mass fractions of halothane

halothane | elemental composition

Find the elemental composition for halothane 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: BrCH(Cl)CF_3 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 C (carbon) | 2 Cl (chlorine) | 1 F (fluorine) | 3 H (hydrogen) | 1 N_atoms = 1 + 2 + 1 + 3 + 1 = 8 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 | 1/8 C (carbon) | 2 | 2/8 Cl (chlorine) | 1 | 1/8 F (fluorine) | 3 | 3/8 H (hydrogen) | 1 | 1/8 Check: 1/8 + 2/8 + 1/8 + 3/8 + 1/8 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent Br (bromine) | 1 | 1/8 × 100% = 12.5% C (carbon) | 2 | 2/8 × 100% = 25.0% Cl (chlorine) | 1 | 1/8 × 100% = 12.5% F (fluorine) | 3 | 3/8 × 100% = 37.5% H (hydrogen) | 1 | 1/8 × 100% = 12.5% 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 | 12.5% | 79.904 C (carbon) | 2 | 25.0% | 12.011 Cl (chlorine) | 1 | 12.5% | 35.45 F (fluorine) | 3 | 37.5% | 18.998403163 H (hydrogen) | 1 | 12.5% | 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 | 12.5% | 79.904 | 1 × 79.904 = 79.904 C (carbon) | 2 | 25.0% | 12.011 | 2 × 12.011 = 24.022 Cl (chlorine) | 1 | 12.5% | 35.45 | 1 × 35.45 = 35.45 F (fluorine) | 3 | 37.5% | 18.998403163 | 3 × 18.998403163 = 56.995209489 H (hydrogen) | 1 | 12.5% | 1.008 | 1 × 1.008 = 1.008 m = 79.904 u + 24.022 u + 35.45 u + 56.995209489 u + 1.008 u = 197.379209489 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 | 12.5% | 79.904/197.379209489 C (carbon) | 2 | 25.0% | 24.022/197.379209489 Cl (chlorine) | 1 | 12.5% | 35.45/197.379209489 F (fluorine) | 3 | 37.5% | 56.995209489/197.379209489 H (hydrogen) | 1 | 12.5% | 1.008/197.379209489 Check: 79.904/197.379209489 + 24.022/197.379209489 + 35.45/197.379209489 + 56.995209489/197.379209489 + 1.008/197.379209489 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent Br (bromine) | 1 | 12.5% | 79.904/197.379209489 × 100% = 40.48% C (carbon) | 2 | 25.0% | 24.022/197.379209489 × 100% = 12.17% Cl (chlorine) | 1 | 12.5% | 35.45/197.379209489 × 100% = 17.96% F (fluorine) | 3 | 37.5% | 56.995209489/197.379209489 × 100% = 28.88% H (hydrogen) | 1 | 12.5% | 1.008/197.379209489 × 100% = 0.5107%

Mass fraction pie chart