Chapter 5, Problem 1ASA

A student attempted to identify an unknown compound by the method used in this experiment. She found that when she heated a sample weighing $0.5032\text{ g}$, the mass dropped to $0.2663\text{ g}$. When the product was converted to a chloride, the mass went back up, $0.3747\text{ g}$.

a. Is the sample a carbonate?

Yes / no (Circle one)

Please provide reasoning below.

b. What are the two compounds that might be in the unknown?

$\underset{\_}{}$ or $\underset{\_}{}$

c. Write the balanced chemical equation for the overall reaction that occurs when each of these two original compounds is converted to a chloride. If the compound is a hydrogen carbonate, use the sum of Reactions $1$ and $2$. If the sample is a carbonate, use Reaction $2$. Write the equation for a sodium salt and then for a potassium salt.

d. How many moles of the chloride salt would be produced from one mole of the original compound?

$\underset{\_}{}$

e. How many grams of the chloride salt would be produced from one molar mass of the original compound?

$\begin{array}{cccc}\text{Molar masses}:& {\text{NaHCO}}_{\text{3}}\underset{\_}{}\text{g}/\text{mol}& {\text{Na}}_2{\text{CO}}_{\text{3}}\underset{\_}{}\text{g}/\text{mol}& \text{NaCl}\underset{\_}{}\text{g}/\text{mol}\\ & {\text{KHCO}}_{\text{3}}\underset{\_}{}\text{g}/\text{mol}& {\text{K}}_2{\text{CO}}_{\text{3}}\underset{\_}{}\text{g}/\text{mol}& \text{KCl}\underset{\_}{}\text{g}/\text{mol}\end{array}$

If a sodium salt, $\underset{\_}{}\text{g}$ original compound $\to$ $\underset{\_}{}\text{g}$ chloride

If a potassium salt, $\underset{\_}{}\text{g}$ original compound $\to$ $\underset{\_}{}\text{g}$ chloride

f. What is the theoretical value of $\text{Q}$, as found by Equation 3,

if she has the $\text{Na}$ salt? $\underset{\_}{}$ if she has the $\text{K}$ salt? $\underset{\_}{}$

g. What was the observed value of $\text{Q}$?

$\underset{\_}{}$

h. Which compound did she have as an unknown?

$\underset{\_}{}$

Interpretation Introduction

(a)

Interpretation:

Whether the unknown compound is a carbonate or not is to be stated. The explanation for the corresponding answer is to be stated.

Concept introduction:

Ionic compounds are made of cation and ions that are held together by the strong electrostatic force of attraction. Ionic compounds are classified in two categories binary ionic compound and ternary ionic compound on the basis of elements present in them. The total mass of the compound can be calculated by the sum of masses of individual elements present in it.

Answer to Problem 1ASA

The unknown compound is not carbonated salt as the loss and gain of the mass of the original compound indicates that this compound is hydrogen carbonate salt.

Explanation of Solution

The mass of the unknown compound is $0.5032\text{ g}$.

The mass of the unknown compound after heating is $0.2663\text{ g}$.

The mass of the decomposition product of unknown compound after converting into the chloride is $0.3747\text{ g}$.

The hydrogen carbonate and carbonate compound on heating produce carbon dioxide gas. The decomposition reaction of hydrogen carbonate is shown below.

$\begin{array}{l}{\text{2MHCO}}_{\text{3}}\left(\text{s}\right)\to {\text{M}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{s}\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)+{\text{CO}}_{\text{2}}\left(\text{g}\right)\\ {\text{M}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{s}\right)\to {\text{M}}_{\text{2}}\text{O}\left(\text{s}\right)+{\text{CO}}_{\text{2}}\left(\text{g}\right)\end{array}$

The hydrogen carbonates and carbonates lose masses on heating.

The reaction of carbonate with hydrochloric acid is shown below.

${\text{M}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{s}\right)+2{\text{H}}^{+}\left(\text{aq}\right)+2{\text{Cl}}^{-}\left(\text{aq}\right)\to 2\text{MCl}\left(\text{s}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+{\text{CO}}_{\text{2}}\left(\text{g}\right)$

The reaction of oxide with hydrochloric acid is shown below.

