Input interpretation
4-(trifluoromethoxy)phenylboronic acid | molar mass
Result
Find the molar mass, M, for 4-(trifluoromethoxy)phenylboronic 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: CF_3OC_6H_4B(OH)_2 Use the chemical formula, CF_3OC_6H_4B(OH)_2, to count the number of atoms, N_i, for each element: | N_i B (boron) | 1 C (carbon) | 7 F (fluorine) | 3 H (hydrogen) | 6 O (oxygen) | 3 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) | 7 | 12.011 F (fluorine) | 3 | 18.998403163 H (hydrogen) | 6 | 1.008 O (oxygen) | 3 | 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) | 7 | 12.011 | 7 × 12.011 = 84.077 F (fluorine) | 3 | 18.998403163 | 3 × 18.998403163 = 56.995209489 H (hydrogen) | 6 | 1.008 | 6 × 1.008 = 6.048 O (oxygen) | 3 | 15.999 | 3 × 15.999 = 47.997 M = 10.81 g/mol + 84.077 g/mol + 56.995209489 g/mol + 6.048 g/mol + 47.997 g/mol = 205.93 g/mol
Unit conversion
0.20593 kg/mol (kilograms per mole)
Comparisons
≈ 0.29 × molar mass of fullerene ( ≈ 721 g/mol )
≈ 1.1 × molar mass of caffeine ( ≈ 194 g/mol )
≈ 3.5 × molar mass of sodium chloride ( ≈ 58 g/mol )
Corresponding quantities
Mass of a molecule m from m = M/N_A: | 3.4×10^-22 grams | 3.4×10^-25 kg (kilograms) | 206 u (unified atomic mass units) | 206 Da (daltons)
Relative molecular mass M_r from M_r = M_u/M: | 206