mass fractions of dichloromethotrexate

dichloromethotrexate | elemental composition

Find the elemental composition for dichloromethotrexate 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_20H_20Cl_2N_8O_5 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) | 20 Cl (chlorine) | 2 H (hydrogen) | 20 N (nitrogen) | 8 O (oxygen) | 5 N_atoms = 20 + 2 + 20 + 8 + 5 = 55 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) | 20 | 20/55 Cl (chlorine) | 2 | 2/55 H (hydrogen) | 20 | 20/55 N (nitrogen) | 8 | 8/55 O (oxygen) | 5 | 5/55 Check: 20/55 + 2/55 + 20/55 + 8/55 + 5/55 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent C (carbon) | 20 | 20/55 × 100% = 36.4% Cl (chlorine) | 2 | 2/55 × 100% = 3.64% H (hydrogen) | 20 | 20/55 × 100% = 36.4% N (nitrogen) | 8 | 8/55 × 100% = 14.5% O (oxygen) | 5 | 5/55 × 100% = 9.09% 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) | 20 | 36.4% | 12.011 Cl (chlorine) | 2 | 3.64% | 35.45 H (hydrogen) | 20 | 36.4% | 1.008 N (nitrogen) | 8 | 14.5% | 14.007 O (oxygen) | 5 | 9.09% | 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) | 20 | 36.4% | 12.011 | 20 × 12.011 = 240.220 Cl (chlorine) | 2 | 3.64% | 35.45 | 2 × 35.45 = 70.90 H (hydrogen) | 20 | 36.4% | 1.008 | 20 × 1.008 = 20.160 N (nitrogen) | 8 | 14.5% | 14.007 | 8 × 14.007 = 112.056 O (oxygen) | 5 | 9.09% | 15.999 | 5 × 15.999 = 79.995 m = 240.220 u + 70.90 u + 20.160 u + 112.056 u + 79.995 u = 523.331 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) | 20 | 36.4% | 240.220/523.331 Cl (chlorine) | 2 | 3.64% | 70.90/523.331 H (hydrogen) | 20 | 36.4% | 20.160/523.331 N (nitrogen) | 8 | 14.5% | 112.056/523.331 O (oxygen) | 5 | 9.09% | 79.995/523.331 Check: 240.220/523.331 + 70.90/523.331 + 20.160/523.331 + 112.056/523.331 + 79.995/523.331 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent C (carbon) | 20 | 36.4% | 240.220/523.331 × 100% = 45.90% Cl (chlorine) | 2 | 3.64% | 70.90/523.331 × 100% = 13.55% H (hydrogen) | 20 | 36.4% | 20.160/523.331 × 100% = 3.852% N (nitrogen) | 8 | 14.5% | 112.056/523.331 × 100% = 21.41% O (oxygen) | 5 | 9.09% | 79.995/523.331 × 100% = 15.29%

Mass fraction pie chart