element mass fraction of iron(II) chloride

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iron(II) chloride | elemental composition

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$Find the elemental composition for iron(II) chloride 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: FeCl_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 Cl (chlorine) | 2 Fe (iron) | 1 N_atoms = 2 + 1 = 3 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 Cl (chlorine) | 2 | 2/3 Fe (iron) | 1 | 1/3 Check: 2/3 + 1/3 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent Cl (chlorine) | 2 | 2/3 × 100% = 66.7% Fe (iron) | 1 | 1/3 × 100% = 33.3% 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 Cl (chlorine) | 2 | 66.7% | 35.45 Fe (iron) | 1 | 33.3% | 55.845 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 Cl (chlorine) | 2 | 66.7% | 35.45 | 2 × 35.45 = 70.90 Fe (iron) | 1 | 33.3% | 55.845 | 1 × 55.845 = 55.845 m = 70.90 u + 55.845 u = 126.745 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 Cl (chlorine) | 2 | 66.7% | 70.90/126.745 Fe (iron) | 1 | 33.3% | 55.845/126.745 Check: 70.90/126.745 + 55.845/126.745 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent Cl (chlorine) | 2 | 66.7% | 70.90/126.745 × 100% = 55.94% Fe (iron) | 1 | 33.3% | 55.845/126.745 × 100% = 44.06%$

Find the elemental composition for iron(II) chloride 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: FeCl_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 Cl (chlorine) | 2 Fe (iron) | 1 N_atoms = 2 + 1 = 3 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 Cl (chlorine) | 2 | 2/3 Fe (iron) | 1 | 1/3 Check: 2/3 + 1/3 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent Cl (chlorine) | 2 | 2/3 × 100% = 66.7% Fe (iron) | 1 | 1/3 × 100% = 33.3% 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 Cl (chlorine) | 2 | 66.7% | 35.45 Fe (iron) | 1 | 33.3% | 55.845 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 Cl (chlorine) | 2 | 66.7% | 35.45 | 2 × 35.45 = 70.90 Fe (iron) | 1 | 33.3% | 55.845 | 1 × 55.845 = 55.845 m = 70.90 u + 55.845 u = 126.745 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 Cl (chlorine) | 2 | 66.7% | 70.90/126.745 Fe (iron) | 1 | 33.3% | 55.845/126.745 Check: 70.90/126.745 + 55.845/126.745 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent Cl (chlorine) | 2 | 66.7% | 70.90/126.745 × 100% = 55.94% Fe (iron) | 1 | 33.3% | 55.845/126.745 × 100% = 44.06%