mass fractions of barium iron oxide

barium iron oxide | elemental composition

Find the elemental composition for barium iron oxide 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: BaFe_12O_19 Use the chemical formula, BaFe_12O_19, to count the number of atoms, N_i, for each element and find the total number of atoms, N_atoms: | number of atoms Ba (barium) | 1 Fe (iron) | 12 O (oxygen) | 19 N_atoms = 1 + 12 + 19 = 32 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 | 1/32 Fe (iron) | 12 | 12/32 O (oxygen) | 19 | 19/32 Check: 1/32 + 12/32 + 19/32 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent Ba (barium) | 1 | 1/32 × 100% = 3.13% Fe (iron) | 12 | 12/32 × 100% = 37.5% O (oxygen) | 19 | 19/32 × 100% = 59.4% 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 Ba (barium) | 1 | 3.13% | 137.327 Fe (iron) | 12 | 37.5% | 55.845 O (oxygen) | 19 | 59.4% | 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 | 3.13% | 137.327 | 1 × 137.327 = 137.327 Fe (iron) | 12 | 37.5% | 55.845 | 12 × 55.845 = 670.140 O (oxygen) | 19 | 59.4% | 15.999 | 19 × 15.999 = 303.981 m = 137.327 u + 670.140 u + 303.981 u = 1111.448 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 | 3.13% | 137.327/1111.448 Fe (iron) | 12 | 37.5% | 670.140/1111.448 O (oxygen) | 19 | 59.4% | 303.981/1111.448 Check: 137.327/1111.448 + 670.140/1111.448 + 303.981/1111.448 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent Ba (barium) | 1 | 3.13% | 137.327/1111.448 × 100% = 12.36% Fe (iron) | 12 | 37.5% | 670.140/1111.448 × 100% = 60.29% O (oxygen) | 19 | 59.4% | 303.981/1111.448 × 100% = 27.35%

Mass fraction pie chart