Input interpretation
SMILES: BrNO | elemental composition
Result
Find the elemental composition for SMILES:BrNO 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: BrH_2NO 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 Br (bromine) | 1 N (nitrogen) | 1 O (oxygen) | 1 H (hydrogen) | 2 N_atoms = 1 + 1 + 1 + 2 = 5 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 Br (bromine) | 1 | 1/5 N (nitrogen) | 1 | 1/5 O (oxygen) | 1 | 1/5 H (hydrogen) | 2 | 2/5 Check: 1/5 + 1/5 + 1/5 + 2/5 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent Br (bromine) | 1 | 1/5 × 100% = 20.0% N (nitrogen) | 1 | 1/5 × 100% = 20.0% O (oxygen) | 1 | 1/5 × 100% = 20.0% H (hydrogen) | 2 | 2/5 × 100% = 40.0% 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 Br (bromine) | 1 | 20.0% | 79.904 N (nitrogen) | 1 | 20.0% | 14.007 O (oxygen) | 1 | 20.0% | 15.999 H (hydrogen) | 2 | 40.0% | 1.008 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 Br (bromine) | 1 | 20.0% | 79.904 | 1 × 79.904 = 79.904 N (nitrogen) | 1 | 20.0% | 14.007 | 1 × 14.007 = 14.007 O (oxygen) | 1 | 20.0% | 15.999 | 1 × 15.999 = 15.999 H (hydrogen) | 2 | 40.0% | 1.008 | 2 × 1.008 = 2.016 m = 79.904 u + 14.007 u + 15.999 u + 2.016 u = 111.926 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 Br (bromine) | 1 | 20.0% | 79.904/111.926 N (nitrogen) | 1 | 20.0% | 14.007/111.926 O (oxygen) | 1 | 20.0% | 15.999/111.926 H (hydrogen) | 2 | 40.0% | 2.016/111.926 Check: 79.904/111.926 + 14.007/111.926 + 15.999/111.926 + 2.016/111.926 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent Br (bromine) | 1 | 20.0% | 79.904/111.926 × 100% = 71.39% N (nitrogen) | 1 | 20.0% | 14.007/111.926 × 100% = 12.51% O (oxygen) | 1 | 20.0% | 15.999/111.926 × 100% = 14.29% H (hydrogen) | 2 | 40.0% | 2.016/111.926 × 100% = 1.801%