Chapter 4, Problem 4.149P

(a)

Interpretation Introduction

Interpretation:

The empirical formula of the compound is to be determined.

Concept introduction:

An empirical formula gives the simplest whole number ratio of atoms of each element present in a molecule. The molecular formula tells the exact number of atoms of each element present in a molecule.

Following are the steps to determine the empirical formula of a compound when the masses of ${\text{CO}}_2$ and ${\text{H}}_2\text{O}$ are given,

Step 1: One mole of ${\text{CO}}_2$ gives one mole of carbon $\left(\text{C}\right)$. Calculate the mass and moles of carbon in the original sample by using the mass of ${\text{CO}}_2$ and molar masses of $\text{C}$ and ${\text{CO}}_2$. One mole of ${\text{H}}_2\text{O}$ gives two moles of hydrogen $\left(\text{H}\right)$. Calculate the moles and mass of hydrogen in the original sample by using the mass of ${\text{H}}_2\text{O}$ and molar masses of $\text{H}$ and ${\text{H}}_2\text{O}$. The formula to calculate the moles of carbon in the sample when moles of ${\text{CO}}_2$ is given is as follows:

$\text{Amount of C}\left(\text{mol}\right)=\left({\text{mass of CO}}_2\left(\text{g}\right)\right)\left(\frac{{\text{1 mol CO}}_2}{{\text{Mass of 1 mol CO}}_2\left(\text{g}\right)}\right)\left(\frac{1\text{ mol C}}{1{\text{ mol CO}}_2}\right)$        (1)

The formula to calculate the mass of carbon in the sample is as follows:

$\text{Mass of C}\left(\text{g}\right)=\left(\text{Amount of C}\left(\text{mol}\right)\right)\left(\begin{array}{l}\text{Molar mass of }\\ \text{C}\end{array}\right)$        (2)

The formula to calculate the moles of hydrogen in the sample when the mass of ${\text{H}}_2\text{O}$ is given is as follows:

$\text{Amount of H}\left(\text{mol}\right)=\left({\text{mass of H}}_2\text{O}\left(\text{g}\right)\right)\left(\frac{{\text{1 mol H}}_2\text{O }}{\begin{array}{l}\text{Mass of 1 mol}\\ {\text{H}}_2\text{O}\left(\text{g}\right)\end{array}}\right)\left(\frac{2\text{ mol H}}{1{\text{ mol H}}_2\text{O}}\right)$        (3)

The formula to calculate the mass of hydrogen in the sample is as follows:

$\text{Mass of H}\left(\text{g}\right)=\left(\text{Amount of H}\left(\text{mol}\right)\right)\left(\begin{array}{l}\text{Molar mass of }\\ \text{H}\end{array}\right)$        (4)

Step 2: If the sample contains any other element X, then calculate the mass of element X as follows:

$\text{Mass of element X}\left(\text{g}\right)=\text{Mass of sample}\left(\text{g}\right)-\left(\text{mass of C}+\text{mass of H}\right)$        (5)

Step 3: Divide mass of element X by its molar mass to convert the mass to moles. The formula to calculate moles from the mass is as follows:

$\text{Amount}\left(\text{mol}\right)=\left(\text{Given mass}\left(\text{g}\right)\right)\left(\frac{1\text{ mol}}{\text{No}\text{. of grams}}\right)$        (6)

Step 4: The number of moles of the elements is the fractional amounts, thus, write the calculated amount $\left(\text{mol}\right)$ of each element as the subscript of the element’s symbol to obtain a preliminary formula for the compound.

Step 5: Convert the moles of each element to the whole number subscripts. The steps for this math conversion are as follows:

(a) Each subscript is divided by the smallest subscript.

(b) If the whole number is not obtained after division, multiply the obtained subscripts by the smallest integer. This gives the empirical formula of the compound.

Answer to Problem 4.149P

The empirical formula of the compound is ${\text{C}}_7{\text{H}}_5{\text{O}}_4\text{Bi}$.

Explanation of Solution

Substitute $0.1880\text{g}$ for the mass of ${\text{CO}}_2$ and $44.01\text{ g}$ for the mass of ${\text{1 mol CO}}_2$ in the equation (1).

$\begin{array}{c}\text{Amount of C}\left(\text{mol}\right)=\left(0.1880\text{g}\right)\left(\frac{{\text{1 mol CO}}_2}{\text{44}{\text{.01 g CO}}_2}\right)\left(\frac{1\text{ mol C}}{1{\text{ mol CO}}_2}\right)\\ =4.271756\times {10}^{-3}\text{ mol}\end{array}$

Substitute $4.271756\times {10}^{-3}\text{ mol}$ for the amount of $\text{C}$ in the equation (2) to calculate the mass of carbon.

