Problem 33 - Chemistry and Chemical Reactivity

Question

A sample of limestone and other soil materials is heated, and the limestone decomposes to give calcium oxide and carbon dioxide. $$ \mathrm{CaCO}_{3}(\mathrm{s}) \longrightarrow \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g}) $$ A \(1.506-\mathrm{g}\) sample of limestone-containing material gives \(0.558 \mathrm{g}\) of \(\mathrm{CO}_{2},\) in addition to \(\mathrm{CaO},\) after being heated at a high temperature. What is the mass percent of \(\mathrm{CaCO}_{3}\) in the original sample?

Step-by-Step Solution

Answer

The mass percent of \( \mathrm{CaCO}_{3} \) is approximately 84.36\%.

Step 1: An Overview of the Reaction

The balanced chemical equation for decomposition is \( \mathrm{CaCO}_{3}(\mathrm{s}) \rightarrow \mathrm{CaO}(\mathrm{s}) + \mathrm{CO}_{2}(\mathrm{g}) \). This means each mole of \( \mathrm{CaCO}_{3} \) decomposes into one mole of \( \mathrm{CaO} \) and one mole of \( \mathrm{CO}_{2} \). We need to find the mass percentage of \( \mathrm{CaCO}_{3} \) in the sample using this stoichiometric relationship.

Step 2: Calculating the Molar Mass

Calculate the molar masses: \( \mathrm{CaCO}_{3} \) has a molar mass of approximately 100.09 g/mol, and \( \mathrm{CO}_{2} \) has a molar mass of approximately 44.01 g/mol. We will use these values in stoichiometric conversions.

Step 3: Determine Moles of CO2

Convert the mass of \( \mathrm{CO}_{2} \) produced (0.558 g) to moles using its molar mass: \( n_{\mathrm{CO}_{2}} = \frac{0.558 \, \mathrm{g}}{44.01 \, \mathrm{g/mol}} \approx 0.01268 \, \mathrm{mol} \).

Step 4: Use Stoichiometry to Find Moles of CaCO3

Use the stoichiometric ratio from the balanced equation: given 1 mole of \( \mathrm{CaCO}_{3} \) produces 1 mole of \( \mathrm{CO}_{2} \), the moles of \( \mathrm{CaCO}_{3} \) is the same as \( \mathrm{CO}_{2} \): \( n_{\mathrm{CaCO}_{3}} = 0.01268 \, \mathrm{mol} \).

Step 5: Convert Moles of CaCO3 to Mass

Convert moles of \( \mathrm{CaCO}_{3} \) back to grams: \( m_{\mathrm{CaCO}_{3}} = 0.01268 \, \mathrm{mol} \times 100.09 \, \mathrm{g/mol} \approx 1.270 \, \mathrm{g} \).

Step 6: Calculate Mass Percent of CaCO3

Finally, calculate the mass percent of \( \mathrm{CaCO}_{3} \) in the original sample using \( \frac{1.270 \, \mathrm{g}}{1.506 \, \mathrm{g}} \times 100\% \approx 84.36\% \).

Key Concepts

Molar Mass Calculation

Understanding how to calculate molar mass is essential in stoichiometry. Every chemical compound has a molar mass, which is the total mass of one mole of that compound. This is calculated by adding together the atomic masses of each element present in the molecule, based on the periodic table. For example, in the compound calcium carbonate, \( \mathrm{CaCO}_{3} \), its molar mass is derived from these atomic masses:

• Calcium (Ca) has an atomic mass of approximately 40.08 g/mol.
• Carbon (C) is approximately 12.01 g/mol.
• Oxygen (O) is about 16.00 g/mol. Since there are three oxygen atoms in \( \mathrm{CaCO}_{3} \), you multiply 16.00 by 3.

Putting these together, the molar mass of \( \mathrm{CaCO}_{3} \) is approximately \( 40.08 + 12.01 + (16.00 \times 3) = 100.09 \) g/mol.
These values are crucial for stoichiometric calculations, allowing conversion between mass and moles.

Mass Percent Composition

In chemistry, determining the mass percent composition of a compound in a sample is a crucial skill. Mass percent tells us what portion of a compound's total mass is made up by a particular element or compound. To find the mass percent of a substance, we use the formula: \[ \text{Mass Percent} = \left( \frac{\text{mass of component}}{\text{total mass of sample}} \right) \times 100\% \]
For instance, in our limestone problem, we are calculating the mass percent of \( \mathrm{CaCO}_{3} \) in the original sample. We determined the mass of \( \mathrm{CaCO}_{3} \) to be approximately 1.270 g. The total mass of the sample was 1.506 g. Applying the formula gives:
\[ \text{Mass Percent of} \, \mathrm{CaCO}_{3} = \left( \frac{1.270 \, \text{g}}{1.506 \, \text{g}} \right) \times 100\% \approx 84.36\% \]
This shows that a substantial portion of the sample is composed of \( \mathrm{CaCO}_{3} \).

Limestone Decomposition

Limestone decomposition is an important reaction in industrial and natural processes. Limestone, primarily composed of calcium carbonate (\( \mathrm{CaCO}_{3} \)), decomposes to form calcium oxide (\( \mathrm{CaO} \)) and carbon dioxide (\( \mathrm{CO}_{2} \)) gas when heated. The reaction is represented by the chemical equation: \[ \mathrm{CaCO}_{3} \rightarrow \mathrm{CaO} + \mathrm{CO}_{2} \]
This process is not only vital for producing cement and lime in industrial settings but also plays a role in carbon cycling in nature.

• In the cement industry: the decomposition of limestone provides lime, which is an essential ingredient in cement production.
• In atmospheric science: carbonate decomposition releases carbon dioxide, contributing to the carbonate-silicate cycle that regulates carbon dioxide levels over geological timescales.

Understanding this decomposition is key to both industrial applications and ecological balance.