Input interpretation
potassium 3-formylphenyltrifluoroborate | molar mass
Result
Find the molar mass, M, for potassium 3-formylphenyltrifluoroborate: 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_7H_5BF_3KO Use the chemical formula, C_7H_5BF_3KO, to count the number of atoms, N_i, for each element: | N_i B (boron) | 1 C (carbon) | 7 F (fluorine) | 3 H (hydrogen) | 5 K (potassium) | 1 O (oxygen) | 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) B (boron) | 1 | 10.81 C (carbon) | 7 | 12.011 F (fluorine) | 3 | 18.998403163 H (hydrogen) | 5 | 1.008 K (potassium) | 1 | 39.0983 O (oxygen) | 1 | 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) | 5 | 1.008 | 5 × 1.008 = 5.040 K (potassium) | 1 | 39.0983 | 1 × 39.0983 = 39.0983 O (oxygen) | 1 | 15.999 | 1 × 15.999 = 15.999 M = 10.81 g/mol + 84.077 g/mol + 56.995209489 g/mol + 5.040 g/mol + 39.0983 g/mol + 15.999 g/mol = 212.02 g/mol
Unit conversion
0.21202 kg/mol (kilograms per mole)
Comparisons
≈ 0.29 × molar mass of fullerene ( ≈ 721 g/mol )
≈ 1.1 × molar mass of caffeine ( ≈ 194 g/mol )
≈ 3.6 × molar mass of sodium chloride ( ≈ 58 g/mol )
Corresponding quantities
Mass of a molecule m from m = M/N_A: | 3.5×10^-22 grams | 3.5×10^-25 kg (kilograms) | 212 u (unified atomic mass units) | 212 Da (daltons)
Relative molecular mass M_r from M_r = M_u/M: | 212