mass fractions of terbium(III) sulfate octahydrate

terbium(III) sulfate octahydrate | elemental composition

Find the elemental composition for terbium(III) sulfate octahydrate 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: Tb_2(SO_4)_3·8H_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 H (hydrogen) | 16 O (oxygen) | 20 S (sulfur) | 3 Tb (terbium) | 2 N_atoms = 16 + 20 + 3 + 2 = 41 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 H (hydrogen) | 16 | 16/41 O (oxygen) | 20 | 20/41 S (sulfur) | 3 | 3/41 Tb (terbium) | 2 | 2/41 Check: 16/41 + 20/41 + 3/41 + 2/41 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent H (hydrogen) | 16 | 16/41 × 100% = 39.0% O (oxygen) | 20 | 20/41 × 100% = 48.8% S (sulfur) | 3 | 3/41 × 100% = 7.32% Tb (terbium) | 2 | 2/41 × 100% = 4.88% 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 H (hydrogen) | 16 | 39.0% | 1.008 O (oxygen) | 20 | 48.8% | 15.999 S (sulfur) | 3 | 7.32% | 32.06 Tb (terbium) | 2 | 4.88% | 158.92535 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 H (hydrogen) | 16 | 39.0% | 1.008 | 16 × 1.008 = 16.128 O (oxygen) | 20 | 48.8% | 15.999 | 20 × 15.999 = 319.980 S (sulfur) | 3 | 7.32% | 32.06 | 3 × 32.06 = 96.18 Tb (terbium) | 2 | 4.88% | 158.92535 | 2 × 158.92535 = 317.85070 m = 16.128 u + 319.980 u + 96.18 u + 317.85070 u = 750.13870 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 H (hydrogen) | 16 | 39.0% | 16.128/750.13870 O (oxygen) | 20 | 48.8% | 319.980/750.13870 S (sulfur) | 3 | 7.32% | 96.18/750.13870 Tb (terbium) | 2 | 4.88% | 317.85070/750.13870 Check: 16.128/750.13870 + 319.980/750.13870 + 96.18/750.13870 + 317.85070/750.13870 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent H (hydrogen) | 16 | 39.0% | 16.128/750.13870 × 100% = 2.150% O (oxygen) | 20 | 48.8% | 319.980/750.13870 × 100% = 42.66% S (sulfur) | 3 | 7.32% | 96.18/750.13870 × 100% = 12.82% Tb (terbium) | 2 | 4.88% | 317.85070/750.13870 × 100% = 42.37%

Mass fraction pie chart