mass fractions of 5-(2-hydroxyethyl)-3,4-dimethylthiazolium iodide

5-(2-hydroxyethyl)-3, 4-dimethylthiazolium iodide | elemental composition

Find the elemental composition for 5-(2-hydroxyethyl)-3, 4-dimethylthiazolium iodide 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_7H_12INOS 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 I (iodine) | 1 O (oxygen) | 1 C (carbon) | 7 N (nitrogen) | 1 S (sulfur) | 1 H (hydrogen) | 12 N_atoms = 1 + 1 + 7 + 1 + 1 + 12 = 23 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 I (iodine) | 1 | 1/23 O (oxygen) | 1 | 1/23 C (carbon) | 7 | 7/23 N (nitrogen) | 1 | 1/23 S (sulfur) | 1 | 1/23 H (hydrogen) | 12 | 12/23 Check: 1/23 + 1/23 + 7/23 + 1/23 + 1/23 + 12/23 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent I (iodine) | 1 | 1/23 × 100% = 4.35% O (oxygen) | 1 | 1/23 × 100% = 4.35% C (carbon) | 7 | 7/23 × 100% = 30.4% N (nitrogen) | 1 | 1/23 × 100% = 4.35% S (sulfur) | 1 | 1/23 × 100% = 4.35% H (hydrogen) | 12 | 12/23 × 100% = 52.2% 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 I (iodine) | 1 | 4.35% | 126.90447 O (oxygen) | 1 | 4.35% | 15.999 C (carbon) | 7 | 30.4% | 12.011 N (nitrogen) | 1 | 4.35% | 14.007 S (sulfur) | 1 | 4.35% | 32.06 H (hydrogen) | 12 | 52.2% | 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 I (iodine) | 1 | 4.35% | 126.90447 | 1 × 126.90447 = 126.90447 O (oxygen) | 1 | 4.35% | 15.999 | 1 × 15.999 = 15.999 C (carbon) | 7 | 30.4% | 12.011 | 7 × 12.011 = 84.077 N (nitrogen) | 1 | 4.35% | 14.007 | 1 × 14.007 = 14.007 S (sulfur) | 1 | 4.35% | 32.06 | 1 × 32.06 = 32.06 H (hydrogen) | 12 | 52.2% | 1.008 | 12 × 1.008 = 12.096 m = 126.90447 u + 15.999 u + 84.077 u + 14.007 u + 32.06 u + 12.096 u = 285.14347 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 I (iodine) | 1 | 4.35% | 126.90447/285.14347 O (oxygen) | 1 | 4.35% | 15.999/285.14347 C (carbon) | 7 | 30.4% | 84.077/285.14347 N (nitrogen) | 1 | 4.35% | 14.007/285.14347 S (sulfur) | 1 | 4.35% | 32.06/285.14347 H (hydrogen) | 12 | 52.2% | 12.096/285.14347 Check: 126.90447/285.14347 + 15.999/285.14347 + 84.077/285.14347 + 14.007/285.14347 + 32.06/285.14347 + 12.096/285.14347 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent I (iodine) | 1 | 4.35% | 126.90447/285.14347 × 100% = 44.51% O (oxygen) | 1 | 4.35% | 15.999/285.14347 × 100% = 5.611% C (carbon) | 7 | 30.4% | 84.077/285.14347 × 100% = 29.49% N (nitrogen) | 1 | 4.35% | 14.007/285.14347 × 100% = 4.912% S (sulfur) | 1 | 4.35% | 32.06/285.14347 × 100% = 11.24% H (hydrogen) | 12 | 52.2% | 12.096/285.14347 × 100% = 4.242%