mass fractions of barium perchlorate trihydrate

barium perchlorate trihydrate | elemental composition

Find the elemental composition for barium perchlorate trihydrate 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(ClO_4)_2·3H_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 Cl (chlorine) | 2 H (hydrogen) | 6 O (oxygen) | 11 N_atoms = 1 + 2 + 6 + 11 = 20 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 | 1/20 Cl (chlorine) | 2 | 2/20 H (hydrogen) | 6 | 6/20 O (oxygen) | 11 | 11/20 Check: 1/20 + 2/20 + 6/20 + 11/20 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent Ba (barium) | 1 | 1/20 × 100% = 5.00% Cl (chlorine) | 2 | 2/20 × 100% = 10.00% H (hydrogen) | 6 | 6/20 × 100% = 30.0% O (oxygen) | 11 | 11/20 × 100% = 55.0% 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 | 5.00% | 137.327 Cl (chlorine) | 2 | 10.00% | 35.45 H (hydrogen) | 6 | 30.0% | 1.008 O (oxygen) | 11 | 55.0% | 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 | 5.00% | 137.327 | 1 × 137.327 = 137.327 Cl (chlorine) | 2 | 10.00% | 35.45 | 2 × 35.45 = 70.90 H (hydrogen) | 6 | 30.0% | 1.008 | 6 × 1.008 = 6.048 O (oxygen) | 11 | 55.0% | 15.999 | 11 × 15.999 = 175.989 m = 137.327 u + 70.90 u + 6.048 u + 175.989 u = 390.264 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 | 5.00% | 137.327/390.264 Cl (chlorine) | 2 | 10.00% | 70.90/390.264 H (hydrogen) | 6 | 30.0% | 6.048/390.264 O (oxygen) | 11 | 55.0% | 175.989/390.264 Check: 137.327/390.264 + 70.90/390.264 + 6.048/390.264 + 175.989/390.264 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent Ba (barium) | 1 | 5.00% | 137.327/390.264 × 100% = 35.19% Cl (chlorine) | 2 | 10.00% | 70.90/390.264 × 100% = 18.17% H (hydrogen) | 6 | 30.0% | 6.048/390.264 × 100% = 1.550% O (oxygen) | 11 | 55.0% | 175.989/390.264 × 100% = 45.09%

Mass fraction pie chart