mass fractions of barium chromate

barium chromate | elemental composition

Find the elemental composition for barium chromate 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: BaCrO_4 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 Cr (chromium) | 1 O (oxygen) | 4 N_atoms = 1 + 1 + 4 = 6 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 | 1/6 Cr (chromium) | 1 | 1/6 O (oxygen) | 4 | 4/6 Check: 1/6 + 1/6 + 4/6 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent Ba (barium) | 1 | 1/6 × 100% = 16.7% Cr (chromium) | 1 | 1/6 × 100% = 16.7% O (oxygen) | 4 | 4/6 × 100% = 66.7% 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 | 16.7% | 137.327 Cr (chromium) | 1 | 16.7% | 51.9961 O (oxygen) | 4 | 66.7% | 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 | 16.7% | 137.327 | 1 × 137.327 = 137.327 Cr (chromium) | 1 | 16.7% | 51.9961 | 1 × 51.9961 = 51.9961 O (oxygen) | 4 | 66.7% | 15.999 | 4 × 15.999 = 63.996 m = 137.327 u + 51.9961 u + 63.996 u = 253.3191 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 | 16.7% | 137.327/253.3191 Cr (chromium) | 1 | 16.7% | 51.9961/253.3191 O (oxygen) | 4 | 66.7% | 63.996/253.3191 Check: 137.327/253.3191 + 51.9961/253.3191 + 63.996/253.3191 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent Ba (barium) | 1 | 16.7% | 137.327/253.3191 × 100% = 54.21% Cr (chromium) | 1 | 16.7% | 51.9961/253.3191 × 100% = 20.53% O (oxygen) | 4 | 66.7% | 63.996/253.3191 × 100% = 25.26%

Mass fraction pie chart