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Answers to More Chapter 03 Study Questions

  • Page ID
    12604
      1. In 1 mole \(\ce{Na2SO4}\): \(\mathrm{2\: mole\: Na = 2(23.0\: g) = 46.0\: g}\); \(\mathrm{1\: mole\: S = 32.1\: g}\); \(\mathrm{4\: mol\: O = 4(16.0\: g) = 64.0\: g}\). \(\mathrm{Molar\: mass = 46.0 + 32.1 + 64.0 = 142.1\: g/mole}\)

    \(\ce{Na}\): \(\mathrm{\dfrac{46.0}{142.1} = 32.4\%\: Na}\). \(\ce{S}\): \(\mathrm{\dfrac{32.1}{142.1} = 22.6\%\: S}\). \(\ce{O}\): \(\mathrm{\dfrac{64.0}{142.1} = 45.0\%\: O}\).

    32.4% \(\ce{Na}\), 22.6% \(\ce{S}\), 45.0% \(\ce{O}\)

    1. \(\mathrm{2.00\: g\: S\times\dfrac{100\:g\:cpd}{22.6\:g\:S}=8.85\: g\: cpd}\)

    1. \(\mathrm{\textrm{Mass of Sn} = 33.40\: g - 31.50\: g = 1.90\: g\: Sn}\)

    \(\mathrm{\textrm{Mass of O} = 33.91\: g - 33.40\: g = 0.51\: g\: O}\)

    \(\mathrm{\#\:moles\:\textrm{Sn:}\:\:1.90\: g\: Sn \times\dfrac{1\:mole\:Sn}{118.7\:g\:Sn}=0.0160\: moles\: Sn}\) \(\mathrm{\dfrac{0.0160}{0.160}=1.0}\)

    \(\mathrm{\#\:moles\: \textrm{O:}\:\:0.51\: g\: O \times\dfrac{1\:mole\:O}{16.0\:g\:O}=0.032\: moles\: O}\) \(\mathrm{\dfrac{0.32}{0.160} =2.0}\)

    \(\mathrm{formula = SnO_2}\)

      1. In 100 g of this compound, there are 30.4 g \(\ce{N}\) and 69.6 g \(\ce{O}\)

    \(\mathrm{30.4\: g\: N\times\dfrac{1\:mole\:N}{14.0\:g\:N}=2.17\: moles\: N}\) \(\mathrm{\dfrac{2.17}{2.17} = 1}\)

    \(\mathrm{69.6\: g\: O\times\dfrac{1\:mole\:O}{16.0\:g\:O}=4.35\: moles\: O}\) \(\mathrm{\dfrac{4.35}{2.17}=2}\) \(\mathrm{empirical\: formula = NO_2}\)

    1. \(\mathrm{\textrm{Molar mass of }NO_2 = 14.0 + 2(16.0) = 46.0\: g/mole}\)

    \(\mathrm{\dfrac{92.0}{46.0}=2}\) \(\mathrm{\rightarrow molecular\: formula = N_2O_4}\)

      1. \(\mathrm{2 C_8H_{18}(l) + 25 O_2(g) \rightarrow 16 CO_2(g) + 18 H_2O(l)}\)
      2. \(\mathrm{1.00\: mol\: C_8H_{18}\times\dfrac{25\:mol\:O_2}{2\:mol\:C_8H_{18}}=12.5\:mol\: O_2}\)
      3. \(\mathrm{6.63\: mol\: H_2O\times\dfrac{2\:mol\:C_8H_{18}}{18\:mol\:H_2O}\times\dfrac{114\:g\:C_8H_{18}}{1\:mol\:C_8H_{18}}=84.0\: g\: C_8H_{18}}\)
      4. \(\mathrm{101\: g\: C_8H_{18}\times\dfrac{1\:mol\:C_8H_{18}}{114\:g\:C_8H_{18}}\times\dfrac{16\:mol\:CO_2}{2\:mol\:C_8H_{18}}=7.09\: mol\: CO_2}\)
      5. \(\mathrm{4.77\: g\: O_2\times\dfrac{1\:mol\:O_2}{32.0\:g\:O_2}\times\dfrac{16\:mol\:CO_2}{25\:mol\:O_2}\times\dfrac{44.0\:g\:CO_2}{1\:mol\:CO_2}=4.20\: g\: CO_2}\)
      6. \(\mathrm{2.1\: g\: C_8H_{18}\times\dfrac{1\:mol\:C_8H_{18}}{114\:g\:C_8H_{18}}\times\dfrac{18\:mol\:H_2O}{2\:mol\:C_8H_{18}}\times\dfrac{6.022\times10^{23}\:molecules}{1\:mol\:H_2O}=1.0 \times 10^{23}\: molecules}\)
      7. \(\mathrm{5.00\: g\: O_2\times\dfrac{1\:mol\:O_2}{32.0\:g\:O_2}\times\dfrac{16\:mol\:CO_2}{25\:mol\:O_2}\times\dfrac{44.0\:g\:CO_2}{1\:mol\:CO_2}=4.40\: g\: CO_2}\)

    \(\mathrm{1.62\: g\: C_8H_{18}\times\dfrac{1\:mol\:C_8H_{18}}{114\:g\:C_8H_{18}}\times\dfrac{16\:mol\:CO_2}{2\:mol\:C_8H_{18}}\times\dfrac{44.0\:g\:CO_2}{1\:mol\:CO_2}=5.00\: g\: CO_2}\)

    \(\ce{O2}\) is limiting; 4.40 g \(\ce{CO2}\) is produced

    1. \(\mathrm{1.62\: g\: C_8H_{18}\times\dfrac{1\:mol\:C_8H_{18}}{114\:g\:C_8H_{18}}\times\dfrac{16\:mol\:CO_2}{2\:mol\:C_8H_{18}}\times\dfrac{44.0\:g\:CO_2}{1\:mol\:CO_2}=5.00\: g\: CO_2}\)

    \(\mathrm{\%\: yield=\dfrac{actual\:yield}{theoretical\:yield}\times100\%=\dfrac{3.70\:g\:CO_2}{5.00\:g\:CO_2}\times100\% =74.0\%}\)

    1. In 100 g of this compound, there are 45.0 g \(\ce{Pb}\) and 55.0 g \(\ce{I}\)

    \(\mathrm{45.0\: g\: Pb\times\dfrac{1\:mol\:Pb}{207\:g\:Pb}=0.217\: moles\: Pb}\) \(\mathrm{\dfrac{0.217}{0.217}= 1}\)

    \(\mathrm{55.0\: g\: I\times\dfrac{1\:mol\:I}{127\:g\:I} =0.433\: moles\: I}\) \(\mathrm{\dfrac{0.433}{0.217}=2}\)

    \(\mathrm{empirical\: formula = PbI_2}\) lead(II) iodide