${\text{M}}_{\text{2}}\text{O}\left(\text{s}\right)+2{\text{H}}^{+}\left(\text{aq}\right)+2{\text{Cl}}^{-}\left(\text{aq}\right)\to 2\text{MCl}\left(\text{s}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

The mass of the chloride is more than the mass of the oxide. One mole of hydrogen carbonate salt produces one mole of chloride salt but one mole of carbonate salt produces two moles of chloride salt. The overall decreased mass from the original compound to chloride salt is justified by hydrogen carbonate salt. Therefore, the compound is a hydrogen carbonate.

Conclusion

The unknown compound has hydrogen carbonate ion in it.

Interpretation Introduction

(b)

Interpretation:

The two compounds that could be present in the unknown sample are to be stated.

Concept introduction:

Ionic compounds are made of cation and ions that are held together by the strong electrostatic force of attraction. Ionic compounds are classified in two categories binary ionic compound and ternary ionic compound on the basis of elements present in them. The total mass of the compound can be calculated by the sum of masses of individual elements present in it.

Answer to Problem 1ASA

Potassium hydrogen carbonate and sodium hydrogen carbonate can be present in the unknown sample.

Explanation of Solution

The unknown compound is a hydrogen carbonate salt.

The hydrogen carbonates of alkali metal are very reactive in nature. The two alkali metals are potassium and sodium. The decomposition reaction of potassium hydrogen carbonate and sodium hydrogen carbonate are shown below.

$\begin{array}{l}{\text{2NaHCO}}_{\text{3}}\left(\text{s}\right)\to {\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{s}\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)+{\text{CO}}_{\text{2}}\left(\text{g}\right)\\ {\text{2KHCO}}_{\text{3}}\left(\text{s}\right)\to {\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{s}\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)+{\text{CO}}_{\text{2}}\left(\text{g}\right)\end{array}$

Therefore, the unknown sample can be ${\text{NaHCO}}_{\text{3}}$ or ${\text{KHCO}}_{\text{3}}$.

Conclusion

The two compounds that could be present in the unknown sample are ${\text{KHCO}}_{\text{3}}$ and ${\text{NaHCO}}_{\text{3}}$.

Interpretation Introduction

(c)

Interpretation:

The chemical equations for the conversion reaction of hydrogen carbonate of sodium and potassium into their corresponding chloride are to be stated.

Concept introduction:

A chemical reaction is a process in which rearrangement of atom or ions takes place between two reacting species. A balanced chemical equation represents an equation in which all the reactants and products are written with their stoichiometric coefficient and physical states. The number of atoms of an element on both sides of the equation is equal.

Answer to Problem 1ASA

The chemical equations for the conversion reaction of carbonate of sodium and potassium into their corresponding chloride are shown below.

$\begin{array}{l}{\text{2NaHCO}}_{\text{3}}\left(\text{s}\right)\to {\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{s}\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)+{\text{CO}}_{\text{2}}\left(\text{g}\right)\\ {\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{s}\right)+2{\text{H}}^{+}\left(\text{aq}\right)+2{\text{Cl}}^{-}\left(\text{aq}\right)\to 2\text{NaCl}\left(\text{s}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+{\text{CO}}_{\text{2}}\left(\text{g}\right)\\ {\text{2KHCO}}_{\text{3}}\left(\text{s}\right)\to {\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{s}\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)+{\text{CO}}_{\text{2}}\left(\text{g}\right)\\ {\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{s}\right)+2{\text{H}}^{+}\left(\text{aq}\right)+2{\text{Cl}}^{-}\left(\text{aq}\right)\to 2\text{KCl}\left(\text{s}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+{\text{CO}}_{\text{2}}\left(\text{g}\right)\end{array}$

Explanation of Solution

The general reaction sequence of the conversion reactions of hydrogen carbonate salt into chloride salt is shown below.

$\begin{array}{l}{\text{2MHCO}}_{\text{3}}\left(\text{s}\right)\to {\text{M}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{s}\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)+{\text{CO}}_{\text{2}}\left(\text{g}\right)\\ {\text{M}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{s}\right)+2{\text{H}}^{+}\left(\text{aq}\right)+2{\text{Cl}}^{-}\left(\text{aq}\right)\to 2\text{MCl}\left(\text{s}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+{\text{CO}}_{\text{2}}\left(\text{g}\right)\end{array}$

The conversion reactions of sodium hydrogen carbonate into sodium chloride is shown below.

