mass fractions of 4-acetamidophenylboronic acid

4-acetamidophenylboronic acid | elemental composition

Find the elemental composition for 4-acetamidophenylboronic 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: CH_3CONHC_6H_4B(OH)_2 Use the chemical formula to count the number of atoms, N_i, for each element and find the total number of atoms, N_atoms, per molecule: | number of atoms B (boron) | 1 C (carbon) | 8 H (hydrogen) | 10 N (nitrogen) | 1 O (oxygen) | 3 N_atoms = 1 + 8 + 10 + 1 + 3 = 23 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 B (boron) | 1 | 1/23 C (carbon) | 8 | 8/23 H (hydrogen) | 10 | 10/23 N (nitrogen) | 1 | 1/23 O (oxygen) | 3 | 3/23 Check: 1/23 + 8/23 + 10/23 + 1/23 + 3/23 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent B (boron) | 1 | 1/23 × 100% = 4.35% C (carbon) | 8 | 8/23 × 100% = 34.8% H (hydrogen) | 10 | 10/23 × 100% = 43.5% N (nitrogen) | 1 | 1/23 × 100% = 4.35% O (oxygen) | 3 | 3/23 × 100% = 13.0% 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 B (boron) | 1 | 4.35% | 10.81 C (carbon) | 8 | 34.8% | 12.011 H (hydrogen) | 10 | 43.5% | 1.008 N (nitrogen) | 1 | 4.35% | 14.007 O (oxygen) | 3 | 13.0% | 15.999 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 B (boron) | 1 | 4.35% | 10.81 | 1 × 10.81 = 10.81 C (carbon) | 8 | 34.8% | 12.011 | 8 × 12.011 = 96.088 H (hydrogen) | 10 | 43.5% | 1.008 | 10 × 1.008 = 10.080 N (nitrogen) | 1 | 4.35% | 14.007 | 1 × 14.007 = 14.007 O (oxygen) | 3 | 13.0% | 15.999 | 3 × 15.999 = 47.997 m = 10.81 u + 96.088 u + 10.080 u + 14.007 u + 47.997 u = 178.982 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 B (boron) | 1 | 4.35% | 10.81/178.982 C (carbon) | 8 | 34.8% | 96.088/178.982 H (hydrogen) | 10 | 43.5% | 10.080/178.982 N (nitrogen) | 1 | 4.35% | 14.007/178.982 O (oxygen) | 3 | 13.0% | 47.997/178.982 Check: 10.81/178.982 + 96.088/178.982 + 10.080/178.982 + 14.007/178.982 + 47.997/178.982 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent B (boron) | 1 | 4.35% | 10.81/178.982 × 100% = 6.040% C (carbon) | 8 | 34.8% | 96.088/178.982 × 100% = 53.69% H (hydrogen) | 10 | 43.5% | 10.080/178.982 × 100% = 5.632% N (nitrogen) | 1 | 4.35% | 14.007/178.982 × 100% = 7.826% O (oxygen) | 3 | 13.0% | 47.997/178.982 × 100% = 26.82%

Mass fraction pie chart