mass fractions of lithium bromide

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lithium bromide | elemental composition

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$Find the elemental composition for lithium bromide 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: LiBr 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 Li (lithium) | 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 Br (bromine) | 1 | 1/2 Li (lithium) | 1 | 1/2 Check: 1/2 + 1/2 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent Br (bromine) | 1 | 1/2 × 100% = 50.0% Li (lithium) | 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 Br (bromine) | 1 | 50.0% | 79.904 Li (lithium) | 1 | 50.0% | 6.94 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 | 50.0% | 79.904 | 1 × 79.904 = 79.904 Li (lithium) | 1 | 50.0% | 6.94 | 1 × 6.94 = 6.94 m = 79.904 u + 6.94 u = 86.844 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 Br (bromine) | 1 | 50.0% | 79.904/86.844 Li (lithium) | 1 | 50.0% | 6.94/86.844 Check: 79.904/86.844 + 6.94/86.844 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent Br (bromine) | 1 | 50.0% | 79.904/86.844 × 100% = 92.01% Li (lithium) | 1 | 50.0% | 6.94/86.844 × 100% = 7.99%$

Find the elemental composition for lithium bromide 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: LiBr 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 Li (lithium) | 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 Br (bromine) | 1 | 1/2 Li (lithium) | 1 | 1/2 Check: 1/2 + 1/2 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent Br (bromine) | 1 | 1/2 × 100% = 50.0% Li (lithium) | 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 Br (bromine) | 1 | 50.0% | 79.904 Li (lithium) | 1 | 50.0% | 6.94 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 | 50.0% | 79.904 | 1 × 79.904 = 79.904 Li (lithium) | 1 | 50.0% | 6.94 | 1 × 6.94 = 6.94 m = 79.904 u + 6.94 u = 86.844 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 Br (bromine) | 1 | 50.0% | 79.904/86.844 Li (lithium) | 1 | 50.0% | 6.94/86.844 Check: 79.904/86.844 + 6.94/86.844 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent Br (bromine) | 1 | 50.0% | 79.904/86.844 × 100% = 92.01% Li (lithium) | 1 | 50.0% | 6.94/86.844 × 100% = 7.99%