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molar mass of ferroceneboronic acid

Input interpretation

ferroceneboronic acid | molar mass
ferroceneboronic acid | molar mass

Result

Find the molar mass, M, for ferroceneboronic acid: 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_10H_11BFeO_2 Use the chemical formula, C_10H_11BFeO_2, to count the number of atoms, N_i, for each element:  | N_i  B (boron) | 1  C (carbon) | 10  Fe (iron) | 1  H (hydrogen) | 11  O (oxygen) | 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) | 10 | 12.011  Fe (iron) | 1 | 55.845  H (hydrogen) | 11 | 1.008  O (oxygen) | 2 | 15.999 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) | 10 | 12.011 | 10 × 12.011 = 120.110  Fe (iron) | 1 | 55.845 | 1 × 55.845 = 55.845  H (hydrogen) | 11 | 1.008 | 11 × 1.008 = 11.088  O (oxygen) | 2 | 15.999 | 2 × 15.999 = 31.998  M = 10.81 g/mol + 120.110 g/mol + 55.845 g/mol + 11.088 g/mol + 31.998 g/mol = 229.85 g/mol
Find the molar mass, M, for ferroceneboronic acid: 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_10H_11BFeO_2 Use the chemical formula, C_10H_11BFeO_2, to count the number of atoms, N_i, for each element: | N_i B (boron) | 1 C (carbon) | 10 Fe (iron) | 1 H (hydrogen) | 11 O (oxygen) | 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) | 10 | 12.011 Fe (iron) | 1 | 55.845 H (hydrogen) | 11 | 1.008 O (oxygen) | 2 | 15.999 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) | 10 | 12.011 | 10 × 12.011 = 120.110 Fe (iron) | 1 | 55.845 | 1 × 55.845 = 55.845 H (hydrogen) | 11 | 1.008 | 11 × 1.008 = 11.088 O (oxygen) | 2 | 15.999 | 2 × 15.999 = 31.998 M = 10.81 g/mol + 120.110 g/mol + 55.845 g/mol + 11.088 g/mol + 31.998 g/mol = 229.85 g/mol

Unit conversion

0.22985 kg/mol (kilograms per mole)
0.22985 kg/mol (kilograms per mole)

Comparisons

 ≈ 0.32 × molar mass of fullerene ( ≈ 721 g/mol )
≈ 0.32 × molar mass of fullerene ( ≈ 721 g/mol )
 ≈ 1.2 × molar mass of caffeine ( ≈ 194 g/mol )
≈ 1.2 × molar mass of caffeine ( ≈ 194 g/mol )
 ≈ 3.9 × molar mass of sodium chloride ( ≈ 58 g/mol )
≈ 3.9 × molar mass of sodium chloride ( ≈ 58 g/mol )

Corresponding quantities

Mass of a molecule m from m = M/N_A:  | 3.8×10^-22 grams  | 3.8×10^-25 kg (kilograms)  | 230 u (unified atomic mass units)  | 230 Da (daltons)
Mass of a molecule m from m = M/N_A: | 3.8×10^-22 grams | 3.8×10^-25 kg (kilograms) | 230 u (unified atomic mass units) | 230 Da (daltons)
Relative molecular mass M_r from M_r = M_u/M:  | 230
Relative molecular mass M_r from M_r = M_u/M: | 230