Question

Dissolving \(3.00 \mathrm{~g}\) of an impure sample of calcium carbonate in hydrochloric acid produced \(0.656 \mathrm{~L}\) of carbon dioxide (measured at \(20.0^{\circ} \mathrm{C}\) and \(792 \mathrm{mmHg}\) ). Calculate the percent by mass of calcium carbonate in the sample. State any assumptions.

Short answer

The sample contains approximately 93.42% calcium carbonate by mass.

Step-by-step solution

Write the Balanced Chemical Equation

The reaction between calcium carbonate (\(\text{CaCO}_3\)) and hydrochloric acid (\(\text{HCl}\)) produces calcium chloride (\(\text{CaCl}_2\)), water (\(\text{H}_2\text{O}\)), and carbon dioxide (\(\text{CO}_2\)). The balanced equation is: \(\text{CaCO}_3 + 2 \text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2\).

Use Ideal Gas Law to calculate moles of CO2

To find the moles of \(\text{CO}_2\) produced, we use the ideal gas law \(PV = nRT\). First, we convert the pressure from mmHg to atm: \(792 \, \text{mmHg} \times \frac{1 \, \text{atm}}{760 \, \text{mmHg}} = 1.0421 \, \text{atm}\). The volume is \(0.656 \, \text{L}\), the temperature in Kelvin is \(20.0^\circ \text{C} + 273.15 = 293.15 \, \text{K}\), and \(R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1}\). Solve for \(n\) (moles of \(\text{CO}_2\)): \(n = \frac{PV}{RT} = \frac{1.0421 \, \text{atm} \times 0.656 \, \text{L}}{0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \times 293.15 \, \text{K}} \approx 0.0280 \, \text{mol}\).

Relate Moles of CO2 to CaCO3

According to the balanced chemical equation, 1 mole of \(\text{CaCO}_3\) produces 1 mole of \(\text{CO}_2\). Therefore, the moles of \(\text{CaCO}_3\) is equal to the moles of \(\text{CO}_2\), which is \(0.0280 \, \text{mol}\).

Calculate Mass of Pure CaCO3

Calculate the mass of \(\text{CaCO}_3\) using its molar mass. The molar mass of \(\text{CaCO}_3\) is \(40.08 + 12.01 + 3\times16.00 = 100.09 \, \text{g/mol}\). The mass of pure \(\text{CaCO}_3\) is \(0.0280 \, \text{mol} \times 100.09 \, \text{g/mol} = 2.8025 \, \text{g}\).

Calculate Percent by Mass of CaCO3 in the Sample

Calculate the percentage by mass of \(\text{CaCO}_3\) in the original sample: \( \frac{2.8025 \, \text{g of CaCO}_3}{3.00 \, \text{g of sample}} \times 100 \approx 93.42\%\).

Assumptions Made

We assumed the gas behaves ideally, and the temperature and pressure measurements are accurate. We also assume all \(\text{CaCO}_3\) in the sample reacts completely with \(\text{HCl}\).

Key concepts

Ideal Gas Law

The Ideal Gas Law is an essential concept for understanding the behavior of gases. It is expressed as the equation \( PV = nRT \). Here, \(P\) stands for pressure, \(V\) is volume, \(n\) represents the moles of gas, \(R\) is the ideal gas constant, and \(T\) is the temperature in Kelvin. This equation helps determine the properties of a gas under specific conditions.
In the given problem, the Ideal Gas Law is used to calculate the number of moles of carbon dioxide (\(CO_2\)) produced in a chemical reaction. The pressure is converted from mmHg to atm, which is more compatible with the gas constant \(R\), typically valued at \(0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1}\). Temperature must be in Kelvin, achieved by adding 273.15 to the Celsius temperature.
Using the values \(P = 1.0421 \, \text{atm}\), \(V = 0.656 \, \text{L}\), and \(T = 293.15 \, \text{K}\), the number of moles \(n\) of \(CO_2\) is calculated as \(0.0280 \, \text{mol}\). Thus, understanding each component of the Ideal Gas Law is crucial for solving problems involving gases.

Chemical Reactions

Chemical reactions are processes where reactants transform into products. This is represented by balanced chemical equations. Balancing is vital as it ensures the same number of atoms for each element on both sides of the equation, following the Law of Conservation of Mass.
In our scenario, the reaction of calcium carbonate (\(\text{CaCO}_3\)) with hydrochloric acid (\(\text{HCl}\)) is described by the equation: \(\text{CaCO}_3 + 2 \text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2\).
This equation tells us that one mole of \(\text{CaCO}_3\) will produce one mole of \(\text{CO}_2\). Balancing the equation ensures an accurate representation of how reactants convert into products. Understanding chemical reactions helps in determining the quantities and relationships between various substances within a reaction.

Molar Mass

Molar Mass is a concept that represents the mass of one mole of a substance. It is measured in grams per mole (g/mol) and is vital for converting between the mass of a substance and the number of moles.
In the exercise, the molar mass of calcium carbonate \((\text{CaCO}_3)\) is determined by adding the atomic masses of its constituent elements: \(40.08\) (calcium) + \(12.01\) (carbon) + \(48.00\) (three oxygens) = \(100.09 \, \text{g/mol}\).
The molar mass allows conversion of moles of a substance to its mass and vice versa. Here, with \(0.0280\) moles of \(\text{CaCO}_3\), we calculate the mass as \(0.0280 \, \text{mol} \times 100.09 \, \text{g/mol} = 2.8025 \, \text{g}\). Molar mass is a foundational concept in stoichiometry, helping us understand and calculate chemical transformations.

Percent Composition

Percent composition is a way to express the amount of a specific substance within a mixture, typically by mass. It is calculated using the formula: \(\frac{\text{mass of component}}{\text{total mass of mixture}} \times 100\).
In this exercise, the percent by mass of \(\text{CaCO}_3\) in the impure sample is determined. After finding that \(2.8025\) grams of \(\text{CaCO}_3\) are present in the impurity sample of \(3.00\) grams, the calculation is as follows:
\[\frac{2.8025 \, \text{g (CaCO}_3\text{)}}{3.00 \, \text{g (sample)}} \times 100\% \approx 93.42\%\]
Percent composition helps determine purity and quality of the sample. It's essential for analyzing mixtures in various fields, such as chemistry, pharmacology, and materials science.