mass fractions of bromo difluoroaniline

bromo difluoroaniline | elemental composition

Find the elemental composition for bromo difluoroaniline 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_6H_4BrF_2N 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) | 6 N (nitrogen) | 1 F (fluorine) | 2 H (hydrogen) | 4 N_atoms = 1 + 6 + 1 + 2 + 4 = 14 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 | 1/14 C (carbon) | 6 | 6/14 N (nitrogen) | 1 | 1/14 F (fluorine) | 2 | 2/14 H (hydrogen) | 4 | 4/14 Check: 1/14 + 6/14 + 1/14 + 2/14 + 4/14 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent Br (bromine) | 1 | 1/14 × 100% = 7.14% C (carbon) | 6 | 6/14 × 100% = 42.9% N (nitrogen) | 1 | 1/14 × 100% = 7.14% F (fluorine) | 2 | 2/14 × 100% = 14.3% H (hydrogen) | 4 | 4/14 × 100% = 28.6% 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 | 7.14% | 79.904 C (carbon) | 6 | 42.9% | 12.011 N (nitrogen) | 1 | 7.14% | 14.007 F (fluorine) | 2 | 14.3% | 18.998403163 H (hydrogen) | 4 | 28.6% | 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 | 7.14% | 79.904 | 1 × 79.904 = 79.904 C (carbon) | 6 | 42.9% | 12.011 | 6 × 12.011 = 72.066 N (nitrogen) | 1 | 7.14% | 14.007 | 1 × 14.007 = 14.007 F (fluorine) | 2 | 14.3% | 18.998403163 | 2 × 18.998403163 = 37.996806326 H (hydrogen) | 4 | 28.6% | 1.008 | 4 × 1.008 = 4.032 m = 79.904 u + 72.066 u + 14.007 u + 37.996806326 u + 4.032 u = 208.005806326 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 | 7.14% | 79.904/208.005806326 C (carbon) | 6 | 42.9% | 72.066/208.005806326 N (nitrogen) | 1 | 7.14% | 14.007/208.005806326 F (fluorine) | 2 | 14.3% | 37.996806326/208.005806326 H (hydrogen) | 4 | 28.6% | 4.032/208.005806326 Check: 79.904/208.005806326 + 72.066/208.005806326 + 14.007/208.005806326 + 37.996806326/208.005806326 + 4.032/208.005806326 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent Br (bromine) | 1 | 7.14% | 79.904/208.005806326 × 100% = 38.41% C (carbon) | 6 | 42.9% | 72.066/208.005806326 × 100% = 34.65% N (nitrogen) | 1 | 7.14% | 14.007/208.005806326 × 100% = 6.734% F (fluorine) | 2 | 14.3% | 37.996806326/208.005806326 × 100% = 18.27% H (hydrogen) | 4 | 28.6% | 4.032/208.005806326 × 100% = 1.938%