$\begin{array}{c}\text{Mass of C}\left(\text{g}\right)=\left(4.271756\times {10}^{-3}\text{ mol}\right)\left(\frac{12.01\text{ g C}}{1\text{ mol C}}\right)\\ =0.0513038\text{ g}\end{array}$

Substitute $0.02750\text{ g}$ for the mass of ${\text{H}}_2\text{O}$, $18.02\text{ g}$ for the mass of ${\text{1 mol H}}_2\text{O}$ in the equation (3).

$\begin{array}{c}\text{Amount of H}\left(\text{g}\right)=\left(0.02750\text{ g}\right)\left(\frac{{\text{1 mol H}}_2\text{O }}{\text{18}{\text{.2 g H}}_2\text{O}}\right)\left(\frac{2\text{ mol H}}{1{\text{ mol H}}_2\text{O}}\right)\\ =3.052164\times {10}^{-3}\text{ mol}\end{array}$

Substitute $3.052164\times {10}^{-3}\text{ mol}$ for the amount of $\text{H}$ in the equation (4) to calculate the mass of hydrogen.

$\begin{array}{c}\text{Mass of H}\left(\text{g}\right)=\left(3.052164\times {10}^{-3}\text{ mol}\right)\left(\frac{1.008\text{ g H}}{1\text{ mol H}}\right)\\ =0.0030766\text{g}\end{array}$

The formula to calculate the moles of bismuth in the sample when moles of ${\text{Bi}}_{\text{2}}{\text{O}}_{\text{3}}$ is given is as follows:

$\text{Amount of Bi}\left(\text{mol}\right)=\left({\text{mass of Bi}}_{\text{2}}{\text{O}}_{\text{3}}\right)\left(\frac{{\text{1 mol Bi}}_{\text{2}}{\text{O}}_{\text{3}}}{{\text{Mass of 1 mol Bi}}_{\text{2}}{\text{O}}_{\text{3}}}\right)\left(\frac{\text{2 mol B}}{1{\text{ mol Bi}}_{\text{2}}{\text{O}}_{\text{3}}}\right)$        (7)

Substitute $0.1422\text{g}$ for the mass of ${\text{Bi}}_{\text{2}}{\text{O}}_{\text{3}}$ and $466.0\text{ g}$ for the mass of ${\text{1 mol Bi}}_{\text{2}}{\text{O}}_{\text{3}}$ in the equation (1).

$\begin{array}{c}\text{Amount of Bi}\left(\text{mol}\right)=\left(0.1422\text{g}\right)\left(\frac{{\text{1 mol Bi}}_{\text{2}}{\text{O}}_{\text{3}}}{{\text{Mass of 1 mol Bi}}_{\text{2}}{\text{O}}_{\text{3}}}\right)\left(\frac{\text{2 mol B}}{1{\text{ mol Bi}}_{\text{2}}{\text{O}}_{\text{3}}}\right)\\ =6.103004\times {10}^{-4}\text{ mol}\end{array}$

The formula to calculate the mass of bismuth in the sample is as follows:

$\text{Mass of Bi}\left(\text{g}\right)=\left(\text{Amount of Bi}\right)\left(\begin{array}{l}\text{Molar mass of }\\ \text{Bi}\end{array}\right)$        (8)

Substitute $6.103004\times {10}^{-4}\text{ mol}$ for the amount of $\text{Bi}$ and $209.0\text{ g/mol}$ for molar mass of $\text{Bi}$ in the equation (8) to calculate the mass of bismuth as follows:

$\begin{array}{c}\text{Mass of Bi}\left(\text{g}\right)=\left(6.103004\times {10}^{-4}\text{ mol}\right)\left(209.0\text{ g/mol}\right)\\ =0.127553\text{ g}\end{array}$

The formula to calculate the mass of oxygen $\left(\text{O}\right)$ is as follows:

$\text{Mass of O}\left(\text{g}\right)=\text{Mass of sample}\left(\text{g}\right)-\left(\text{mass of C}+\text{mass of H}+\text{mass of Bi}\right)$        (9)

Substitute $0.22105\text{ g}$ for the mass of the sample, $0.0513038\text{ g}$ for the mass of $\text{C}$, $0.127553\text{ g}$ for the mass of $\text{Bi}$ and $0.0030766\text{g}$ for the mass of $\text{H}$ in equation (9). $\begin{array}{c}\text{Mass of O}\left(\text{g}\right)=0.22105\text{ g}-\left(0.0513038\text{ g}+0.0030766\text{g}+0.127553\text{ g}\right)\\ =0.0391166\text{ g}\end{array}$

The molar mass of $\text{O}$ is $16.00\text{ g/mol}$. Calculate the amount of $\text{O}$ as follows:

