Input interpretation
10-(4-methylpiperazino)-10, 11-dihydrodibenzo(b, f)selenepin maleate | elemental composition
Result
Find the elemental composition for 10-(4-methylpiperazino)-10, 11-dihydrodibenzo(b, f)selenepin maleate 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_23H_26N_2O_4Se 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) | 23 O (oxygen) | 4 N (nitrogen) | 2 Se (selenium) | 1 H (hydrogen) | 26 N_atoms = 23 + 4 + 2 + 1 + 26 = 56 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) | 23 | 23/56 O (oxygen) | 4 | 4/56 N (nitrogen) | 2 | 2/56 Se (selenium) | 1 | 1/56 H (hydrogen) | 26 | 26/56 Check: 23/56 + 4/56 + 2/56 + 1/56 + 26/56 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent C (carbon) | 23 | 23/56 × 100% = 41.1% O (oxygen) | 4 | 4/56 × 100% = 7.14% N (nitrogen) | 2 | 2/56 × 100% = 3.57% Se (selenium) | 1 | 1/56 × 100% = 1.79% H (hydrogen) | 26 | 26/56 × 100% = 46.4% 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) | 23 | 41.1% | 12.011 O (oxygen) | 4 | 7.14% | 15.999 N (nitrogen) | 2 | 3.57% | 14.007 Se (selenium) | 1 | 1.79% | 78.971 H (hydrogen) | 26 | 46.4% | 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 C (carbon) | 23 | 41.1% | 12.011 | 23 × 12.011 = 276.253 O (oxygen) | 4 | 7.14% | 15.999 | 4 × 15.999 = 63.996 N (nitrogen) | 2 | 3.57% | 14.007 | 2 × 14.007 = 28.014 Se (selenium) | 1 | 1.79% | 78.971 | 1 × 78.971 = 78.971 H (hydrogen) | 26 | 46.4% | 1.008 | 26 × 1.008 = 26.208 m = 276.253 u + 63.996 u + 28.014 u + 78.971 u + 26.208 u = 473.442 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) | 23 | 41.1% | 276.253/473.442 O (oxygen) | 4 | 7.14% | 63.996/473.442 N (nitrogen) | 2 | 3.57% | 28.014/473.442 Se (selenium) | 1 | 1.79% | 78.971/473.442 H (hydrogen) | 26 | 46.4% | 26.208/473.442 Check: 276.253/473.442 + 63.996/473.442 + 28.014/473.442 + 78.971/473.442 + 26.208/473.442 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent C (carbon) | 23 | 41.1% | 276.253/473.442 × 100% = 58.35% O (oxygen) | 4 | 7.14% | 63.996/473.442 × 100% = 13.52% N (nitrogen) | 2 | 3.57% | 28.014/473.442 × 100% = 5.917% Se (selenium) | 1 | 1.79% | 78.971/473.442 × 100% = 16.68% H (hydrogen) | 26 | 46.4% | 26.208/473.442 × 100% = 5.536%