mass fractions of fluorotryptophane

fluorotryptophane | elemental composition

Find the elemental composition for fluorotryptophane 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_11H_11FN_2O_2 Use the chemical formula, C_11H_11FN_2O_2, to count the number of atoms, N_i, for each element and find the total number of atoms, N_atoms: | number of atoms C (carbon) | 11 F (fluorine) | 1 H (hydrogen) | 11 N (nitrogen) | 2 O (oxygen) | 2 N_atoms = 11 + 1 + 11 + 2 + 2 = 27 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 C (carbon) | 11 | 11/27 F (fluorine) | 1 | 1/27 H (hydrogen) | 11 | 11/27 N (nitrogen) | 2 | 2/27 O (oxygen) | 2 | 2/27 Check: 11/27 + 1/27 + 11/27 + 2/27 + 2/27 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent C (carbon) | 11 | 11/27 × 100% = 40.7% F (fluorine) | 1 | 1/27 × 100% = 3.70% H (hydrogen) | 11 | 11/27 × 100% = 40.7% N (nitrogen) | 2 | 2/27 × 100% = 7.41% O (oxygen) | 2 | 2/27 × 100% = 7.41% 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 C (carbon) | 11 | 40.7% | 12.011 F (fluorine) | 1 | 3.70% | 18.998403163 H (hydrogen) | 11 | 40.7% | 1.008 N (nitrogen) | 2 | 7.41% | 14.007 O (oxygen) | 2 | 7.41% | 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 C (carbon) | 11 | 40.7% | 12.011 | 11 × 12.011 = 132.121 F (fluorine) | 1 | 3.70% | 18.998403163 | 1 × 18.998403163 = 18.998403163 H (hydrogen) | 11 | 40.7% | 1.008 | 11 × 1.008 = 11.088 N (nitrogen) | 2 | 7.41% | 14.007 | 2 × 14.007 = 28.014 O (oxygen) | 2 | 7.41% | 15.999 | 2 × 15.999 = 31.998 m = 132.121 u + 18.998403163 u + 11.088 u + 28.014 u + 31.998 u = 222.219403163 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 C (carbon) | 11 | 40.7% | 132.121/222.219403163 F (fluorine) | 1 | 3.70% | 18.998403163/222.219403163 H (hydrogen) | 11 | 40.7% | 11.088/222.219403163 N (nitrogen) | 2 | 7.41% | 28.014/222.219403163 O (oxygen) | 2 | 7.41% | 31.998/222.219403163 Check: 132.121/222.219403163 + 18.998403163/222.219403163 + 11.088/222.219403163 + 28.014/222.219403163 + 31.998/222.219403163 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent C (carbon) | 11 | 40.7% | 132.121/222.219403163 × 100% = 59.46% F (fluorine) | 1 | 3.70% | 18.998403163/222.219403163 × 100% = 8.549% H (hydrogen) | 11 | 40.7% | 11.088/222.219403163 × 100% = 4.990% N (nitrogen) | 2 | 7.41% | 28.014/222.219403163 × 100% = 12.61% O (oxygen) | 2 | 7.41% | 31.998/222.219403163 × 100% = 14.40%

Mass fraction pie chart