mass fractions of ytterbium(III) chloride hexahydrate

ytterbium(III) chloride hexahydrate | elemental composition

Find the elemental composition for ytterbium(III) chloride hexahydrate 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: YbCl_3·6H_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 Cl (chlorine) | 3 H (hydrogen) | 12 O (oxygen) | 6 Yb (ytterbium) | 1 N_atoms = 3 + 12 + 6 + 1 = 22 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 Cl (chlorine) | 3 | 3/22 H (hydrogen) | 12 | 12/22 O (oxygen) | 6 | 6/22 Yb (ytterbium) | 1 | 1/22 Check: 3/22 + 12/22 + 6/22 + 1/22 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent Cl (chlorine) | 3 | 3/22 × 100% = 13.6% H (hydrogen) | 12 | 12/22 × 100% = 54.5% O (oxygen) | 6 | 6/22 × 100% = 27.3% Yb (ytterbium) | 1 | 1/22 × 100% = 4.55% 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 Cl (chlorine) | 3 | 13.6% | 35.45 H (hydrogen) | 12 | 54.5% | 1.008 O (oxygen) | 6 | 27.3% | 15.999 Yb (ytterbium) | 1 | 4.55% | 173.045 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 Cl (chlorine) | 3 | 13.6% | 35.45 | 3 × 35.45 = 106.35 H (hydrogen) | 12 | 54.5% | 1.008 | 12 × 1.008 = 12.096 O (oxygen) | 6 | 27.3% | 15.999 | 6 × 15.999 = 95.994 Yb (ytterbium) | 1 | 4.55% | 173.045 | 1 × 173.045 = 173.045 m = 106.35 u + 12.096 u + 95.994 u + 173.045 u = 387.485 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 Cl (chlorine) | 3 | 13.6% | 106.35/387.485 H (hydrogen) | 12 | 54.5% | 12.096/387.485 O (oxygen) | 6 | 27.3% | 95.994/387.485 Yb (ytterbium) | 1 | 4.55% | 173.045/387.485 Check: 106.35/387.485 + 12.096/387.485 + 95.994/387.485 + 173.045/387.485 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent Cl (chlorine) | 3 | 13.6% | 106.35/387.485 × 100% = 27.45% H (hydrogen) | 12 | 54.5% | 12.096/387.485 × 100% = 3.122% O (oxygen) | 6 | 27.3% | 95.994/387.485 × 100% = 24.77% Yb (ytterbium) | 1 | 4.55% | 173.045/387.485 × 100% = 44.66%

Mass fraction pie chart