mass fractions of diazane

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diazane | elemental composition

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$Find the elemental composition for diazane 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: NH_2NH_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 H (hydrogen) | 4 N (nitrogen) | 2 N_atoms = 4 + 2 = 6 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 H (hydrogen) | 4 | 4/6 N (nitrogen) | 2 | 2/6 Check: 4/6 + 2/6 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent H (hydrogen) | 4 | 4/6 × 100% = 66.7% N (nitrogen) | 2 | 2/6 × 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 H (hydrogen) | 4 | 66.7% | 1.008 N (nitrogen) | 2 | 33.3% | 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 H (hydrogen) | 4 | 66.7% | 1.008 | 4 × 1.008 = 4.032 N (nitrogen) | 2 | 33.3% | 14.007 | 2 × 14.007 = 28.014 m = 4.032 u + 28.014 u = 32.046 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 H (hydrogen) | 4 | 66.7% | 4.032/32.046 N (nitrogen) | 2 | 33.3% | 28.014/32.046 Check: 4.032/32.046 + 28.014/32.046 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent H (hydrogen) | 4 | 66.7% | 4.032/32.046 × 100% = 12.58% N (nitrogen) | 2 | 33.3% | 28.014/32.046 × 100% = 87.42%$

Find the elemental composition for diazane 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: NH_2NH_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 H (hydrogen) | 4 N (nitrogen) | 2 N_atoms = 4 + 2 = 6 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 H (hydrogen) | 4 | 4/6 N (nitrogen) | 2 | 2/6 Check: 4/6 + 2/6 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent H (hydrogen) | 4 | 4/6 × 100% = 66.7% N (nitrogen) | 2 | 2/6 × 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 H (hydrogen) | 4 | 66.7% | 1.008 N (nitrogen) | 2 | 33.3% | 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 H (hydrogen) | 4 | 66.7% | 1.008 | 4 × 1.008 = 4.032 N (nitrogen) | 2 | 33.3% | 14.007 | 2 × 14.007 = 28.014 m = 4.032 u + 28.014 u = 32.046 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 H (hydrogen) | 4 | 66.7% | 4.032/32.046 N (nitrogen) | 2 | 33.3% | 28.014/32.046 Check: 4.032/32.046 + 28.014/32.046 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent H (hydrogen) | 4 | 66.7% | 4.032/32.046 × 100% = 12.58% N (nitrogen) | 2 | 33.3% | 28.014/32.046 × 100% = 87.42%