$\begin{array}{l}{\text{2NaHCO}}_{\text{3}}\left(\text{s}\right)\to {\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{s}\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)+{\text{CO}}_{\text{2}}\left(\text{g}\right)\\ {\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{s}\right)+2{\text{H}}^{+}\left(\text{aq}\right)+2{\text{Cl}}^{-}\left(\text{aq}\right)\to 2\text{NaCl}\left(\text{s}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+{\text{CO}}_{\text{2}}\left(\text{g}\right)\end{array}$

The conversion reactions of potassium hydrogen carbonate into potassium chloride salt is shown below.

$\begin{array}{l}{\text{2KHCO}}_{\text{3}}\left(\text{s}\right)\to {\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{s}\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)+{\text{CO}}_{\text{2}}\left(\text{g}\right)\\ {\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{s}\right)+2{\text{H}}^{+}\left(\text{aq}\right)+2{\text{Cl}}^{-}\left(\text{aq}\right)\to 2\text{KCl}\left(\text{s}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+{\text{CO}}_{\text{2}}\left(\text{g}\right)\end{array}$

The numbers of atoms of each element on the left side of the reactions are equal to the right side of the reactions. Therefore, the reactions are balanced.

Conclusion

The balanced reaction of for conversion of sodium hydrogen carbonate into chloride salt is shown below.

$\begin{array}{l}{\text{2NaHCO}}_{\text{3}}\left(\text{s}\right)\to {\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{s}\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)+{\text{CO}}_{\text{2}}\left(\text{g}\right)\\ {\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{s}\right)+2{\text{H}}^{+}\left(\text{aq}\right)+2{\text{Cl}}^{-}\left(\text{aq}\right)\to 2\text{NaCl}\left(\text{s}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+{\text{CO}}_{\text{2}}\left(\text{g}\right)\end{array}$

The balanced reaction of for conversion of potassium hydrogen carbonate into chloride salt is shown below.

$\begin{array}{l}{\text{2KHCO}}_{\text{3}}\left(\text{s}\right)\to {\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{s}\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)+{\text{CO}}_{\text{2}}\left(\text{g}\right)\\ {\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{s}\right)+2{\text{H}}^{+}\left(\text{aq}\right)+2{\text{Cl}}^{-}\left(\text{aq}\right)\to 2\text{KCl}\left(\text{s}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+{\text{CO}}_{\text{2}}\left(\text{g}\right)\end{array}$

Interpretation Introduction

(d)

Interpretation:

The number of moles of chloride salt produced by one mole of metal hydrogen carbonate is to be stated.

Concept introduction:

A chemical reaction is a process in which rearrangement of atom or ions takes place between two reacting species. A balanced chemical equation represents an equation in which all the reactants and products are written with their stoichiometric coefficient and physical states. The number of atoms of an element on both sides of the equation is equal.

Answer to Problem 1ASA

The number of moles of chloride salt produced by one mole of metal hydrogen carbonate is $1\text{mol}$

Explanation of Solution

The general reaction sequence of the conversion reactions of hydrogen carbonate into chloride salt is shown below.

$\begin{array}{l}{\text{2MHCO}}_{\text{3}}\left(\text{s}\right)\to {\text{M}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{s}\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)+{\text{CO}}_{\text{2}}\left(\text{g}\right)\\ {\text{M}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{s}\right)+2{\text{H}}^{+}\left(\text{aq}\right)+2{\text{Cl}}^{-}\left(\text{aq}\right)\to 2\text{MCl}\left(\text{s}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+{\text{CO}}_{\text{2}}\left(\text{g}\right)\end{array}$

Therefore, one mole of hydrogen carbonate salt produces one mole of chloride salt.

Conclusion

One mole of chloride salts is produced by one mole of hydrogen carbonate salt.

Interpretation Introduction

(e)

Interpretation:

The mass of chloride salt of sodium and potassium produced by one mole of their corresponding hydrogen carbonate salt is to be stated.

