Q136CP - Chemistry: Molecular Nature Of Matter And Change

Question

The active compound in Pepto-Bismol contains C, H, O, and Bi.

(a) When 0.22105 g of it was burned in excess O2, 0.1422 g of bismuth(III) oxide, 0.1880 g of carbon dioxide, and 0.02750 g of water were formed. What is the empirical formula of this compound?

(b) Given a molar mass of 1086 g/mol, determine the molecular formula.

(c) Complete and balance the acid-base reaction between bismuth(III) hydroxide and salicylic acid (HC7H5O3), which is used to form this compound.

(d) A dose of Pepto-Bismol contains 0.600 mg of the active ingredient. If the yield of the reaction in part (c) is 88.0%, what mass (in mg) of bismuth(III) hydroxide is required to prepare one dose?

Step-by-Step Solution

Answer

a)   The empirical formula of the compound is C₇H₅O₄Bi.

b)   The molecular formula is C₂₁H₁₅O₁₂Bi₃.

c)   The balanced reaction is,

      $\mathbf{3}{\mathbf{HC}}_{\mathbf{7}}{\mathbf{H}}_{\mathbf{5}}{\mathbf{O}}_{\mathbf{3}}\left(\mathrm{aq}\right)\mathbf{+}\mathbf{BI}{\left(\mathrm{OH}\right)}_{\mathbf{3}}\left(S\right)\mathbf{\to }\mathbf{Bi}{\left(C_7{H}_5{O}_3\right)}_{\mathbf{3}}\left(S\right)\mathbf{+}\mathbf{3}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(I\right)$

d)   The mass of bismuth(III) hydroxide is 0.490 mg.

Step 1: (a) Determination of empirical formula

The weight of the sample containing C, H, O and Bi is 0.22105 g.

 

The number of moles of CO₂ are,

 $$\begin{gathered} \mathbf{Moles}\mathbf{=}\frac{\mathbf{mass}}{\mathbf{molar}\mathbf{ }\mathbf{mass}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{1880}\mathbf{g}}{\mathbf{44}\mathbf{.}\mathbf{01}\mathbf{g}\mathbf{/}\mathbf{mol}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{00427}\mathbf{mol} \end{gathered}$$ 

There is 1 mol of C in 1 mol of CO₂. So, the number of moles of C is 0.00427 mol.

 

Mass of C is,

 $$\begin{gathered} \mathbf{Mass}\mathbf{ }\mathbf{=}\mathbf{moles}\mathbf{\times }\mathbf{molar}\mathbf{ }\mathbf{mass} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{00427}\mathbf{mol}\mathbf{\times }\mathbf{12}\mathbf{.}\mathbf{01}\mathbf{g}\mathbf{/}\mathbf{mol} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0513}\mathbf{g} \end{gathered}$$ 

The number of moles of Bi₂O₃ are,

$$\begin{gathered} \mathbf{Moles}\mathbf{=}\frac{\mathbf{mass}}{\mathbf{molar}\mathbf{ }\mathbf{mass}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{1422}\mathbf{g}}{\mathbf{446}\mathbf{.}\mathbf{0}\mathbf{g}\mathbf{/}\mathbf{mol}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{19}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{mol} \end{gathered}$$

There are 2 moles of Bi in 1 mol of Bi₂O₃. So, the number of moles of Bi are,

 $\mathbf{3}\mathbf{.}\mathbf{19}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{mol}\mathbf{\times }\frac{\mathbf{2}\mathbf{molBi}}{\mathbf{1}\mathbf{mol}\mathbf{ }{\mathbf{Bi}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{38}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{mol}$

The mass of Bi is,

 $$\begin{gathered} \mathbf{Mass}\mathbf{=}\mathbf{moles}\mathbf{\times }\mathbf{molar}\mathbf{ }\mathbf{mass} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{38}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{ }\mathbf{mol}\mathbf{\times }\mathbf{209}\mathbf{.}\mathbf{0}\mathbf{g}\mathbf{/}\mathbf{mol} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{133}\mathbf{g} \end{gathered}$$

The number of moles of H₂O are,

 $$\begin{gathered} \mathbf{Moles}\mathbf{=}\frac{\mathbf{mass}}{\mathbf{molar}\mathbf{ }\mathbf{mass}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{02750}\mathbf{g}}{\mathbf{18}\mathbf{.}\mathbf{02}\mathbf{g}\mathbf{/}\mathbf{mol}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{00153}\mathbf{mol} \end{gathered}$$

There are 2 moles of H in 1 mol of H₂O. So, the number of moles of H are,

$\mathbf{0}\mathbf{.}\mathbf{00153}\mathbf{mol}\mathbf{\times }\frac{\mathbf{2}\mathbf{mol}\mathbf{ }\mathbf{H}}{\mathbf{1}\mathbf{mol}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{00306}\mathbf{mol}$

The mass of H is,

$$\begin{gathered} \mathbf{Mass}\mathbf{=}\mathbf{moles}\mathbf{\times }\mathbf{molar}\mathbf{ }\mathbf{mass} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{00306}\mathbf{mol}\mathbf{\times }\mathbf{1}\mathbf{.}\mathbf{008}\mathbf{g}\mathbf{/}\mathbf{mol} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{00308}\mathbf{g} \end{gathered}$$ 

