mass fractions of nickel(II) fluoborate

nickel(II) fluoborate | elemental composition

Find the elemental composition for nickel(II) fluoborate 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: Ni(BF_4)_2 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 B (boron) | 2 F (fluorine) | 8 Ni (nickel) | 1 N_atoms = 2 + 8 + 1 = 11 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 B (boron) | 2 | 2/11 F (fluorine) | 8 | 8/11 Ni (nickel) | 1 | 1/11 Check: 2/11 + 8/11 + 1/11 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent B (boron) | 2 | 2/11 × 100% = 18.2% F (fluorine) | 8 | 8/11 × 100% = 72.7% Ni (nickel) | 1 | 1/11 × 100% = 9.09% 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 B (boron) | 2 | 18.2% | 10.81 F (fluorine) | 8 | 72.7% | 18.998403163 Ni (nickel) | 1 | 9.09% | 58.6934 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 B (boron) | 2 | 18.2% | 10.81 | 2 × 10.81 = 21.62 F (fluorine) | 8 | 72.7% | 18.998403163 | 8 × 18.998403163 = 151.987225304 Ni (nickel) | 1 | 9.09% | 58.6934 | 1 × 58.6934 = 58.6934 m = 21.62 u + 151.987225304 u + 58.6934 u = 232.300625304 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 B (boron) | 2 | 18.2% | 21.62/232.300625304 F (fluorine) | 8 | 72.7% | 151.987225304/232.300625304 Ni (nickel) | 1 | 9.09% | 58.6934/232.300625304 Check: 21.62/232.300625304 + 151.987225304/232.300625304 + 58.6934/232.300625304 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent B (boron) | 2 | 18.2% | 21.62/232.300625304 × 100% = 9.307% F (fluorine) | 8 | 72.7% | 151.987225304/232.300625304 × 100% = 65.43% Ni (nickel) | 1 | 9.09% | 58.6934/232.300625304 × 100% = 25.27%

Mass fraction pie chart