Concept introduction:

A chemical reaction is a process in which rearrangement of atom or ions takes place between two reacting species. A balanced chemical equation represents an equation in which all the reactants and products are written with their stoichiometric coefficient and physical states. The number of atoms of an element on both sides of the equation is equal.

Answer to Problem 1ASA

The mass of chloride salt of sodium and potassium produced by one mole of their corresponding hydrogen carbonate salt are $58.44\text{g}$ and $74.55\text{g}$ respectively.

Explanation of Solution

The balanced reaction of for conversion of sodium hydrogen carbonate into chloride salt is shown below.

$\begin{array}{l}{\text{2NaHCO}}_{\text{3}}\left(\text{s}\right)\to {\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{s}\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)+{\text{CO}}_{\text{2}}\left(\text{g}\right)\\ {\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{s}\right)+2{\text{H}}^{+}\left(\text{aq}\right)+2{\text{Cl}}^{-}\left(\text{aq}\right)\to 2\text{NaCl}\left(\text{s}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+{\text{CO}}_{\text{2}}\left(\text{g}\right)\end{array}$

The balanced reaction of for conversion of potassium hydrogen carbonate into chloride salt is shown below.

$\begin{array}{l}{\text{2KHCO}}_{\text{3}}\left(\text{s}\right)\to {\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{s}\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)+{\text{CO}}_{\text{2}}\left(\text{g}\right)\\ {\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{s}\right)+2{\text{H}}^{+}\left(\text{aq}\right)+2{\text{Cl}}^{-}\left(\text{aq}\right)\to 2\text{KCl}\left(\text{s}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+{\text{CO}}_{\text{2}}\left(\text{g}\right)\end{array}$

The molar mass of $\text{NaCl}$ is $58.44\text{g}/\text{mol}$.

The molar mass of $\text{KCl}$ is $74.55\text{g}/\text{mol}$.

One mole of chloride salts is produced by one mole of hydrogen carbonate salt.

The relation between the number of moles and molar mass of a substance can be given by the formula as shown below.

$m=nM$ ……(1)

Where,

• $m$ is the mass of the substance.

• $n$ is the molar mass of the substance.

• $M$ is the molar mass of the substance.

Substitute the value of the number of moles of sodium chloride and molar mass of the sodium chloride in the equation (1).

$\begin{array}{rll}m& =& \left(1\text{mol}\right)\left(58.44\text{g}/\text{mol}\right)\\ & =& 58.44\text{g}\end{array}$

Therefore, one mole of sodium carbonate will produce $58.44\text{g}$ of sodium chloride.

Substitute the value of the number of moles of potassium chloride and molar mass of the potassium chloride in the equation (1).

$\begin{array}{rll}m& =& \left(1\text{mol}\right)\left(74.55\text{g}/\text{mol}\right)\\ & =& 74.55\text{g}\end{array}$

Therefore, one mole of potassium carbonate will produce $74.55\text{g}$ of potassium chloride.

Conclusion

The mass of sodium chloride produced by one mole of sodium hydrogen carbonate is $58.44\text{g}$. The mass of sodium chloride produced by one mole of sodium hydrogen carbonate is $74.55\text{g}$.

Interpretation Introduction

(f)

Interpretation:

The values of $\text{Q}$ for ${\text{NaHCO}}_{\text{3}}$ and ${\text{KHCO}}_{\text{3}}$ are to be calculated.

Concept introduction:

Ionic compounds are made of cation and ions that are held together by the strong electrostatic force of attraction. The value of $\text{Q}$ of a compound is used to identify the molecular formula of the compound. The value of $\text{Q}$ is calculated by the formula as shown below.

$\text{Q}=\frac{\text{mass}\text{of}\text{chloride}-\text{original}\text{mass}}{\text{original}\text{mass}}$

Answer to Problem 1ASA

The values of $\text{Q}$ for ${\text{NaHCO}}_{\text{3}}$ and ${\text{KHCO}}_{\text{3}}$ are $-0.2412$ and $0.2554\text{g}$ respectively.

Explanation of Solution

The molar mass of one mole of ${\text{NaHCO}}_{\text{3}}$ is $84.007\text{g}$.

The molar mass of one mole of ${\text{KHCO}}_{\text{3}}$ is $100.115\text{g}$.

