Question

A mixture contained calcium carbonate, \(\mathrm{CaCO}_{3}\), and magnesium carbonate, \(\mathrm{MgCO}_{3}\). A sample of this mixture weighing \(7.85 \mathrm{~g}\) was reacted with excess hydrochloric acid. The reactions are $$ \mathrm{CaCO}_{3}(g)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{CaCl}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g) $$ \(\mathrm{MgCO}_{3}(s)+2 \mathrm{HCl}(a q) \longrightarrow\) $$ \mathrm{MgCl}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(t)+\mathrm{CO}_{2}(g) $$ If the sample reacted completely and produced \(1.94 \mathrm{~L}\) of carbon dioxide, \(\mathrm{CO}_{2}\), at \(25^{\circ} \mathrm{C}\) and \(785 \mathrm{mmHg}\), what were the percentages of \(\mathrm{CaCO}_{3}\) and \(\mathrm{MgCO}_{3}\) in the mixture?

Short answer

The mixture is approximately 80.38% \(\text{CaCO}_3\) and 19.62% \(\text{MgCO}_3\).

Step-by-step solution

Convert pressure to atm

The given pressure is 785 mmHg. To convert this to atmospheres (atm), use the conversion factor \(1 \text{ atm} = 760 \text{ mmHg}\). So, the pressure in atm is \(\frac{785 \text{ mmHg}}{760 \text{ mmHg/atm}}\approx 1.033\text{ atm}\).

Convert volume to liters

The volume of \(\text{CO}_2\) given is already in liters: 1.94 L.

Use Ideal Gas Law to find moles of CO2

Using the ideal gas law \(PV = nRT\), where \(P\) is pressure in atm, \(V\) is volume in liters, \(n\) is number of moles, \(R\) is the ideal gas constant \(0.0821 \text{ L atm/mol K}\), and \(T\) is temperature in Kelvin. Convert \(25^{\circ} \text{C}\) to Kelvin: \(298 \text{ K}\). Substitute the known values: \[ n = \frac{PV}{RT} = \frac{1.033 \times 1.94}{0.0821 \times 298} \approx 0.0813 \text{ moles of CO}_2 \].

Relate moles of CO2 to moles of carbonates

Each mole of \(\text{CaCO}_3\) and \(\text{MgCO}_3\) produces 1 mole of \(\text{CO}_2\). Let \(x\) be the moles of \(\text{CaCO}_3\) and \(y\) be the moles of \(\text{MgCO}_3\). Then: \(x + y = 0.0813\).

Set up mass equation for the mixture

The molar mass of \(\text{CaCO}_3\) is 100.09 g/mol, and \(\text{MgCO}_3\) is 84.31 g/mol. The total mass equation based on the mixture is: \(100.09x + 84.31y = 7.85\).

Solve the simultaneous equations

Using the equations: \(x + y = 0.0813\) and \(100.09x + 84.31y = 7.85\), solve for \(x\) and \(y\). First, solve the first equation for \(y\): \(y = 0.0813 - x\). Substitute this into the second equation: \(100.09x + 84.31(0.0813 - x) = 7.85\). Simplifying: \(100.09x - 84.31x + 6.856 = 7.85\), \(15.78x = 0.994\), \(x \approx 0.0630\) mol. Substituting \(x\) back into \(y = 0.0813 - x\), \(y \approx 0.0183\) mol.

Calculate the percentages of CaCO3 and MgCO3

Convert moles to mass: \(\text{mass of CaCO}_3 = 0.0630 \times 100.09 = 6.31 \text{ g}\) and \(\text{mass of MgCO}_3 = 0.0183 \times 84.31 = 1.54 \text{ g}\). Calculate the percentage: \(\frac{6.31}{7.85} \times 100\approx 80.38\%\ text{ CaCO}_3\) and \(\frac{1.54}{7.85} \times 100\approx 19.62\%\text{ MgCO}_3\).

Key concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry used to describe the behavior of an ideal gas. It is represented by the formula \( PV = nRT \), where:

• \( P \): Pressure measured in atmospheres (atm)
• \( V \): Volume measured in liters (L)
• \( n \): Number of moles of the gas
• \( R \): Ideal gas constant, approximately 0.0821 L atm/mol K
• \( T \): Temperature measured in Kelvin (K)

The Ideal Gas Law allows us to relate these four properties and solve for any unknown quantity, making it a powerful tool for solving gas-related problems. In the context of the exercise, the Ideal Gas Law was used to calculate the number of moles of carbon dioxide (\( \text{CO}_2 \)) produced. By converting the given temperature to Kelvin and pressure to atmospheres, students can apply the Ideal Gas Law to find \( n \), the moles of gas. This calculation plays a crucial role in determining the composition of the sample in terms of calcium carbonate and magnesium carbonate.

Mole Concept

The mole concept is a fundamental chemical principle that provides a method for quantifying substances at the atomic level. A "mole" is a standard chemical unit that contains Avogadro's number (\( 6.022 \times 10^{23} \)) of entities, typically atoms, ions, or molecules. It establishes a direct relationship between any given mass of a substance and its amount in terms of moles.

In the exercise, utilizing the mole concept allows the comparison between the moles of \( \text{CaCO}_3 \), \( \text{MgCO}_3 \), and the resulting \( \text{CO}_2 \) gas. Each mole of \( \text{CaCO}_3 \) and \( \text{MgCO}_3 \) produces one mole of \( \text{CO}_2 \), making the mole calculations straightforward.

This concept simplifies the step-by-step solution by providing a consistent method to transform mass into moles, facilitating stoichiometric calculations needed to determine the composition percentage of each compound in the mixture.

Chemical Reactions

Chemical reactions represent the process where reactants are transformed into products. They are typically described by chemical equations that must be balanced to reflect the conservation of mass. In chemical equations, the coefficients in front of compounds indicate the number of moles necessary for the reaction to occur.

In this exercise, two chemical reactions occur: the reaction of calcium carbonate \( (\text{CaCO}_3) \) and magnesium carbonate \( (\text{MgCO}_3) \) with hydrochloric acid (\( \text{HCl} \)). Both reactions produce carbon dioxide, water, and their respective chlorides. Here are the balanced reactions:

• \( \text{CaCO}_3 + 2 \text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2 \text{O} + \text{CO}_2 \)
• \( \text{MgCO}_3 + 2 \text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2 \text{O} + \text{CO}_2 \)

Understanding these reactions helps in identifying how much \( \text{CO}_2 \) is produced from given quantities of \( \text{CaCO}_3 \) and \( \text{MgCO}_3 \). These balanced equations are central to the calculation of the moles and then the mass percent of each compound in the original mixture.