mass fractions of barium iodate monohydrate

barium iodate monohydrate | elemental composition

Find the elemental composition for barium iodate monohydrate 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: Ba(IO_3)_2·H_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 Ba (barium) | 1 H (hydrogen) | 2 I (iodine) | 2 O (oxygen) | 7 N_atoms = 1 + 2 + 2 + 7 = 12 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 Ba (barium) | 1 | 1/12 H (hydrogen) | 2 | 2/12 I (iodine) | 2 | 2/12 O (oxygen) | 7 | 7/12 Check: 1/12 + 2/12 + 2/12 + 7/12 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent Ba (barium) | 1 | 1/12 × 100% = 8.33% H (hydrogen) | 2 | 2/12 × 100% = 16.7% I (iodine) | 2 | 2/12 × 100% = 16.7% O (oxygen) | 7 | 7/12 × 100% = 58.3% 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 Ba (barium) | 1 | 8.33% | 137.327 H (hydrogen) | 2 | 16.7% | 1.008 I (iodine) | 2 | 16.7% | 126.90447 O (oxygen) | 7 | 58.3% | 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 Ba (barium) | 1 | 8.33% | 137.327 | 1 × 137.327 = 137.327 H (hydrogen) | 2 | 16.7% | 1.008 | 2 × 1.008 = 2.016 I (iodine) | 2 | 16.7% | 126.90447 | 2 × 126.90447 = 253.80894 O (oxygen) | 7 | 58.3% | 15.999 | 7 × 15.999 = 111.993 m = 137.327 u + 2.016 u + 253.80894 u + 111.993 u = 505.14494 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 Ba (barium) | 1 | 8.33% | 137.327/505.14494 H (hydrogen) | 2 | 16.7% | 2.016/505.14494 I (iodine) | 2 | 16.7% | 253.80894/505.14494 O (oxygen) | 7 | 58.3% | 111.993/505.14494 Check: 137.327/505.14494 + 2.016/505.14494 + 253.80894/505.14494 + 111.993/505.14494 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent Ba (barium) | 1 | 8.33% | 137.327/505.14494 × 100% = 27.19% H (hydrogen) | 2 | 16.7% | 2.016/505.14494 × 100% = 0.3991% I (iodine) | 2 | 16.7% | 253.80894/505.14494 × 100% = 50.24% O (oxygen) | 7 | 58.3% | 111.993/505.14494 × 100% = 22.17%

Mass fraction pie chart