element mass fraction of lead tetrachloride

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lead tetrachloride | elemental composition

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$Find the elemental composition for lead tetrachloride 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: Cl_4Pb 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) | 4 Pb (lead) | 1 N_atoms = 4 + 1 = 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 Cl (chlorine) | 4 | 4/5 Pb (lead) | 1 | 1/5 Check: 4/5 + 1/5 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent Cl (chlorine) | 4 | 4/5 × 100% = 80.0% Pb (lead) | 1 | 1/5 × 100% = 20.0% 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 Cl (chlorine) | 4 | 80.0% | 35.45 Pb (lead) | 1 | 20.0% | 207.2 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) | 4 | 80.0% | 35.45 | 4 × 35.45 = 141.80 Pb (lead) | 1 | 20.0% | 207.2 | 1 × 207.2 = 207.2 m = 141.80 u + 207.2 u = 349.00 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) | 4 | 80.0% | 141.80/349.00 Pb (lead) | 1 | 20.0% | 207.2/349.00 Check: 141.80/349.00 + 207.2/349.00 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent Cl (chlorine) | 4 | 80.0% | 141.80/349.00 × 100% = 40.63% Pb (lead) | 1 | 20.0% | 207.2/349.00 × 100% = 59.37%$

Find the elemental composition for lead tetrachloride 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: Cl_4Pb 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) | 4 Pb (lead) | 1 N_atoms = 4 + 1 = 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 Cl (chlorine) | 4 | 4/5 Pb (lead) | 1 | 1/5 Check: 4/5 + 1/5 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent Cl (chlorine) | 4 | 4/5 × 100% = 80.0% Pb (lead) | 1 | 1/5 × 100% = 20.0% 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 Cl (chlorine) | 4 | 80.0% | 35.45 Pb (lead) | 1 | 20.0% | 207.2 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) | 4 | 80.0% | 35.45 | 4 × 35.45 = 141.80 Pb (lead) | 1 | 20.0% | 207.2 | 1 × 207.2 = 207.2 m = 141.80 u + 207.2 u = 349.00 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) | 4 | 80.0% | 141.80/349.00 Pb (lead) | 1 | 20.0% | 207.2/349.00 Check: 141.80/349.00 + 207.2/349.00 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent Cl (chlorine) | 4 | 80.0% | 141.80/349.00 × 100% = 40.63% Pb (lead) | 1 | 20.0% | 207.2/349.00 × 100% = 59.37%