Input interpretation
2-(trifluoromethyl)pyridine-3-carboxylic acid | elemental composition
Result
Find the elemental composition for 2-(trifluoromethyl)pyridine-3-carboxylic acid 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_4F_3NO_2 Use the chemical formula, C_7H_4F_3NO_2, to count the number of atoms, N_i, for each element and find the total number of atoms, N_atoms: | number of atoms F (fluorine) | 3 C (carbon) | 7 N (nitrogen) | 1 O (oxygen) | 2 H (hydrogen) | 4 N_atoms = 3 + 7 + 1 + 2 + 4 = 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) | 3 | 3/17 C (carbon) | 7 | 7/17 N (nitrogen) | 1 | 1/17 O (oxygen) | 2 | 2/17 H (hydrogen) | 4 | 4/17 Check: 3/17 + 7/17 + 1/17 + 2/17 + 4/17 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent F (fluorine) | 3 | 3/17 × 100% = 17.6% C (carbon) | 7 | 7/17 × 100% = 41.2% N (nitrogen) | 1 | 1/17 × 100% = 5.88% O (oxygen) | 2 | 2/17 × 100% = 11.8% H (hydrogen) | 4 | 4/17 × 100% = 23.5% 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) | 3 | 17.6% | 18.998403163 C (carbon) | 7 | 41.2% | 12.011 N (nitrogen) | 1 | 5.88% | 14.007 O (oxygen) | 2 | 11.8% | 15.999 H (hydrogen) | 4 | 23.5% | 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 F (fluorine) | 3 | 17.6% | 18.998403163 | 3 × 18.998403163 = 56.995209489 C (carbon) | 7 | 41.2% | 12.011 | 7 × 12.011 = 84.077 N (nitrogen) | 1 | 5.88% | 14.007 | 1 × 14.007 = 14.007 O (oxygen) | 2 | 11.8% | 15.999 | 2 × 15.999 = 31.998 H (hydrogen) | 4 | 23.5% | 1.008 | 4 × 1.008 = 4.032 m = 56.995209489 u + 84.077 u + 14.007 u + 31.998 u + 4.032 u = 191.109209489 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) | 3 | 17.6% | 56.995209489/191.109209489 C (carbon) | 7 | 41.2% | 84.077/191.109209489 N (nitrogen) | 1 | 5.88% | 14.007/191.109209489 O (oxygen) | 2 | 11.8% | 31.998/191.109209489 H (hydrogen) | 4 | 23.5% | 4.032/191.109209489 Check: 56.995209489/191.109209489 + 84.077/191.109209489 + 14.007/191.109209489 + 31.998/191.109209489 + 4.032/191.109209489 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent F (fluorine) | 3 | 17.6% | 56.995209489/191.109209489 × 100% = 29.82% C (carbon) | 7 | 41.2% | 84.077/191.109209489 × 100% = 43.99% N (nitrogen) | 1 | 5.88% | 14.007/191.109209489 × 100% = 7.329% O (oxygen) | 2 | 11.8% | 31.998/191.109209489 × 100% = 16.74% H (hydrogen) | 4 | 23.5% | 4.032/191.109209489 × 100% = 2.110%