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element mass fraction of lead(II) sulfate

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lead(II) sulfate | elemental composition
lead(II) sulfate | elemental composition

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Find the elemental composition for lead(II) sulfate 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: PbSO_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  O (oxygen) | 4  Pb (lead) | 1  S (sulfur) | 1  N_atoms = 4 + 1 + 1 = 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  O (oxygen) | 4 | 4/6  Pb (lead) | 1 | 1/6  S (sulfur) | 1 | 1/6 Check: 4/6 + 1/6 + 1/6 = 1 Compute atom percents using the atom fractions:  | number of atoms | atom percent  O (oxygen) | 4 | 4/6 × 100% = 66.7%  Pb (lead) | 1 | 1/6 × 100% = 16.7%  S (sulfur) | 1 | 1/6 × 100% = 16.7% 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  O (oxygen) | 4 | 66.7% | 15.999  Pb (lead) | 1 | 16.7% | 207.2  S (sulfur) | 1 | 16.7% | 32.06 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  O (oxygen) | 4 | 66.7% | 15.999 | 4 × 15.999 = 63.996  Pb (lead) | 1 | 16.7% | 207.2 | 1 × 207.2 = 207.2  S (sulfur) | 1 | 16.7% | 32.06 | 1 × 32.06 = 32.06  m = 63.996 u + 207.2 u + 32.06 u = 303.256 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  O (oxygen) | 4 | 66.7% | 63.996/303.256  Pb (lead) | 1 | 16.7% | 207.2/303.256  S (sulfur) | 1 | 16.7% | 32.06/303.256 Check: 63.996/303.256 + 207.2/303.256 + 32.06/303.256 = 1 Compute mass percents using the mass fractions: Answer: |   | | number of atoms | atom percent | mass percent  O (oxygen) | 4 | 66.7% | 63.996/303.256 × 100% = 21.10%  Pb (lead) | 1 | 16.7% | 207.2/303.256 × 100% = 68.33%  S (sulfur) | 1 | 16.7% | 32.06/303.256 × 100% = 10.57%
Find the elemental composition for lead(II) sulfate 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: PbSO_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 O (oxygen) | 4 Pb (lead) | 1 S (sulfur) | 1 N_atoms = 4 + 1 + 1 = 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 O (oxygen) | 4 | 4/6 Pb (lead) | 1 | 1/6 S (sulfur) | 1 | 1/6 Check: 4/6 + 1/6 + 1/6 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent O (oxygen) | 4 | 4/6 × 100% = 66.7% Pb (lead) | 1 | 1/6 × 100% = 16.7% S (sulfur) | 1 | 1/6 × 100% = 16.7% 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 O (oxygen) | 4 | 66.7% | 15.999 Pb (lead) | 1 | 16.7% | 207.2 S (sulfur) | 1 | 16.7% | 32.06 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 O (oxygen) | 4 | 66.7% | 15.999 | 4 × 15.999 = 63.996 Pb (lead) | 1 | 16.7% | 207.2 | 1 × 207.2 = 207.2 S (sulfur) | 1 | 16.7% | 32.06 | 1 × 32.06 = 32.06 m = 63.996 u + 207.2 u + 32.06 u = 303.256 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 O (oxygen) | 4 | 66.7% | 63.996/303.256 Pb (lead) | 1 | 16.7% | 207.2/303.256 S (sulfur) | 1 | 16.7% | 32.06/303.256 Check: 63.996/303.256 + 207.2/303.256 + 32.06/303.256 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent O (oxygen) | 4 | 66.7% | 63.996/303.256 × 100% = 21.10% Pb (lead) | 1 | 16.7% | 207.2/303.256 × 100% = 68.33% S (sulfur) | 1 | 16.7% | 32.06/303.256 × 100% = 10.57%

Mass fraction pie chart

Mass fraction pie chart
Mass fraction pie chart