Input interpretation
lead(II) chromate | elemental composition
Result
Find the elemental composition for lead(II) 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: PbCrO_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 Cr (chromium) | 1 O (oxygen) | 4 Pb (lead) | 1 N_atoms = 1 + 4 + 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 Cr (chromium) | 1 | 1/6 O (oxygen) | 4 | 4/6 Pb (lead) | 1 | 1/6 Check: 1/6 + 4/6 + 1/6 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent Cr (chromium) | 1 | 1/6 × 100% = 16.7% O (oxygen) | 4 | 4/6 × 100% = 66.7% Pb (lead) | 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 Cr (chromium) | 1 | 16.7% | 51.9961 O (oxygen) | 4 | 66.7% | 15.999 Pb (lead) | 1 | 16.7% | 207.2 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 Cr (chromium) | 1 | 16.7% | 51.9961 | 1 × 51.9961 = 51.9961 O (oxygen) | 4 | 66.7% | 15.999 | 4 × 15.999 = 63.996 Pb (lead) | 1 | 16.7% | 207.2 | 1 × 207.2 = 207.2 m = 51.9961 u + 63.996 u + 207.2 u = 323.1921 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 Cr (chromium) | 1 | 16.7% | 51.9961/323.1921 O (oxygen) | 4 | 66.7% | 63.996/323.1921 Pb (lead) | 1 | 16.7% | 207.2/323.1921 Check: 51.9961/323.1921 + 63.996/323.1921 + 207.2/323.1921 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent Cr (chromium) | 1 | 16.7% | 51.9961/323.1921 × 100% = 16.09% O (oxygen) | 4 | 66.7% | 63.996/323.1921 × 100% = 19.80% Pb (lead) | 1 | 16.7% | 207.2/323.1921 × 100% = 64.11%
Mass fraction pie chart
Mass fraction pie chart