mass fractions of fluvalinate

fluvalinate | elemental composition

Find the elemental composition for fluvalinate 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_26H_22ClF_3N_2O_3 Use the chemical formula, C_26H_22ClF_3N_2O_3, 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) | 26 Cl (chlorine) | 1 F (fluorine) | 3 H (hydrogen) | 22 N (nitrogen) | 2 O (oxygen) | 3 N_atoms = 26 + 1 + 3 + 22 + 2 + 3 = 57 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) | 26 | 26/57 Cl (chlorine) | 1 | 1/57 F (fluorine) | 3 | 3/57 H (hydrogen) | 22 | 22/57 N (nitrogen) | 2 | 2/57 O (oxygen) | 3 | 3/57 Check: 26/57 + 1/57 + 3/57 + 22/57 + 2/57 + 3/57 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent C (carbon) | 26 | 26/57 × 100% = 45.6% Cl (chlorine) | 1 | 1/57 × 100% = 1.75% F (fluorine) | 3 | 3/57 × 100% = 5.26% H (hydrogen) | 22 | 22/57 × 100% = 38.6% N (nitrogen) | 2 | 2/57 × 100% = 3.51% O (oxygen) | 3 | 3/57 × 100% = 5.26% 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) | 26 | 45.6% | 12.011 Cl (chlorine) | 1 | 1.75% | 35.45 F (fluorine) | 3 | 5.26% | 18.998403163 H (hydrogen) | 22 | 38.6% | 1.008 N (nitrogen) | 2 | 3.51% | 14.007 O (oxygen) | 3 | 5.26% | 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) | 26 | 45.6% | 12.011 | 26 × 12.011 = 312.286 Cl (chlorine) | 1 | 1.75% | 35.45 | 1 × 35.45 = 35.45 F (fluorine) | 3 | 5.26% | 18.998403163 | 3 × 18.998403163 = 56.995209489 H (hydrogen) | 22 | 38.6% | 1.008 | 22 × 1.008 = 22.176 N (nitrogen) | 2 | 3.51% | 14.007 | 2 × 14.007 = 28.014 O (oxygen) | 3 | 5.26% | 15.999 | 3 × 15.999 = 47.997 m = 312.286 u + 35.45 u + 56.995209489 u + 22.176 u + 28.014 u + 47.997 u = 502.918209489 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) | 26 | 45.6% | 312.286/502.918209489 Cl (chlorine) | 1 | 1.75% | 35.45/502.918209489 F (fluorine) | 3 | 5.26% | 56.995209489/502.918209489 H (hydrogen) | 22 | 38.6% | 22.176/502.918209489 N (nitrogen) | 2 | 3.51% | 28.014/502.918209489 O (oxygen) | 3 | 5.26% | 47.997/502.918209489 Check: 312.286/502.918209489 + 35.45/502.918209489 + 56.995209489/502.918209489 + 22.176/502.918209489 + 28.014/502.918209489 + 47.997/502.918209489 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent C (carbon) | 26 | 45.6% | 312.286/502.918209489 × 100% = 62.09% Cl (chlorine) | 1 | 1.75% | 35.45/502.918209489 × 100% = 7.049% F (fluorine) | 3 | 5.26% | 56.995209489/502.918209489 × 100% = 11.33% H (hydrogen) | 22 | 38.6% | 22.176/502.918209489 × 100% = 4.409% N (nitrogen) | 2 | 3.51% | 28.014/502.918209489 × 100% = 5.570% O (oxygen) | 3 | 5.26% | 47.997/502.918209489 × 100% = 9.544%

Mass fraction pie chart