Input interpretation
boron trifluoride phosphoric acid complex | molar mass
Result
Find the molar mass, M, for boron trifluoride phosphoric acid complex: 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: BF_3·H_3PO_4 Use the chemical formula to count the number of atoms, N_i, for each element: | N_i B (boron) | 1 F (fluorine) | 3 H (hydrogen) | 3 O (oxygen) | 4 P (phosphorus) | 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 F (fluorine) | 3 | 18.998403163 H (hydrogen) | 3 | 1.008 O (oxygen) | 4 | 15.999 P (phosphorus) | 1 | 30.973761998 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 F (fluorine) | 3 | 18.998403163 | 3 × 18.998403163 = 56.995209489 H (hydrogen) | 3 | 1.008 | 3 × 1.008 = 3.024 O (oxygen) | 4 | 15.999 | 4 × 15.999 = 63.996 P (phosphorus) | 1 | 30.973761998 | 1 × 30.973761998 = 30.973761998 M = 10.81 g/mol + 56.995209489 g/mol + 3.024 g/mol + 63.996 g/mol + 30.973761998 g/mol = 165.80 g/mol
Unit conversion
0.1658 kg/mol (kilograms per mole)
Comparisons
≈ ( 0.23 ≈ 1/4 ) × molar mass of fullerene ( ≈ 721 g/mol )
≈ 0.85 × molar mass of caffeine ( ≈ 194 g/mol )
≈ 2.8 × 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) | 166 u (unified atomic mass units) | 166 Da (daltons)
Relative molecular mass M_r from M_r = M_u/M: | 166