mass fractions of neocuproine hydrochloride hydrate

neocuproine hydrochloride hydrate | elemental composition

Find the elemental composition for neocuproine hydrochloride hydrate 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_14H_12N_2·HCl·xH_2O 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) | 14 Cl (chlorine) | 1 H (hydrogen) | 15 N (nitrogen) | 2 O (oxygen) | 1 N_atoms = 14 + 1 + 15 + 2 + 1 = 33 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) | 14 | 14/33 Cl (chlorine) | 1 | 1/33 H (hydrogen) | 15 | 15/33 N (nitrogen) | 2 | 2/33 O (oxygen) | 1 | 1/33 Check: 14/33 + 1/33 + 15/33 + 2/33 + 1/33 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent C (carbon) | 14 | 14/33 × 100% = 42.4% Cl (chlorine) | 1 | 1/33 × 100% = 3.03% H (hydrogen) | 15 | 15/33 × 100% = 45.5% N (nitrogen) | 2 | 2/33 × 100% = 6.06% O (oxygen) | 1 | 1/33 × 100% = 3.03% 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) | 14 | 42.4% | 12.011 Cl (chlorine) | 1 | 3.03% | 35.45 H (hydrogen) | 15 | 45.5% | 1.008 N (nitrogen) | 2 | 6.06% | 14.007 O (oxygen) | 1 | 3.03% | 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) | 14 | 42.4% | 12.011 | 14 × 12.011 = 168.154 Cl (chlorine) | 1 | 3.03% | 35.45 | 1 × 35.45 = 35.45 H (hydrogen) | 15 | 45.5% | 1.008 | 15 × 1.008 = 15.120 N (nitrogen) | 2 | 6.06% | 14.007 | 2 × 14.007 = 28.014 O (oxygen) | 1 | 3.03% | 15.999 | 1 × 15.999 = 15.999 m = 168.154 u + 35.45 u + 15.120 u + 28.014 u + 15.999 u = 262.737 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) | 14 | 42.4% | 168.154/262.737 Cl (chlorine) | 1 | 3.03% | 35.45/262.737 H (hydrogen) | 15 | 45.5% | 15.120/262.737 N (nitrogen) | 2 | 6.06% | 28.014/262.737 O (oxygen) | 1 | 3.03% | 15.999/262.737 Check: 168.154/262.737 + 35.45/262.737 + 15.120/262.737 + 28.014/262.737 + 15.999/262.737 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent C (carbon) | 14 | 42.4% | 168.154/262.737 × 100% = 64.00% Cl (chlorine) | 1 | 3.03% | 35.45/262.737 × 100% = 13.49% H (hydrogen) | 15 | 45.5% | 15.120/262.737 × 100% = 5.755% N (nitrogen) | 2 | 6.06% | 28.014/262.737 × 100% = 10.66% O (oxygen) | 1 | 3.03% | 15.999/262.737 × 100% = 6.089%

Mass fraction pie chart