mass fractions of tetraoctylphosphonium bromide

tetraoctylphosphonium bromide | elemental composition

Find the elemental composition for tetraoctylphosphonium bromide 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(CH_2)_7]_4P(Br) Use the chemical formula, [CH_3(CH_2)_7]_4P(Br), 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) | 32 H (hydrogen) | 68 P (phosphorus) | 1 N_atoms = 1 + 32 + 68 + 1 = 102 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 | 1/102 C (carbon) | 32 | 32/102 H (hydrogen) | 68 | 68/102 P (phosphorus) | 1 | 1/102 Check: 1/102 + 32/102 + 68/102 + 1/102 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent Br (bromine) | 1 | 1/102 × 100% = 0.980% C (carbon) | 32 | 32/102 × 100% = 31.4% H (hydrogen) | 68 | 68/102 × 100% = 66.7% P (phosphorus) | 1 | 1/102 × 100% = 0.980% 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 | 0.980% | 79.904 C (carbon) | 32 | 31.4% | 12.011 H (hydrogen) | 68 | 66.7% | 1.008 P (phosphorus) | 1 | 0.980% | 30.973761998 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 | 0.980% | 79.904 | 1 × 79.904 = 79.904 C (carbon) | 32 | 31.4% | 12.011 | 32 × 12.011 = 384.352 H (hydrogen) | 68 | 66.7% | 1.008 | 68 × 1.008 = 68.544 P (phosphorus) | 1 | 0.980% | 30.973761998 | 1 × 30.973761998 = 30.973761998 m = 79.904 u + 384.352 u + 68.544 u + 30.973761998 u = 563.773761998 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 | 0.980% | 79.904/563.773761998 C (carbon) | 32 | 31.4% | 384.352/563.773761998 H (hydrogen) | 68 | 66.7% | 68.544/563.773761998 P (phosphorus) | 1 | 0.980% | 30.973761998/563.773761998 Check: 79.904/563.773761998 + 384.352/563.773761998 + 68.544/563.773761998 + 30.973761998/563.773761998 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent Br (bromine) | 1 | 0.980% | 79.904/563.773761998 × 100% = 14.17% C (carbon) | 32 | 31.4% | 384.352/563.773761998 × 100% = 68.17% H (hydrogen) | 68 | 66.7% | 68.544/563.773761998 × 100% = 12.16% P (phosphorus) | 1 | 0.980% | 30.973761998/563.773761998 × 100% = 5.494%

Mass fraction pie chart