The mass of O is,

 $$\begin{gathered} \mathbf{Mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{O}\mathbf{=}\mathbf{Mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{sample}\mathbf{-}\left(\mathrm{mass} \mathrm{of} C+\mathrm{mass} \mathrm{of} H+\mathrm{mass} \mathrm{of} \mathrm{Bi}\right) \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{22105}\mathbf{g}\mathbf{-}\left(0.0513g+0.00308g+0.133g\right) \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{03367}\mathbf{g} \end{gathered}$$ 

Moles of O are,

$$\begin{gathered} \mathbf{Moles}\mathbf{=}\frac{\mathbf{mass}}{\mathbf{molar}\mathbf{ }\mathbf{mass}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{03367}\mathbf{g}}{\mathbf{16}\mathbf{.}\mathbf{0}\mathbf{g}\mathbf{/}\mathbf{mol}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{00210}\mathbf{mol} \end{gathered}$$

Now, divide each mole by the smallest number of moles,

$$\begin{gathered} \mathbf{C}\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{00427}\mathbf{mol}}{\mathbf{6}\mathbf{.}\mathbf{38}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{mol}}\mathbf{=}\mathbf{7} \\ \mathbf{O}\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{00210}\mathbf{mol}}{\mathbf{6}\mathbf{.}\mathbf{38}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{mol}}\mathbf{=}\mathbf{4} \\ \mathbf{Bi}\mathbf{=}\frac{\mathbf{6}\mathbf{.}\mathbf{38}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{mol}}{\mathbf{6}\mathbf{.}\mathbf{38}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{mol}}\mathbf{=}\mathbf{1} \\ \mathbf{H}\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{00306}\mathbf{mol}}{\mathbf{6}\mathbf{.}\mathbf{38}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{mol}}\mathbf{=}\mathbf{5} \end{gathered}$$ 

So, the formula is C₇H₅O₄Bi.

 

Thus, the empirical formula is C₇H₅O₄Bi.  

Step 2: (b) Determination of molecular formula

The molar mass of empirical formula is 362 g/mol.

Given molar mass is 1086 g/mol.

 

The smallest whole number is,

 $$\begin{gathered} \mathbf{n}\mathbf{=}\frac{\mathbf{molar}\mathbf{ }\mathbf{mass}}{\mathbf{empirical}\mathbf{ }\mathbf{mas}} \\ \mathbf{ }\mathbf{ }\mathbf{=}\frac{\mathbf{1086}\mathbf{g}\mathbf{/}\mathbf{mol}}{\mathbf{362}\mathbf{g}\mathbf{/}\mathbf{mol}} \\ \mathbf{ }\mathbf{ }\mathbf{=}\mathbf{3} \end{gathered}$$

Multiply the empirical formula by 3 to get the molecular formula.

 $$\begin{gathered} \mathbf{=}\mathbf{3}\left(C_7{H}_5{O}_4\mathrm{Bi}\right) \\ \mathbf{=}{\mathbf{C}}_{\mathbf{21}}{\mathbf{H}}_{\mathbf{15}}{\mathbf{O}}_{\mathbf{12}}{\mathbf{Bi}}_{\mathbf{3}} \end{gathered}$$

Thus, the molecular formula is C₂₁H₁₅O₁₂Bi₃.

Step 3: (c) Balanced reaction

 The balanced chemical reaction for the given process is,

$\mathbf{3}{\mathbf{HC}}_{\mathbf{7}}{\mathbf{H}}_{\mathbf{5}}{\mathbf{O}}_{\mathbf{3}}\left(\mathrm{aq}\right)\mathbf{+}\mathbf{Bi}{\left(\mathrm{OH}\right)}_{\mathbf{3}}\left(S\right)\mathbf{\to }\mathbf{Bi}{\left(C_7{H}_5{O}_3\right)}_{\mathbf{3}}\left(S\right)\mathbf{+}\mathbf{3}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(I\right)$

Step 4: (d) Determination of mass of Bismuth(III) hydroxide

The percent yield of the reaction is 88%.

 

Now, the mass of Bi(OH)₃ required to prepare 1 dose of the compound is,

$\mathbf{0}\mathbf{.}\mathbf{600}\mathbf{mg}\mathbf{\times }\left(\frac{{10}^{-3}g}{1\mathrm{mg}}\right)\mathbf{\times }\left(\frac{1\mathrm{mol} \mathrm{compound}}{1086g}\right)\mathbf{\times }\left(\frac{3\mathrm{mol} \mathrm{Bi}}{1\mathrm{mol} \mathrm{compound}}\right)\mathbf{\times }\left(\frac{100\%}{88\%}\right)\mathbf{\times }\left(\frac{1\mathrm{mol} \mathrm{Bi}{\left(\mathrm{OH}\right)}_2}{1\mathrm{mol} \mathrm{Bi}}\right)\mathbf{\times }\left(\frac{260.0\mathrm{gBi}{\left(\mathrm{OH}\right)}_3}{1\mathrm{mol} \mathrm{Bi}{\left(\mathrm{OH}\right)}_2}\right)\mathbf{\times }\left(\frac{1\mathrm{mg}}{{10}^{-3}g}\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{490}\mathbf{mg}$  

Thus, the mass of bismuth(III) hydroxide is 0.490 mg.