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molar mass of 1-ethyl-3-methylimidazolium trifluoromethanesulfonate

Input interpretation

1-ethyl-3-methylimidazolium tetrafluoroborate | molar mass
1-ethyl-3-methylimidazolium tetrafluoroborate | molar mass

Result

Find the molar mass, M, for 1-ethyl-3-methylimidazolium tetrafluoroborate: 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_6H_11BF_4N_2 Use the chemical formula, C_6H_11BF_4N_2, to count the number of atoms, N_i, for each element:  | N_i  B (boron) | 1  C (carbon) | 6  F (fluorine) | 4  H (hydrogen) | 11  N (nitrogen) | 2 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)  B (boron) | 1 | 10.81  C (carbon) | 6 | 12.011  F (fluorine) | 4 | 18.998403163  H (hydrogen) | 11 | 1.008  N (nitrogen) | 2 | 14.007 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)  B (boron) | 1 | 10.81 | 1 × 10.81 = 10.81  C (carbon) | 6 | 12.011 | 6 × 12.011 = 72.066  F (fluorine) | 4 | 18.998403163 | 4 × 18.998403163 = 75.993612652  H (hydrogen) | 11 | 1.008 | 11 × 1.008 = 11.088  N (nitrogen) | 2 | 14.007 | 2 × 14.007 = 28.014  M = 10.81 g/mol + 72.066 g/mol + 75.993612652 g/mol + 11.088 g/mol + 28.014 g/mol = 197.97 g/mol
Find the molar mass, M, for 1-ethyl-3-methylimidazolium tetrafluoroborate: 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_6H_11BF_4N_2 Use the chemical formula, C_6H_11BF_4N_2, to count the number of atoms, N_i, for each element: | N_i B (boron) | 1 C (carbon) | 6 F (fluorine) | 4 H (hydrogen) | 11 N (nitrogen) | 2 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) B (boron) | 1 | 10.81 C (carbon) | 6 | 12.011 F (fluorine) | 4 | 18.998403163 H (hydrogen) | 11 | 1.008 N (nitrogen) | 2 | 14.007 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) B (boron) | 1 | 10.81 | 1 × 10.81 = 10.81 C (carbon) | 6 | 12.011 | 6 × 12.011 = 72.066 F (fluorine) | 4 | 18.998403163 | 4 × 18.998403163 = 75.993612652 H (hydrogen) | 11 | 1.008 | 11 × 1.008 = 11.088 N (nitrogen) | 2 | 14.007 | 2 × 14.007 = 28.014 M = 10.81 g/mol + 72.066 g/mol + 75.993612652 g/mol + 11.088 g/mol + 28.014 g/mol = 197.97 g/mol

Unit conversion

0.19797 kg/mol (kilograms per mole)
0.19797 kg/mol (kilograms per mole)

Comparisons

 ≈ 0.27 × molar mass of fullerene ( ≈ 721 g/mol )
≈ 0.27 × molar mass of fullerene ( ≈ 721 g/mol )
 ≈ molar mass of caffeine ( ≈ 194 g/mol )
≈ molar mass of caffeine ( ≈ 194 g/mol )
 ≈ 3.4 × molar mass of sodium chloride ( ≈ 58 g/mol )
≈ 3.4 × molar mass of sodium chloride ( ≈ 58 g/mol )

Corresponding quantities

Mass of a molecule m from m = M/N_A:  | 3.3×10^-22 grams  | 3.3×10^-25 kg (kilograms)  | 198 u (unified atomic mass units)  | 198 Da (daltons)
Mass of a molecule m from m = M/N_A: | 3.3×10^-22 grams | 3.3×10^-25 kg (kilograms) | 198 u (unified atomic mass units) | 198 Da (daltons)
Relative molecular mass M_r from M_r = M_u/M:  | 198
Relative molecular mass M_r from M_r = M_u/M: | 198