mass fractions of iron(II) oxide

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iron(II) oxide | elemental composition

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$Find the elemental composition for iron(II) oxide 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: FeO 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 Fe (iron) | 1 O (oxygen) | 1 N_atoms = 1 + 1 = 2 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 Fe (iron) | 1 | 1/2 O (oxygen) | 1 | 1/2 Check: 1/2 + 1/2 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent Fe (iron) | 1 | 1/2 × 100% = 50.0% O (oxygen) | 1 | 1/2 × 100% = 50.0% 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 Fe (iron) | 1 | 50.0% | 55.845 O (oxygen) | 1 | 50.0% | 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 Fe (iron) | 1 | 50.0% | 55.845 | 1 × 55.845 = 55.845 O (oxygen) | 1 | 50.0% | 15.999 | 1 × 15.999 = 15.999 m = 55.845 u + 15.999 u = 71.844 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 Fe (iron) | 1 | 50.0% | 55.845/71.844 O (oxygen) | 1 | 50.0% | 15.999/71.844 Check: 55.845/71.844 + 15.999/71.844 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent Fe (iron) | 1 | 50.0% | 55.845/71.844 × 100% = 77.73% O (oxygen) | 1 | 50.0% | 15.999/71.844 × 100% = 22.27%$

Find the elemental composition for iron(II) oxide 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: FeO 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 Fe (iron) | 1 O (oxygen) | 1 N_atoms = 1 + 1 = 2 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 Fe (iron) | 1 | 1/2 O (oxygen) | 1 | 1/2 Check: 1/2 + 1/2 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent Fe (iron) | 1 | 1/2 × 100% = 50.0% O (oxygen) | 1 | 1/2 × 100% = 50.0% 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 Fe (iron) | 1 | 50.0% | 55.845 O (oxygen) | 1 | 50.0% | 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 Fe (iron) | 1 | 50.0% | 55.845 | 1 × 55.845 = 55.845 O (oxygen) | 1 | 50.0% | 15.999 | 1 × 15.999 = 15.999 m = 55.845 u + 15.999 u = 71.844 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 Fe (iron) | 1 | 50.0% | 55.845/71.844 O (oxygen) | 1 | 50.0% | 15.999/71.844 Check: 55.845/71.844 + 15.999/71.844 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent Fe (iron) | 1 | 50.0% | 55.845/71.844 × 100% = 77.73% O (oxygen) | 1 | 50.0% | 15.999/71.844 × 100% = 22.27%