mass fractions of 1,3-butadiene

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1, 3-butadiene | elemental composition

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$Find the elemental composition for 1, 3-butadiene 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: CH_2=CHCH=CH_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) | 4 H (hydrogen) | 6 N_atoms = 4 + 6 = 10 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) | 4 | 4/10 H (hydrogen) | 6 | 6/10 Check: 4/10 + 6/10 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent C (carbon) | 4 | 4/10 × 100% = 40.0% H (hydrogen) | 6 | 6/10 × 100% = 60.0% 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 C (carbon) | 4 | 40.0% | 12.011 H (hydrogen) | 6 | 60.0% | 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 C (carbon) | 4 | 40.0% | 12.011 | 4 × 12.011 = 48.044 H (hydrogen) | 6 | 60.0% | 1.008 | 6 × 1.008 = 6.048 m = 48.044 u + 6.048 u = 54.092 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) | 4 | 40.0% | 48.044/54.092 H (hydrogen) | 6 | 60.0% | 6.048/54.092 Check: 48.044/54.092 + 6.048/54.092 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent C (carbon) | 4 | 40.0% | 48.044/54.092 × 100% = 88.82% H (hydrogen) | 6 | 60.0% | 6.048/54.092 × 100% = 11.18%$

Find the elemental composition for 1, 3-butadiene 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: CH_2=CHCH=CH_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) | 4 H (hydrogen) | 6 N_atoms = 4 + 6 = 10 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) | 4 | 4/10 H (hydrogen) | 6 | 6/10 Check: 4/10 + 6/10 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent C (carbon) | 4 | 4/10 × 100% = 40.0% H (hydrogen) | 6 | 6/10 × 100% = 60.0% 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 C (carbon) | 4 | 40.0% | 12.011 H (hydrogen) | 6 | 60.0% | 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 C (carbon) | 4 | 40.0% | 12.011 | 4 × 12.011 = 48.044 H (hydrogen) | 6 | 60.0% | 1.008 | 6 × 1.008 = 6.048 m = 48.044 u + 6.048 u = 54.092 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) | 4 | 40.0% | 48.044/54.092 H (hydrogen) | 6 | 60.0% | 6.048/54.092 Check: 48.044/54.092 + 6.048/54.092 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent C (carbon) | 4 | 40.0% | 48.044/54.092 × 100% = 88.82% H (hydrogen) | 6 | 60.0% | 6.048/54.092 × 100% = 11.18%