Problem 96 - General Chemistry: Principles and Modern Applications

Question

Chalkboard chalk is made from calcium carbonate and calcium sulfate, with minor impurities such as \(\mathrm{SiO}_{2} .\) Only the \(\mathrm{CaCO}_{3}\) reacts with dilute \(\mathrm{HCl}(\mathrm{aq})\) What is the mass percent \(\mathrm{CaCO}_{3}\) in a piece of chalk if a 3.28 -g sample yields \(0.981 \mathrm{g} \mathrm{CO}_{2}(\mathrm{g}) ?\) $$\begin{aligned} \mathrm{CaCO}_{3}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}) & \longrightarrow \mathrm{CaCl}_{2}(\mathrm{aq}) +\mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{CO}_{2}(\mathrm{g}) \end{aligned}$$

Step-by-Step Solution

Answer

The mass percent of \(\mathrm{CaCO}_{3}\) in the chalk sample is 68.0%.

Step 1: Calculate Molar Masses

The first step is to calculate the molar mass of \(\mathrm{CO}_{2}\) and \(\mathrm{CaCO}_{3}\). The molar mass of \(\mathrm{CO}_{2}\) is \(12.01 g/mol + 2(16.00 g/mol) = 44.01 g/mol\). The molar mass of \(\mathrm{CaCO}_{3}\) is \(40.08 g/mol + 12.01 g/mol + 3(16.00 g/mol) = 100.09 g/mol\).

Step 2: Calculate Moles of Produced CO2

Next, calculate the number of moles of \(\mathrm{CO}_{2}\) produced from the reaction. This can be done by dividing the given mass of \(\mathrm{CO}_{2}\) by its molar mass. In this case, \(0.981 g ÷ 44.01 g/mol = 0.0223 mol\).

Step 3: Calculate Mass of Reacted CaCO3

With the balanced chemical equation, we can conclude that since 1 mole of \(\mathrm{CaCO}_{3}\) results in 1 mole of \(\mathrm{CO}_{2}\), the moles of \(\mathrm{CaCO}_{3}\) reacted will be the same as the moles of \(\mathrm{CO}_{2}\) produced. Therefore, 0.0223 mol of \(\mathrm{CaCO}_{3}\) reacted. The mass can therefore be calculated as: \(0.0223 mol × 100.09 g/mol = 2.23 g\).

Step 4: Calculate Mass Percent CaCO3

The mass percent of \(\mathrm{CaCO}_{3}\) in the chalk sample can be calculated using the formula: mass percent = mass of component in sample / total mass of sample × 100%. Therefore, the mass percent of \(\mathrm{CaCO}_{3}\) is \(2.23 g / 3.28 g × 100% = 68.0% .\)

Key Concepts

Calcium Carbonate

Calcium carbonate, represented by the chemical formula \( \mathrm{CaCO}_3 \), is a compound that forms an essential part of many everyday substances. It's notably present in chalkboard chalk, limestone, and even some antacids. This compound comprises calcium, carbon, and oxygen atoms.Arrange these elements in a crystalline lattice, where calcium bonds with carbonate ions. This structure provides calcium carbonate with its characteristic solidity and density.
It's important due to its interactions in chemical reactions and its prevalence in natural environments. In the context of the given problem, calcium carbonate reacts with hydrochloric acid (\( \mathrm{HCl} \)) to produce carbon dioxide, which can then be measured. The production of \( \mathrm{CO}_2 \) is utilized to determine the mass percent of \( \mathrm{CaCO}_3 \) in a sample. Understanding this process requires grasping both its physical and chemical properties.

Chemical Reactions

Chemical reactions involve the re-arrangement of atoms to form new substances. They are represented by chemical equations. These equations illustrate the reactants that start a reaction and the products that form as a result. For example, the reaction of calcium carbonate with hydrochloric acid can be represented as follows:

• \( \mathrm{CaCO}_3(\mathrm{s}) + 2 \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{CaCl}_2(\mathrm{aq}) + \mathrm{H}_2 \mathrm{O}(\mathrm{l}) + \mathrm{CO}_2(\mathrm{g}) \)

In this equation, solid calcium carbonate combines with aqueous hydrochloric acid to yield aqueous calcium chloride, water, and gaseous carbon dioxide. The equation is balanced, meaning there is an equal number of each type of atom on both sides.
The key point in this reaction is the production of \( \mathrm{CO}_2 \) gas as the main product. Measuring the amount of \( \mathrm{CO}_2 \) produced can provide insights into the quantity of \( \mathrm{CaCO}_3 \) present in the original sample. This reaction forms the basis for mass percent calculation, where the quantity of one substance helps determine the proportion of another.

Stoichiometry

Stoichiometry deals with the quantitative relationships between reactants and products in a chemical reaction. It uses balanced chemical equations to understand how reactants transform into products, and it is central to calculations in chemistry. The problem described utilizes stoichiometry to find the mass percent of \( \mathrm{CaCO}_3 \) in a chalk sample.
The process involves several critical steps:

• Calculating molar masses of the substances involved, such as \( \mathrm{CO}_2 \) and \( \mathrm{CaCO}_3 \).
• Determining the number of moles of \( \mathrm{CO}_2 \) produced by dividing the mass of the \( \mathrm{CO}_2 \) by its molar mass.
• Using the balanced chemical equation to relate moles of \( \mathrm{CO}_2 \) to moles of \( \mathrm{CaCO}_3 \). In this case, the mole ratio is 1:1.
• Converting this mole quantity back to the mass of \( \mathrm{CaCO}_3 \) using its molar mass.
• Finally, using this mass to calculate the mass percent of \( \mathrm{CaCO}_3 \) in the sample by comparing it to the total sample mass.

Each step relies on the previous one and illustrates the importance of precise stoichiometric calculations to derive meaningful chemical insights.