mass fractions of (ethyl benzoate)tricarbonylchromium(0)

(ethyl benzoate)tricarbonylchromium(0) | elemental composition

Find the elemental composition for (ethyl benzoate)tricarbonylchromium(0) 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: C_6H_5CO_2C_2H_5Cr(CO)_3 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 C (carbon) | 12 Cr (chromium) | 1 H (hydrogen) | 10 O (oxygen) | 5 N_atoms = 12 + 1 + 10 + 5 = 28 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 C (carbon) | 12 | 12/28 Cr (chromium) | 1 | 1/28 H (hydrogen) | 10 | 10/28 O (oxygen) | 5 | 5/28 Check: 12/28 + 1/28 + 10/28 + 5/28 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent C (carbon) | 12 | 12/28 × 100% = 42.9% Cr (chromium) | 1 | 1/28 × 100% = 3.57% H (hydrogen) | 10 | 10/28 × 100% = 35.7% O (oxygen) | 5 | 5/28 × 100% = 17.9% 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 C (carbon) | 12 | 42.9% | 12.011 Cr (chromium) | 1 | 3.57% | 51.9961 H (hydrogen) | 10 | 35.7% | 1.008 O (oxygen) | 5 | 17.9% | 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 C (carbon) | 12 | 42.9% | 12.011 | 12 × 12.011 = 144.132 Cr (chromium) | 1 | 3.57% | 51.9961 | 1 × 51.9961 = 51.9961 H (hydrogen) | 10 | 35.7% | 1.008 | 10 × 1.008 = 10.080 O (oxygen) | 5 | 17.9% | 15.999 | 5 × 15.999 = 79.995 m = 144.132 u + 51.9961 u + 10.080 u + 79.995 u = 286.2031 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 C (carbon) | 12 | 42.9% | 144.132/286.2031 Cr (chromium) | 1 | 3.57% | 51.9961/286.2031 H (hydrogen) | 10 | 35.7% | 10.080/286.2031 O (oxygen) | 5 | 17.9% | 79.995/286.2031 Check: 144.132/286.2031 + 51.9961/286.2031 + 10.080/286.2031 + 79.995/286.2031 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent C (carbon) | 12 | 42.9% | 144.132/286.2031 × 100% = 50.36% Cr (chromium) | 1 | 3.57% | 51.9961/286.2031 × 100% = 18.17% H (hydrogen) | 10 | 35.7% | 10.080/286.2031 × 100% = 3.522% O (oxygen) | 5 | 17.9% | 79.995/286.2031 × 100% = 27.95%

Mass fraction pie chart