Skip to main content
Chemistry LibreTexts

Answers to More Chapter 03 Study Questions

1. 

  1. In 1 mole Na2SO4: 2 mole Na = 2(23.0 g) = 46.0 g; 1 mole S = 32.1 g; 4 mol O = 4(16.0 g) = 64.0 g.        Molar mass = 46.0 + 32.1 + 64.0 = 142.1 g/mole

Na:  46.0/142.1  =  32.4% Na.  S:  32.1/142.1 = 22.6% S.  O:  64.0/142.1 = 45.0% O.

                        32.4% Na, 22.6% S, 45.0% O

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

 

2.  Mass of Sn = 33.40 g – 31.50 g = 1.90 g Sn

      Mass of O = 33.91 g – 33.40 g = 0.51 g O

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

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

                        formula   =   SnO2

 

3. 

  1. In 100 g of this compound, there are 30.4 g N and 69.6 g O

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

        \(\mathrm{69.6\: g\: O\times\dfrac{1\:mole\:O}{16.0\:g\:O}=4.35\: moles\: O}\)          \(\mathrm{\dfrac{4.35}{2.17}=2}\)       empirical formula  =  NO2

  1. Molar mass of NO2  =  14.0 +  2(16.0)   =  46.0 g/mole

            \(\mathrm{\dfrac{92.0}{46.0}=2}\)  →  molecular formula  = N2O4

 

4. 

  1. 2 C8H18(l)   +   25 O2(g)    →     16  CO2(g)   +    18  H2O(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. \(\mathrm{5.00\: g\: O_2\times\dfrac{1\:mol\:O_2}{32.0\:g\:O_2}\times\dfrac{16\:mol\:CO_2}{25\:mol\:O_2}\times\dfrac{44.0\:g\:CO_2}{1\:mol\:CO_2}=4.40\: g\: CO_2}\)
     

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

        O2 is limiting;  4.40 g CO2 is produced
 

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

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

 

5.  In 100 g of this compound, there are 45.0 g Pb and 55.0 g I

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

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

                        empirical formula  =  PbI2       lead(II) iodide