mass fractions of 3-chloro-4-fluoroaniline

3-chloro-4-fluoroaniline | elemental composition

Find the elemental composition for 3-chloro-4-fluoroaniline 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: ClC_6H_3(F)NH_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 C (carbon) | 6 Cl (chlorine) | 1 F (fluorine) | 1 H (hydrogen) | 5 N (nitrogen) | 1 N_atoms = 6 + 1 + 1 + 5 + 1 = 14 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 C (carbon) | 6 | 6/14 Cl (chlorine) | 1 | 1/14 F (fluorine) | 1 | 1/14 H (hydrogen) | 5 | 5/14 N (nitrogen) | 1 | 1/14 Check: 6/14 + 1/14 + 1/14 + 5/14 + 1/14 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent C (carbon) | 6 | 6/14 × 100% = 42.9% Cl (chlorine) | 1 | 1/14 × 100% = 7.14% F (fluorine) | 1 | 1/14 × 100% = 7.14% H (hydrogen) | 5 | 5/14 × 100% = 35.7% N (nitrogen) | 1 | 1/14 × 100% = 7.14% 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 C (carbon) | 6 | 42.9% | 12.011 Cl (chlorine) | 1 | 7.14% | 35.45 F (fluorine) | 1 | 7.14% | 18.998403163 H (hydrogen) | 5 | 35.7% | 1.008 N (nitrogen) | 1 | 7.14% | 14.007 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 C (carbon) | 6 | 42.9% | 12.011 | 6 × 12.011 = 72.066 Cl (chlorine) | 1 | 7.14% | 35.45 | 1 × 35.45 = 35.45 F (fluorine) | 1 | 7.14% | 18.998403163 | 1 × 18.998403163 = 18.998403163 H (hydrogen) | 5 | 35.7% | 1.008 | 5 × 1.008 = 5.040 N (nitrogen) | 1 | 7.14% | 14.007 | 1 × 14.007 = 14.007 m = 72.066 u + 35.45 u + 18.998403163 u + 5.040 u + 14.007 u = 145.561403163 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 C (carbon) | 6 | 42.9% | 72.066/145.561403163 Cl (chlorine) | 1 | 7.14% | 35.45/145.561403163 F (fluorine) | 1 | 7.14% | 18.998403163/145.561403163 H (hydrogen) | 5 | 35.7% | 5.040/145.561403163 N (nitrogen) | 1 | 7.14% | 14.007/145.561403163 Check: 72.066/145.561403163 + 35.45/145.561403163 + 18.998403163/145.561403163 + 5.040/145.561403163 + 14.007/145.561403163 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent C (carbon) | 6 | 42.9% | 72.066/145.561403163 × 100% = 49.51% Cl (chlorine) | 1 | 7.14% | 35.45/145.561403163 × 100% = 24.35% F (fluorine) | 1 | 7.14% | 18.998403163/145.561403163 × 100% = 13.05% H (hydrogen) | 5 | 35.7% | 5.040/145.561403163 × 100% = 3.462% N (nitrogen) | 1 | 7.14% | 14.007/145.561403163 × 100% = 9.623%

Mass fraction pie chart