Problem 25 - Chemistry and Chemical Reactivity

Question

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

Step-by-Step Solution

Answer

The mass percent of \(\mathrm{CaCO}_{3}\) in the sample is 84.39\%.

Step 1: Calculate Molar Masses

First, determine the molar masses of the reactants and products. The molar mass of \(\mathrm{CaCO}_{3}\) is calculated as: \(\text{Ca: 40.08 g/mol, C: 12.01 g/mol, O: 16.00 g/mol x 3}\), so that equals \(100.09\ \text{g/mol}\). The molar mass of \(\mathrm{CO}_{2}\) is \(\text{(C: 12.01 g/mol + O: 16.00 g/mol x 2)}\), which results in \(44.01\ \text{g/mol}\).

Step 2: Determine Moles of CO2 Produced

Calculate the number of moles of \(\mathrm{CO}_{2}\) that were produced using its mass: \(0.558\ \text{g}\ ÷ 44.01\ \text{g/mol} = 0.0127\ \text{moles of }\mathrm{CO}_{2}\).

Step 3: Relate CaCO3 and CO2 Moles

Based on the reaction equation \(\mathrm{CaCO}_{3} \rightarrow \mathrm{CaO} + \mathrm{CO}_{2}\), the moles of \(\mathrm{CaCO}_{3}\) decomposed is equal to the moles of \(\mathrm{CO}_{2}\) produced. Therefore, there are \(0.0127\ \text{moles of } \mathrm{CaCO}_{3}\).

Step 4: Calculate Mass of CaCO3

Find the mass of \(\mathrm{CaCO}_{3}\) that decomposed by multiplying the moles by its molar mass: \(0.0127\ \text{moles} \times 100.09\ \text{g/mol} = 1.271\ \text{g of } \mathrm{CaCO}_{3}\).

Step 5: Find Mass Percent of CaCO3

Calculate the mass percent of \(\mathrm{CaCO}_{3}\) in the original sample using the formula: \( \frac{\text{mass of } \mathrm{CaCO}_{3}}{\text{mass of the original sample}} \times 100\%% \). This gives \( \frac{1.271\ \text{g}}{1.506\ \text{g}} \times 100\%% = 84.39\%%\).

Key Concepts

Mass Percent Calculation

Calculating the mass percent of a component in a mixture helps us understand how much of that component is present relative to the entire mixture. To find the mass percent of a substance like calcium carbonate (\( \text{CaCO}_3 \)) in a sample, follow these steps:

• First, determine the mass of the desired component (e.g.,\( \text{CaCO}_3 \)) in the sample. In our case, a chemical reaction helps identify this mass.
• Second, use the mass of the entire sample. For example, if the original limestone mix sample weighed 1.506 g, this would be the total mass.
• Finally, use the formula:\[ \text{Mass percent} = \left( \frac{\text{mass of component}}{\text{total mass of sample}} \right) \times 100\% \]Plug in the numbers, like the decomposed\( \text{CaCO}_3 \), and the mass of the full sample to calculate the mass percent.

This calculation method is valuable for assessing purity or concentration in various chemical processes.

Limestone Decomposition

Limestone decomposition involves breaking down calcium carbonate (\( \text{CaCO}_3 \)) into calcium oxide (\( \text{CaO} \)) and carbon dioxide (\( \text{CO}_2 \)) gas. This reaction is essential in industries and occurs when heating limestone to high temperatures:

\[ \text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g) \]Understanding this process helps in applications like cement production, where calcium oxide is a critical component. The reaction is a simple example of thermal decomposition, involving only one reactant that breaks into two products.

• The decomposing reaction is significant as it suggests equal moles of\( \text{CaCO}_3 \)are required to produce equal moles of\( \text{CO}_2 \).
• Any practical applications must also consider the capture of\( \text{CO}_2 \), due to its role as a greenhouse gas.

Being familiar with limestone decomposition and its equation is not only critical for solving chemistry problems but also for understanding industrial processes and environmental implications.

Molar Mass Determination

Molar mass is a fundamental concept to quantify the amount of a certain element or compound in chemistry. It refers to the mass of one mole of a given substance, usually expressed in grams per mole (g/mol).

Calculating the molar mass involves adding up the atomic masses of all the atoms present in a molecule. For instance:

• The molar mass of\( \text{CaCO}_3 \)can be determined by summing the atomic masses of calcium (Ca, 40.08), carbon (C, 12.01), and three oxygen atoms (O, 16.00 each):\[ 40.08 + 12.01 + (16.00 \times 3) = 100.09 \text{ g/mol} \]
• Similarly, for carbon dioxide (\( \text{CO}_2 \)), you add the atomic masses for carbon and two oxygens:\[ 12.01 + (16.00 \times 2) = 44.01 \text{ g/mol} \]

Understanding molar mass is crucial for stoichiometry, allowing the conversion of mass to moles and vice versa. This calculation assists in determining the amount of reactants and products involved in chemical reactions, laying the groundwork for quantitative chemistry. Using molar mass correctly is essential for precise experimental outcomes and chemistry calculations.