mass fractions of tetraconazole

tetraconazole | elemental composition

Find the elemental composition for tetraconazole 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_13H_11Cl_2F_4N_3O 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 C (carbon) | 13 Cl (chlorine) | 2 F (fluorine) | 4 H (hydrogen) | 11 N (nitrogen) | 3 O (oxygen) | 1 N_atoms = 13 + 2 + 4 + 11 + 3 + 1 = 34 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) | 13 | 13/34 Cl (chlorine) | 2 | 2/34 F (fluorine) | 4 | 4/34 H (hydrogen) | 11 | 11/34 N (nitrogen) | 3 | 3/34 O (oxygen) | 1 | 1/34 Check: 13/34 + 2/34 + 4/34 + 11/34 + 3/34 + 1/34 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent C (carbon) | 13 | 13/34 × 100% = 38.2% Cl (chlorine) | 2 | 2/34 × 100% = 5.88% F (fluorine) | 4 | 4/34 × 100% = 11.8% H (hydrogen) | 11 | 11/34 × 100% = 32.4% N (nitrogen) | 3 | 3/34 × 100% = 8.82% O (oxygen) | 1 | 1/34 × 100% = 2.94% 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) | 13 | 38.2% | 12.011 Cl (chlorine) | 2 | 5.88% | 35.45 F (fluorine) | 4 | 11.8% | 18.998403163 H (hydrogen) | 11 | 32.4% | 1.008 N (nitrogen) | 3 | 8.82% | 14.007 O (oxygen) | 1 | 2.94% | 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) | 13 | 38.2% | 12.011 | 13 × 12.011 = 156.143 Cl (chlorine) | 2 | 5.88% | 35.45 | 2 × 35.45 = 70.90 F (fluorine) | 4 | 11.8% | 18.998403163 | 4 × 18.998403163 = 75.993612652 H (hydrogen) | 11 | 32.4% | 1.008 | 11 × 1.008 = 11.088 N (nitrogen) | 3 | 8.82% | 14.007 | 3 × 14.007 = 42.021 O (oxygen) | 1 | 2.94% | 15.999 | 1 × 15.999 = 15.999 m = 156.143 u + 70.90 u + 75.993612652 u + 11.088 u + 42.021 u + 15.999 u = 372.144612652 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) | 13 | 38.2% | 156.143/372.144612652 Cl (chlorine) | 2 | 5.88% | 70.90/372.144612652 F (fluorine) | 4 | 11.8% | 75.993612652/372.144612652 H (hydrogen) | 11 | 32.4% | 11.088/372.144612652 N (nitrogen) | 3 | 8.82% | 42.021/372.144612652 O (oxygen) | 1 | 2.94% | 15.999/372.144612652 Check: 156.143/372.144612652 + 70.90/372.144612652 + 75.993612652/372.144612652 + 11.088/372.144612652 + 42.021/372.144612652 + 15.999/372.144612652 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent C (carbon) | 13 | 38.2% | 156.143/372.144612652 × 100% = 41.96% Cl (chlorine) | 2 | 5.88% | 70.90/372.144612652 × 100% = 19.05% F (fluorine) | 4 | 11.8% | 75.993612652/372.144612652 × 100% = 20.42% H (hydrogen) | 11 | 32.4% | 11.088/372.144612652 × 100% = 2.979% N (nitrogen) | 3 | 8.82% | 42.021/372.144612652 × 100% = 11.29% O (oxygen) | 1 | 2.94% | 15.999/372.144612652 × 100% = 4.299%

Mass fraction pie chart