mass fractions of 3-bromo-2-(3'-fluorobenzyloxy)-5-methylphenylboronic acid

3-bromo-2-(3'-fluorobenzyloxy)-5-methylphenylboronic acid | elemental composition

Find the elemental composition for 3-bromo-2-(3'-fluorobenzyloxy)-5-methylphenylboronic acid in terms of the atom and mass percents: atom percent = N_i/N_atoms × 100% mass percent = (N_im_i)/m × 100% Plan: • Write the chemical formula and gather atomic masses from the periodic table. • Determine values for N_i, m_i, N_atoms and m using these items. • Finally, compute the percents and check the results. Write the chemical formula: C_14H_13BBrFO_3 Use the chemical formula, C_14H_13BBrFO_3, to count the number of atoms, N_i, for each element and find the total number of atoms, N_atoms: | number of atoms Br (bromine) | 1 C (carbon) | 14 B (boron) | 1 O (oxygen) | 3 F (fluorine) | 1 H (hydrogen) | 13 N_atoms = 1 + 14 + 1 + 3 + 1 + 13 = 33 Divide each N_i by N_atoms to calculate atom fractions. Then use the property that atom fractions must sum to one to check the work: | number of atoms | atom fraction Br (bromine) | 1 | 1/33 C (carbon) | 14 | 14/33 B (boron) | 1 | 1/33 O (oxygen) | 3 | 3/33 F (fluorine) | 1 | 1/33 H (hydrogen) | 13 | 13/33 Check: 1/33 + 14/33 + 1/33 + 3/33 + 1/33 + 13/33 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent Br (bromine) | 1 | 1/33 × 100% = 3.03% C (carbon) | 14 | 14/33 × 100% = 42.4% B (boron) | 1 | 1/33 × 100% = 3.03% O (oxygen) | 3 | 3/33 × 100% = 9.09% F (fluorine) | 1 | 1/33 × 100% = 3.03% H (hydrogen) | 13 | 13/33 × 100% = 39.4% Look up the atomic mass, m_i, in unified atomic mass units, u, for each element in the periodic table: | number of atoms | atom percent | atomic mass/u Br (bromine) | 1 | 3.03% | 79.904 C (carbon) | 14 | 42.4% | 12.011 B (boron) | 1 | 3.03% | 10.81 O (oxygen) | 3 | 9.09% | 15.999 F (fluorine) | 1 | 3.03% | 18.998403163 H (hydrogen) | 13 | 39.4% | 1.008 Multiply N_i by m_i to compute the mass for each element. Then sum those values to compute the molecular mass, m: | number of atoms | atom percent | atomic mass/u | mass/u Br (bromine) | 1 | 3.03% | 79.904 | 1 × 79.904 = 79.904 C (carbon) | 14 | 42.4% | 12.011 | 14 × 12.011 = 168.154 B (boron) | 1 | 3.03% | 10.81 | 1 × 10.81 = 10.81 O (oxygen) | 3 | 9.09% | 15.999 | 3 × 15.999 = 47.997 F (fluorine) | 1 | 3.03% | 18.998403163 | 1 × 18.998403163 = 18.998403163 H (hydrogen) | 13 | 39.4% | 1.008 | 13 × 1.008 = 13.104 m = 79.904 u + 168.154 u + 10.81 u + 47.997 u + 18.998403163 u + 13.104 u = 338.967403163 u Divide the mass for each element by m to calculate mass fractions. Then use the property that mass fractions must sum to one to check the work: | number of atoms | atom percent | mass fraction Br (bromine) | 1 | 3.03% | 79.904/338.967403163 C (carbon) | 14 | 42.4% | 168.154/338.967403163 B (boron) | 1 | 3.03% | 10.81/338.967403163 O (oxygen) | 3 | 9.09% | 47.997/338.967403163 F (fluorine) | 1 | 3.03% | 18.998403163/338.967403163 H (hydrogen) | 13 | 39.4% | 13.104/338.967403163 Check: 79.904/338.967403163 + 168.154/338.967403163 + 10.81/338.967403163 + 47.997/338.967403163 + 18.998403163/338.967403163 + 13.104/338.967403163 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent Br (bromine) | 1 | 3.03% | 79.904/338.967403163 × 100% = 23.57% C (carbon) | 14 | 42.4% | 168.154/338.967403163 × 100% = 49.61% B (boron) | 1 | 3.03% | 10.81/338.967403163 × 100% = 3.189% O (oxygen) | 3 | 9.09% | 47.997/338.967403163 × 100% = 14.16% F (fluorine) | 1 | 3.03% | 18.998403163/338.967403163 × 100% = 5.605% H (hydrogen) | 13 | 39.4% | 13.104/338.967403163 × 100% = 3.866%