element tally of selenium trioxide

Input interpretation

Image
Text

selenium trioxide | elemental composition

Result

Image
Text

$Find the elemental composition for selenium trioxide 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: SeO_3 Use the chemical formula, SeO_3, to count the number of atoms, N_i, for each element and find the total number of atoms, N_atoms: | number of atoms O (oxygen) | 3 Se (selenium) | 1 N_atoms = 3 + 1 = 4 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) | 3 | 3/4 Se (selenium) | 1 | 1/4 Check: 3/4 + 1/4 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent O (oxygen) | 3 | 3/4 × 100% = 75.0% Se (selenium) | 1 | 1/4 × 100% = 25.0% 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 O (oxygen) | 3 | 75.0% | 15.999 Se (selenium) | 1 | 25.0% | 78.971 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) | 3 | 75.0% | 15.999 | 3 × 15.999 = 47.997 Se (selenium) | 1 | 25.0% | 78.971 | 1 × 78.971 = 78.971 m = 47.997 u + 78.971 u = 126.968 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) | 3 | 75.0% | 47.997/126.968 Se (selenium) | 1 | 25.0% | 78.971/126.968 Check: 47.997/126.968 + 78.971/126.968 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent O (oxygen) | 3 | 75.0% | 47.997/126.968 × 100% = 37.80% Se (selenium) | 1 | 25.0% | 78.971/126.968 × 100% = 62.20%$

Find the elemental composition for selenium trioxide 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: SeO_3 Use the chemical formula, SeO_3, to count the number of atoms, N_i, for each element and find the total number of atoms, N_atoms: | number of atoms O (oxygen) | 3 Se (selenium) | 1 N_atoms = 3 + 1 = 4 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) | 3 | 3/4 Se (selenium) | 1 | 1/4 Check: 3/4 + 1/4 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent O (oxygen) | 3 | 3/4 × 100% = 75.0% Se (selenium) | 1 | 1/4 × 100% = 25.0% 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 O (oxygen) | 3 | 75.0% | 15.999 Se (selenium) | 1 | 25.0% | 78.971 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) | 3 | 75.0% | 15.999 | 3 × 15.999 = 47.997 Se (selenium) | 1 | 25.0% | 78.971 | 1 × 78.971 = 78.971 m = 47.997 u + 78.971 u = 126.968 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) | 3 | 75.0% | 47.997/126.968 Se (selenium) | 1 | 25.0% | 78.971/126.968 Check: 47.997/126.968 + 78.971/126.968 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent O (oxygen) | 3 | 75.0% | 47.997/126.968 × 100% = 37.80% Se (selenium) | 1 | 25.0% | 78.971/126.968 × 100% = 62.20%