# Answers to More Chapter 03 Study Questions

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