element mass fraction of tris[trinitratocerium(IV)]paraperiodate

tris[trinitratocerium(IV)]paraperiodate | elemental composition

Find the elemental composition for tris[trinitratocerium(IV)]paraperiodate 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: [Ce(NO_3)_3]_3(H_2IO_6) 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 Ce (cerium) | 3 H (hydrogen) | 2 I (iodine) | 1 N (nitrogen) | 9 O (oxygen) | 33 N_atoms = 3 + 2 + 1 + 9 + 33 = 48 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 Ce (cerium) | 3 | 3/48 H (hydrogen) | 2 | 2/48 I (iodine) | 1 | 1/48 N (nitrogen) | 9 | 9/48 O (oxygen) | 33 | 33/48 Check: 3/48 + 2/48 + 1/48 + 9/48 + 33/48 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent Ce (cerium) | 3 | 3/48 × 100% = 6.25% H (hydrogen) | 2 | 2/48 × 100% = 4.17% I (iodine) | 1 | 1/48 × 100% = 2.08% N (nitrogen) | 9 | 9/48 × 100% = 18.8% O (oxygen) | 33 | 33/48 × 100% = 68.8% 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 Ce (cerium) | 3 | 6.25% | 140.116 H (hydrogen) | 2 | 4.17% | 1.008 I (iodine) | 1 | 2.08% | 126.90447 N (nitrogen) | 9 | 18.8% | 14.007 O (oxygen) | 33 | 68.8% | 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 Ce (cerium) | 3 | 6.25% | 140.116 | 3 × 140.116 = 420.348 H (hydrogen) | 2 | 4.17% | 1.008 | 2 × 1.008 = 2.016 I (iodine) | 1 | 2.08% | 126.90447 | 1 × 126.90447 = 126.90447 N (nitrogen) | 9 | 18.8% | 14.007 | 9 × 14.007 = 126.063 O (oxygen) | 33 | 68.8% | 15.999 | 33 × 15.999 = 527.967 m = 420.348 u + 2.016 u + 126.90447 u + 126.063 u + 527.967 u = 1203.29847 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 Ce (cerium) | 3 | 6.25% | 420.348/1203.29847 H (hydrogen) | 2 | 4.17% | 2.016/1203.29847 I (iodine) | 1 | 2.08% | 126.90447/1203.29847 N (nitrogen) | 9 | 18.8% | 126.063/1203.29847 O (oxygen) | 33 | 68.8% | 527.967/1203.29847 Check: 420.348/1203.29847 + 2.016/1203.29847 + 126.90447/1203.29847 + 126.063/1203.29847 + 527.967/1203.29847 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent Ce (cerium) | 3 | 6.25% | 420.348/1203.29847 × 100% = 34.93% H (hydrogen) | 2 | 4.17% | 2.016/1203.29847 × 100% = 0.1675% I (iodine) | 1 | 2.08% | 126.90447/1203.29847 × 100% = 10.55% N (nitrogen) | 9 | 18.8% | 126.063/1203.29847 × 100% = 10.48% O (oxygen) | 33 | 68.8% | 527.967/1203.29847 × 100% = 43.88%

Mass fraction pie chart