mass fractions of bromperidol

bromperidol | elemental composition

Find the elemental composition for bromperidol 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_21H_23BrFNO_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 Br (bromine) | 1 C (carbon) | 21 F (fluorine) | 1 H (hydrogen) | 23 N (nitrogen) | 1 O (oxygen) | 2 N_atoms = 1 + 21 + 1 + 23 + 1 + 2 = 49 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 | 1/49 C (carbon) | 21 | 21/49 F (fluorine) | 1 | 1/49 H (hydrogen) | 23 | 23/49 N (nitrogen) | 1 | 1/49 O (oxygen) | 2 | 2/49 Check: 1/49 + 21/49 + 1/49 + 23/49 + 1/49 + 2/49 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent Br (bromine) | 1 | 1/49 × 100% = 2.04% C (carbon) | 21 | 21/49 × 100% = 42.9% F (fluorine) | 1 | 1/49 × 100% = 2.04% H (hydrogen) | 23 | 23/49 × 100% = 46.9% N (nitrogen) | 1 | 1/49 × 100% = 2.04% O (oxygen) | 2 | 2/49 × 100% = 4.08% 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 | 2.04% | 79.904 C (carbon) | 21 | 42.9% | 12.011 F (fluorine) | 1 | 2.04% | 18.998403163 H (hydrogen) | 23 | 46.9% | 1.008 N (nitrogen) | 1 | 2.04% | 14.007 O (oxygen) | 2 | 4.08% | 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 Br (bromine) | 1 | 2.04% | 79.904 | 1 × 79.904 = 79.904 C (carbon) | 21 | 42.9% | 12.011 | 21 × 12.011 = 252.231 F (fluorine) | 1 | 2.04% | 18.998403163 | 1 × 18.998403163 = 18.998403163 H (hydrogen) | 23 | 46.9% | 1.008 | 23 × 1.008 = 23.184 N (nitrogen) | 1 | 2.04% | 14.007 | 1 × 14.007 = 14.007 O (oxygen) | 2 | 4.08% | 15.999 | 2 × 15.999 = 31.998 m = 79.904 u + 252.231 u + 18.998403163 u + 23.184 u + 14.007 u + 31.998 u = 420.322403163 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 | 2.04% | 79.904/420.322403163 C (carbon) | 21 | 42.9% | 252.231/420.322403163 F (fluorine) | 1 | 2.04% | 18.998403163/420.322403163 H (hydrogen) | 23 | 46.9% | 23.184/420.322403163 N (nitrogen) | 1 | 2.04% | 14.007/420.322403163 O (oxygen) | 2 | 4.08% | 31.998/420.322403163 Check: 79.904/420.322403163 + 252.231/420.322403163 + 18.998403163/420.322403163 + 23.184/420.322403163 + 14.007/420.322403163 + 31.998/420.322403163 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent Br (bromine) | 1 | 2.04% | 79.904/420.322403163 × 100% = 19.01% C (carbon) | 21 | 42.9% | 252.231/420.322403163 × 100% = 60.01% F (fluorine) | 1 | 2.04% | 18.998403163/420.322403163 × 100% = 4.520% H (hydrogen) | 23 | 46.9% | 23.184/420.322403163 × 100% = 5.516% N (nitrogen) | 1 | 2.04% | 14.007/420.322403163 × 100% = 3.332% O (oxygen) | 2 | 4.08% | 31.998/420.322403163 × 100% = 7.613%

Mass fraction pie chart