element counts of nonafluorobutane sulfonate

nonafluorobutane sulfonate | elemental composition

Find the elemental composition for nonafluorobutane sulfonate 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_4F_9O_3S)^- 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 F (fluorine) | 9 C (carbon) | 4 S (sulfur) | 1 O (oxygen) | 3 N_atoms = 9 + 4 + 1 + 3 = 17 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 F (fluorine) | 9 | 9/17 C (carbon) | 4 | 4/17 S (sulfur) | 1 | 1/17 O (oxygen) | 3 | 3/17 Check: 9/17 + 4/17 + 1/17 + 3/17 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent F (fluorine) | 9 | 9/17 × 100% = 52.9% C (carbon) | 4 | 4/17 × 100% = 23.5% S (sulfur) | 1 | 1/17 × 100% = 5.88% O (oxygen) | 3 | 3/17 × 100% = 17.6% 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 F (fluorine) | 9 | 52.9% | 18.998403163 C (carbon) | 4 | 23.5% | 12.011 S (sulfur) | 1 | 5.88% | 32.06 O (oxygen) | 3 | 17.6% | 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 F (fluorine) | 9 | 52.9% | 18.998403163 | 9 × 18.998403163 = 170.985628467 C (carbon) | 4 | 23.5% | 12.011 | 4 × 12.011 = 48.044 S (sulfur) | 1 | 5.88% | 32.06 | 1 × 32.06 = 32.06 O (oxygen) | 3 | 17.6% | 15.999 | 3 × 15.999 = 47.997 m = 170.985628467 u + 48.044 u + 32.06 u + 47.997 u = 299.086628467 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 F (fluorine) | 9 | 52.9% | 170.985628467/299.086628467 C (carbon) | 4 | 23.5% | 48.044/299.086628467 S (sulfur) | 1 | 5.88% | 32.06/299.086628467 O (oxygen) | 3 | 17.6% | 47.997/299.086628467 Check: 170.985628467/299.086628467 + 48.044/299.086628467 + 32.06/299.086628467 + 47.997/299.086628467 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent F (fluorine) | 9 | 52.9% | 170.985628467/299.086628467 × 100% = 57.17% C (carbon) | 4 | 23.5% | 48.044/299.086628467 × 100% = 16.06% S (sulfur) | 1 | 5.88% | 32.06/299.086628467 × 100% = 10.72% O (oxygen) | 3 | 17.6% | 47.997/299.086628467 × 100% = 16.05%