S (sulfur) | 1 | 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) | 1 | 12.011 | 1 × 12.011 = 12.011  F (fluorine) | 3 | 18.998403163 | 3 × 18.998403163 = 56.995209489  H (hydrogen) | 4 | 1.008 | 4 × 1.008 = 4.032  N (nitrogen) | 1 | 14.007 | 1 × 14.007 = 14.007  O (oxygen) | 3 | 15.999 | 3 × 15.999 = 47.997  S (sulfur) | 1 | 32.06 | 1 × 32.06 = 32.06  M = 12.011 g/mol + 56.995209489 g/mol + 4.032 g/mol + 14.007 g/mol + 47.997 g/mol + 32.06 g/mol = 167.10 g/mol
Find the molar mass, M, for ammonium 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: CF_3SO_3NH_4 Use the chemical formula to count the number of atoms, N_i, for each element: | N_i C (carbon) | 1 F (fluorine) | 3 H (hydrogen) | 4 N (nitrogen) | 1 O (oxygen) | 3 S (sulfur) | 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) | 1 | 12.011 F (fluorine) | 3 | 18.998403163 H (hydrogen) | 4 | 1.008 N (nitrogen) | 1 | 14.007 O (oxygen) | 3 | 15.999 S (sulfur) | 1 | 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) | 1 | 12.011 | 1 × 12.011 = 12.011 F (fluorine) | 3 | 18.998403163 | 3 × 18.998403163 = 56.995209489 H (hydrogen) | 4 | 1.008 | 4 × 1.008 = 4.032 N (nitrogen) | 1 | 14.007 | 1 × 14.007 = 14.007 O (oxygen) | 3 | 15.999 | 3 × 15.999 = 47.997 S (sulfur) | 1 | 32.06 | 1 × 32.06 = 32.06 M = 12.011 g/mol + 56.995209489 g/mol + 4.032 g/mol + 14.007 g/mol + 47.997 g/mol + 32.06 g/mol = 167.10 g/mol

Unit conversion

0.1671 kg/mol (kilograms per mole)
0.1671 kg/mol (kilograms per mole)

Comparisons

 ≈ ( 0.23 ≈ 1/4 ) × molar mass of fullerene ( ≈ 721 g/mol )
≈ ( 0.23 ≈ 1/4 ) × molar mass of fullerene ( ≈ 721 g/mol )
 ≈ 0.86 × molar mass of caffeine ( ≈ 194 g/mol )
≈ 0.86 × molar mass of caffeine ( ≈ 194 g/mol )
 ≈ 2.9 × molar mass of sodium chloride ( ≈ 58 g/mol )
≈ 2.9 × molar mass of sodium chloride ( ≈ 58 g/mol )

Corresponding quantities

Mass of a molecule m from m = M/N_A:  | 2.8×10^-22 grams  | 2.8×10^-25 kg (kilograms)  | 167 u (unified atomic mass units)  | 167 Da (daltons)
Mass of a molecule m from m = M/N_A: | 2.8×10^-22 grams | 2.8×10^-25 kg (kilograms) | 167 u (unified atomic mass units) | 167 Da (daltons)
Relative molecular mass M_r from M_r = M_u/M:  | 167
Relative molecular mass M_r from M_r = M_u/M: | 167