$\begin{array}{c}\text{Amount of O}\left(\text{mol}\right)=\left(0.0391166\text{ g}\right)\left(\frac{1\text{ mol O}}{16.00\text{ g O}}\right)\\ =2.44482\times {10}^{-4}\text{ mol}\end{array}$

Write the amount of carbon, hydrogen, bismuth, and oxygen as subscripts of their symbols to obtain a preliminary formula as follows:

${\text{C}}_{4.271756\times {10}^{-3}}{\text{H}}_{3.052164\times {10}^{-3}}{\text{O}}_{2.44482\times {10}^{-4}}{\text{Bi}}_{6.103004\times {10}^{-4}}$

The smallest subscript is $6.103004\times {10}^{-4}$. Therefore, divide each subscript by $6.103004\times {10}^{-4}$ as follows:

${\text{C}}_{\frac{4.271756\times {10}^{-3}}{6.103004\times {10}^{-4}}}{\text{H}}_{\frac{3.052164\times {10}^{-3}}{6.103004\times {10}^{-4}}}{\text{O}}_{\frac{2.44482\times {10}^{-4}}{6.103004\times {10}^{-4}}}{\text{Bi}}_{\frac{6.103004\times {10}^{-4}}{6.103004\times {10}^{-4}}}\to {\text{C}}_7{\text{H}}_5{\text{O}}_4\text{Bi}$

The subscripts are in the whole number. Hence, the empirical formula of the compound is ${\text{C}}_7{\text{H}}_5{\text{O}}_4\text{Bi}$.

Conclusion

The empirical formula of the compound is ${\text{C}}_7{\text{H}}_5{\text{O}}_4\text{Bi}$.

(b)

Interpretation Introduction

Interpretation:

The molecular formula of the compound is to be determined.

Concept introduction:

An empirical formula gives the simplest whole number ratio of atoms of each element present in a molecule. The molecular formula tells the exact number of atoms of each element present in a molecule.

Answer to Problem 4.149P

The molecular formula of the compound is ${\text{C}}_{21}{\text{H}}_{15}{\text{O}}_{12}{\text{Bi}}_3$.

Explanation of Solution

The formula to calculate the empirical formula unit is as follows:

$\text{Empirical formula unit}=\left(\frac{\text{Molar mass}}{\text{Empirical}\text{mass}}\right)$        (10)

Substitute $1086\mathrm{g}/mol$ for molar mass and $362\text{ g/mol}$ for empirical mass in the equation (10)

$\begin{array}{c}\text{Empirical formula unit}=\left(\frac{\text{1086 g/mol}}{362\text{ g/mol}}\right)\\ =3\end{array}$

There are 3 empirical formula unit and therefore the molecular mass is ${\text{C}}_{21}{\text{H}}_{15}{\text{O}}_{12}{\text{Bi}}_3$

Conclusion

The molecular formula of the compound is ${\text{C}}_{21}{\text{H}}_{15}{\text{O}}_{12}{\text{Bi}}_3$.

(c)

Interpretation Introduction

Interpretation:

The balanced acid-base reaction between $\text{bismuth}\left(\text{III}\right)\text{hydroxide}$ and salicylic acid is to be written.

Concept introduction:

Strong acids and strong bases are the substance that dissociates completely into its ions when dissolved in the solution. They dissociate completely in water to release ${\text{H}}^{+}$ ions and ${\text{OH}}^{-}$ ions.

Weak acids and weak bases are the substance that does not dissociate completely into its ions when dissolved in the solution. They dissociate partially in water to release ${\text{H}}^{+}$ ions and ${\text{OH}}^{-}$ ions.

The driving force of the acid-base reaction is the formation of a gaseous product or precipitate in the reaction. The formation of a water molecule also acts as a factor to drive the reaction to completion.

Answer to Problem 4.149P

The neutralization reaction of $\text{bismuth}\left(\text{III}\right)\text{hydroxide}$ and salicylic acid is as follows:

$\text{Bi}{\left(\text{OH}\right)}_3\left(s\right)+3{\text{HC}}_7{\text{H}}_5{\text{O}}_3\left(aq\right)\to \text{Bi}{\left({\text{C}}_7{\text{H}}_5{\text{O}}_3\right)}_3\left(s\right)+3{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Explanation of Solution

$\text{Bismuth}\left(\text{III}\right)\text{hydroxide}$ is a base and salicylic acid is an acid. Both react to form salt and water. The neutralization reaction of $\text{bismuth}\left(\text{III}\right)\text{hydroxide}$ and salicylic acid is as follows:

$\text{Bi}{\left(\text{OH}\right)}_3\left(s\right)+3{\text{HC}}_7{\text{H}}_5{\text{O}}_3\left(aq\right)\to \text{Bi}{\left({\text{C}}_7{\text{H}}_5{\text{O}}_3\right)}_3\left(s\right)+3{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Conclusion

The driving force of the acid-base reaction is the formation of a gaseous product or precipitate in the reaction. The formation of a water molecule also acts as a factor to drive the reaction to completion.