The molar mass one mole of $\text{NaCl}$ is $58.44\text{g}$.

The molar mass one mole of $\text{KCl}$ is $74.55\text{g}$.

The value of $\text{Q}$ is calculated by the formula as shown below.

$\text{Q}=\frac{\text{mass}\text{of}\text{chloride}-\text{original}\text{mass}}{\text{original}\text{mass}}$ ……(2)

Substitute the values of the mass sodium chloride and mass of sodium hydrogen carbonate in the equation (2).

$\begin{array}{rll}\text{Q}& =& \frac{58.44\text{g}-84.007\text{g}}{84.007\text{g}}\\ & =& \frac{-25.567}{105.98}\\ & =& -0.2412\end{array}$

Therefore, the value of $\text{Q}$ for ${\text{NaHCO}}_{\text{3}}$ is $-0.2412$.

Substitute the values of the mass potassium chloride and mass of potassium hydrogen carbonate in the equation (2).

$\begin{array}{rll}\text{Q}& =& \frac{74.55\text{g}-100.115\text{g}}{100.115\text{g}}\\ & =& \frac{-25.565}{100.115}\\ & =& 0.2554\text{g}\end{array}$

Therefore, the theoretical value of $\text{Q}$ for ${\text{KHCO}}_{\text{3}}$ is $0.2554\text{g}$.

Conclusion

The values of $\text{Q}$ for ${\text{NaHCO}}_{\text{3}}$ and ${\text{KHCO}}_{\text{3}}$ are calculated as $-0.2412$ and $0.2554\text{g}$ respectively.

Interpretation Introduction

(g)

Interpretation:

The theoretical value of $\text{Q}$ is to be calculated.

Concept introduction:

Ionic compounds are made of cation and ions that are held together by the strong electrostatic force of attraction. The value of $\text{Q}$ of a compound is used to identify the molecular formula of the compound. The value of $\text{Q}$ is calculated by the formula as shown below.

$\text{Q}=\frac{\text{mass}\text{of}\text{chloride}-\text{original}\text{mass}}{\text{original}\text{mass}}$

Answer to Problem 1ASA

The theoretical value of $\text{Q}$ is $-0.2554$.

Explanation of Solution

The mass of the unknown compound is $0.5032\text{ g}$.

The mass of the unknown compound after heating is $0.2663\text{ g}$.

The mass of the decomposition product of unknown compound after converting into the chloride is $0.3747\text{ g}$.

The value of $\text{Q}$ is calculated by the formula as shown below.

$\text{Q}=\frac{\text{mass}\text{of}\text{chloride}-\text{original}\text{mass}}{\text{original}\text{mass}}$

Substitute the values of mass unknown compound and mass of chloride in the above equation.

$\begin{array}{rll}\text{Q}& =& \frac{0.3747\text{ g}-0.5032\text{ g}}{0.5032\text{ g}}\\ & =& \frac{-0.1285}{0.5032}\\ & =& -0.2554\end{array}$

Therefore, the theoretical value of $\text{Q}$ is $-0.2554$.

Conclusion

The theoretical value of $\text{Q}$ is calculated as $-0.2554$.

Interpretation Introduction

(h)

Interpretation:

The unknown compound has to be identified.

Concept introduction:

Ionic compounds are made of cation and ions that are held together by the strong electrostatic force of attraction. The value of $\text{Q}$ of a compound is used to identify the molecular formula of the compound. The value of $\text{Q}$ is calculated by the formula as shown below.

$\text{Q}=\frac{\text{mass}\text{of}\text{chloride}-\text{original}\text{mass}}{\text{original}\text{mass}}$

Answer to Problem 1ASA

The unknown compound is potassium hydrogen carbonate.

Explanation of Solution

The value of $\text{Q}$ for ${\text{NaHCO}}_{\text{3}}$ is $-0.2412$.

The value of $\text{Q}$ for ${\text{KHCO}}_{\text{3}}$ is $0.2554\text{g}$.

The theoretical value of $\text{Q}$ is $-0.2554$.

Therefore, the unknown compound is ${\text{KHCO}}_{\text{3}}$.

Conclusion

The unknown compound is ${\text{KHCO}}_{\text{3}}$.