Input interpretation
![3-bromo-2-(3'-fluorobenzyloxy)-5-methylphenylboronic acid | molar mass](../image_source/506479bbf65c934b093c9e3e17bfcf25.png)
3-bromo-2-(3'-fluorobenzyloxy)-5-methylphenylboronic acid | molar mass
Result
![Find the molar mass, M, for 3-bromo-2-(3'-fluorobenzyloxy)-5-methylphenylboronic 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_14H_13BBrFO_3 Use the chemical formula to count the number of atoms, N_i, for each element: | N_i Br (bromine) | 1 C (carbon) | 14 B (boron) | 1 O (oxygen) | 3 F (fluorine) | 1 H (hydrogen) | 13 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) Br (bromine) | 1 | 79.904 C (carbon) | 14 | 12.011 B (boron) | 1 | 10.81 O (oxygen) | 3 | 15.999 F (fluorine) | 1 | 18.998403163 H (hydrogen) | 13 | 1.008 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) Br (bromine) | 1 | 79.904 | 1 × 79.904 = 79.904 C (carbon) | 14 | 12.011 | 14 × 12.011 = 168.154 B (boron) | 1 | 10.81 | 1 × 10.81 = 10.81 O (oxygen) | 3 | 15.999 | 3 × 15.999 = 47.997 F (fluorine) | 1 | 18.998403163 | 1 × 18.998403163 = 18.998403163 H (hydrogen) | 13 | 1.008 | 13 × 1.008 = 13.104 M = 79.904 g/mol + 168.154 g/mol + 10.81 g/mol + 47.997 g/mol + 18.998403163 g/mol + 13.104 g/mol = 338.97 g/mol](../image_source/fcf9259db0803b93cf38437a6875a5e5.png)
Find the molar mass, M, for 3-bromo-2-(3'-fluorobenzyloxy)-5-methylphenylboronic 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_14H_13BBrFO_3 Use the chemical formula to count the number of atoms, N_i, for each element: | N_i Br (bromine) | 1 C (carbon) | 14 B (boron) | 1 O (oxygen) | 3 F (fluorine) | 1 H (hydrogen) | 13 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) Br (bromine) | 1 | 79.904 C (carbon) | 14 | 12.011 B (boron) | 1 | 10.81 O (oxygen) | 3 | 15.999 F (fluorine) | 1 | 18.998403163 H (hydrogen) | 13 | 1.008 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) Br (bromine) | 1 | 79.904 | 1 × 79.904 = 79.904 C (carbon) | 14 | 12.011 | 14 × 12.011 = 168.154 B (boron) | 1 | 10.81 | 1 × 10.81 = 10.81 O (oxygen) | 3 | 15.999 | 3 × 15.999 = 47.997 F (fluorine) | 1 | 18.998403163 | 1 × 18.998403163 = 18.998403163 H (hydrogen) | 13 | 1.008 | 13 × 1.008 = 13.104 M = 79.904 g/mol + 168.154 g/mol + 10.81 g/mol + 47.997 g/mol + 18.998403163 g/mol + 13.104 g/mol = 338.97 g/mol
Unit conversion
![0.339 kg/mol (kilograms per mole)](../image_source/07e1400e45430fa79af24fa53f4ea568.png)
0.339 kg/mol (kilograms per mole)
Comparisons
![≈ 0.47 × molar mass of fullerene ( ≈ 721 g/mol )](../image_source/96e30ab9a03a2036cd7a771645b63210.png)
≈ 0.47 × molar mass of fullerene ( ≈ 721 g/mol )
![≈ 1.7 × molar mass of caffeine ( ≈ 194 g/mol )](../image_source/ae883b3a92effc1e6438a7ed5f816736.png)
≈ 1.7 × molar mass of caffeine ( ≈ 194 g/mol )
![≈ 5.8 × molar mass of sodium chloride ( ≈ 58 g/mol )](../image_source/599617f3e9f23990ab978e8672ea92a6.png)
≈ 5.8 × molar mass of sodium chloride ( ≈ 58 g/mol )
Corresponding quantities
![Mass of a molecule m from m = M/N_A: | 5.6×10^-22 grams | 5.6×10^-25 kg (kilograms) | 339 u (unified atomic mass units) | 339 Da (daltons)](../image_source/df77139356be6e3fa36c9aaf50625d11.png)
Mass of a molecule m from m = M/N_A: | 5.6×10^-22 grams | 5.6×10^-25 kg (kilograms) | 339 u (unified atomic mass units) | 339 Da (daltons)
![Relative molecular mass M_r from M_r = M_u/M: | 339](../image_source/8562cbd286e0106335f1736e62322214.png)
Relative molecular mass M_r from M_r = M_u/M: | 339