(d)

Interpretation Introduction

Interpretation:

The mass (in mg) of $\text{bismuth}\left(\text{III}\right)\text{hydroxide}$ required to prepare one dose is to be calculated.

Concept introduction:

Stoichiometry of a reaction is utilized to determine the amount of any species in the reaction by the relationship between the reactants and products.

Consider the general reaction,

$\text{A}+2\text{B}\to 3\text{C}$

One mole of $\text{A}$ reacts with two moles of $\text{B}$ to produce three moles of $\text{C}$. The stoichiometric ratio between A and B is 1:2, the stoichiometric ratio between A and C is 1:3 and the stoichiometric ratio between B and C is 2:3.

Answer to Problem 4.149P

The mass (in mg) of $\text{bismuth}\left(\text{III}\right)\text{hydroxide}$ required to prepare one dose is $0.490\text{g}$.

Explanation of Solution

The expression to calculate the moles of ${\text{C}}_{21}{\text{H}}_{15}{\text{O}}_{12}{\text{Bi}}_3$ is as follows:

${\text{Moles of C}}_{21}{\text{H}}_{15}{\text{O}}_{12}{\text{Bi}}_3=\left[\frac{\text{given mass}{\text{of C}}_{21}{\text{H}}_{15}{\text{O}}_{12}{\text{Bi}}_3}{{\text{molecular mass of C}}_{21}{\text{H}}_{15}{\text{O}}_{12}{\text{Bi}}_3}\right]$        (11)

Substitute $0.600\text{ mg}$ for the mass of ${\text{C}}_{21}{\text{H}}_{15}{\text{O}}_{12}{\text{Bi}}_3$ and $1086\text{ g/mol}$ for the molar mass of ${\text{C}}_{21}{\text{H}}_{15}{\text{O}}_{12}{\text{Bi}}_3$ in the equation (3).

$\begin{array}{c}{\text{Moles of C}}_{21}{\text{H}}_{15}{\text{O}}_{12}{\text{Bi}}_3=\left[\frac{0.600\text{ mg}}{1086\text{ g/mol}}\right]\left(\frac{{10}^{-3}\text{g}}{\text{1}\text{mg}}\right)\\ =5.52486\times {10}^{-4}\text{mol}\end{array}$

The formula to calculate the mass of $\text{Bi}{\left(\text{OH}\right)}_3$ is as follows:

$\text{Mass of}\text{Bi}{\left(\text{OH}\right)}_3=\left[\begin{array}{l}\text{moles of}{\text{C}}_{21}{\text{H}}_{15}{\text{O}}_{12}{\text{Bi}}_3\left(\frac{3\text{mol}\text{Bi}}{1\text{mol}{\text{C}}_{21}{\text{H}}_{15}{\text{O}}_{12}{\text{Bi}}_3}\right)\\ \left(\frac{1\text{mol}\text{Bi}{\left(\text{OH}\right)}_3}{1\text{mol}\text{Bi}}\right)\left(\text{molar mass of Bi}{\left(\text{OH}\right)}_3\right)\left(\frac{100\%}{88.0\%}\right)\end{array}\right]$        (12)

Substitute $5.52486\times {10}^{-4}\text{mol}$ for moles of ${\text{C}}_{21}{\text{H}}_{15}{\text{O}}_{12}{\text{Bi}}_3$ and $260.0\text{ g/mol}$ for molar mass of $\text{Bi}{\left(\text{OH}\right)}_3$ in the equation (12).

$\begin{array}{c}\text{Mass of}\text{Bi}{\left(\text{OH}\right)}_3=\left[\begin{array}{l}\left(5.52486\times {10}^{-4}\text{mol}\right)\left(\frac{3\text{mol}\text{Bi}}{1\text{mol}{\text{C}}_{21}{\text{H}}_{15}{\text{O}}_{12}{\text{Bi}}_3}\right)\\ \left(\frac{1\text{mol}\text{Bi}{\left(\text{OH}\right)}_3}{1\text{mol}\text{Bi}}\right)\left(260.0\text{ g/mol}\right)\left(\frac{100\%}{88.0\%}\right)\end{array}\right]\\ =0.48970\text{g}\\ \approx 0.490\text{g}\end{array}$

Conclusion

The mass (in mg) of $\text{bismuth}\left(\text{III}\right)\text{hydroxide}$ required to prepare one dose is $0.490\text{g}$.