mass fractions of 4-ethoxy-3,5-dimethylphenylboronic acid

4-ethoxy-3, 5-dimethylphenylboronic acid | elemental composition

Find the elemental composition for 4-ethoxy-3, 5-dimethylphenylboronic 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_3)_2C_6H_2OCH_2CH_3B(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) | 10 H (hydrogen) | 15 O (oxygen) | 3 N_atoms = 1 + 10 + 15 + 3 = 29 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 | 1/29 C (carbon) | 10 | 10/29 H (hydrogen) | 15 | 15/29 O (oxygen) | 3 | 3/29 Check: 1/29 + 10/29 + 15/29 + 3/29 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent B (boron) | 1 | 1/29 × 100% = 3.45% C (carbon) | 10 | 10/29 × 100% = 34.5% H (hydrogen) | 15 | 15/29 × 100% = 51.7% O (oxygen) | 3 | 3/29 × 100% = 10.3% 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 | 3.45% | 10.81 C (carbon) | 10 | 34.5% | 12.011 H (hydrogen) | 15 | 51.7% | 1.008 O (oxygen) | 3 | 10.3% | 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 | 3.45% | 10.81 | 1 × 10.81 = 10.81 C (carbon) | 10 | 34.5% | 12.011 | 10 × 12.011 = 120.110 H (hydrogen) | 15 | 51.7% | 1.008 | 15 × 1.008 = 15.120 O (oxygen) | 3 | 10.3% | 15.999 | 3 × 15.999 = 47.997 m = 10.81 u + 120.110 u + 15.120 u + 47.997 u = 194.037 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 | 3.45% | 10.81/194.037 C (carbon) | 10 | 34.5% | 120.110/194.037 H (hydrogen) | 15 | 51.7% | 15.120/194.037 O (oxygen) | 3 | 10.3% | 47.997/194.037 Check: 10.81/194.037 + 120.110/194.037 + 15.120/194.037 + 47.997/194.037 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent B (boron) | 1 | 3.45% | 10.81/194.037 × 100% = 5.571% C (carbon) | 10 | 34.5% | 120.110/194.037 × 100% = 61.90% H (hydrogen) | 15 | 51.7% | 15.120/194.037 × 100% = 7.792% O (oxygen) | 3 | 10.3% | 47.997/194.037 × 100% = 24.74%

Mass fraction pie chart