Input interpretation
![2-fluoro-6-iodophenylboronic acid | molar mass](../image_source/0f6b93c59961b5f18b63ea760a9dd005.png)
2-fluoro-6-iodophenylboronic acid | molar mass
Result
![Find the molar mass, M, for 2-fluoro-6-iodophenylboronic 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_6H_5BFIO_2 Use the chemical formula to count the number of atoms, N_i, for each element: | N_i B (boron) | 1 C (carbon) | 6 F (fluorine) | 1 H (hydrogen) | 5 I (iodine) | 1 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) | 6 | 12.011 F (fluorine) | 1 | 18.998403163 H (hydrogen) | 5 | 1.008 I (iodine) | 1 | 126.90447 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) | 6 | 12.011 | 6 × 12.011 = 72.066 F (fluorine) | 1 | 18.998403163 | 1 × 18.998403163 = 18.998403163 H (hydrogen) | 5 | 1.008 | 5 × 1.008 = 5.040 I (iodine) | 1 | 126.90447 | 1 × 126.90447 = 126.90447 O (oxygen) | 2 | 15.999 | 2 × 15.999 = 31.998 M = 10.81 g/mol + 72.066 g/mol + 18.998403163 g/mol + 5.040 g/mol + 126.90447 g/mol + 31.998 g/mol = 265.82 g/mol](../image_source/c77c969706ac67118ac34bb4db5ad2db.png)
Find the molar mass, M, for 2-fluoro-6-iodophenylboronic 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_6H_5BFIO_2 Use the chemical formula to count the number of atoms, N_i, for each element: | N_i B (boron) | 1 C (carbon) | 6 F (fluorine) | 1 H (hydrogen) | 5 I (iodine) | 1 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) | 6 | 12.011 F (fluorine) | 1 | 18.998403163 H (hydrogen) | 5 | 1.008 I (iodine) | 1 | 126.90447 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) | 6 | 12.011 | 6 × 12.011 = 72.066 F (fluorine) | 1 | 18.998403163 | 1 × 18.998403163 = 18.998403163 H (hydrogen) | 5 | 1.008 | 5 × 1.008 = 5.040 I (iodine) | 1 | 126.90447 | 1 × 126.90447 = 126.90447 O (oxygen) | 2 | 15.999 | 2 × 15.999 = 31.998 M = 10.81 g/mol + 72.066 g/mol + 18.998403163 g/mol + 5.040 g/mol + 126.90447 g/mol + 31.998 g/mol = 265.82 g/mol
Unit conversion
![0.26582 kg/mol (kilograms per mole)](../image_source/25cf848537e7440b740d1594e1565eb0.png)
0.26582 kg/mol (kilograms per mole)
Comparisons
![≈ 0.37 × molar mass of fullerene ( ≈ 721 g/mol )](../image_source/07ad46994fbcf44bd1653d089b8ec9b7.png)
≈ 0.37 × molar mass of fullerene ( ≈ 721 g/mol )
![≈ 1.4 × molar mass of caffeine ( ≈ 194 g/mol )](../image_source/3dc37125cca1397d744083dfdcf62825.png)
≈ 1.4 × molar mass of caffeine ( ≈ 194 g/mol )
![≈ 4.5 × molar mass of sodium chloride ( ≈ 58 g/mol )](../image_source/248a17150da8bb536bd0bdd5a6959c5f.png)
≈ 4.5 × molar mass of sodium chloride ( ≈ 58 g/mol )
Corresponding quantities
![Mass of a molecule m from m = M/N_A: | 4.4×10^-22 grams | 4.4×10^-25 kg (kilograms) | 266 u (unified atomic mass units) | 266 Da (daltons)](../image_source/0d803c7c4af7830b84ae426064ce6ed8.png)
Mass of a molecule m from m = M/N_A: | 4.4×10^-22 grams | 4.4×10^-25 kg (kilograms) | 266 u (unified atomic mass units) | 266 Da (daltons)
![Relative molecular mass M_r from M_r = M_u/M: | 266](../image_source/2f1195233e97828357476ee3e5a1392d.png)
Relative molecular mass M_r from M_r = M_u/M: | 266