Problem 144 - General Chemistry: Principles and Modern Applications

Question

The minerals calcite, \(\mathrm{CaCO}_{3},\) magnesite, \(\mathrm{MgCO}_{3}\) and dolomite, \(\mathrm{CaCO}_{3} \cdot \mathrm{MgCO}_{3},\) decompose when strongly heated to form the corresponding metal oxide(s) and carbon dioxide gas. A 1.000 -g sample known to be one of the three minerals was strongly heated and \(0.477 \mathrm{g} \mathrm{CO}_{2}\) was obtained. Which of the three minerals was it?

Step-by-Step Solution

Answer

The mineral is most likely to be calcite, CaCO3.

Step 1: Calculate the molar mass

First, the molar mass of CO2 has to be calculated. Carbon has an atomic mass of approximately 12.01g/mol and oxygen has an atomic mass of about 16.00g/mol. CO2 therefore has a molar mass of \(12.01g/mol + 2*(16.00g/mol) = 44.01 g/mol\).

Step 2: Calculate moles of CO2

Next, the number of moles of the CO2 produced from the sample is calculated using its given mass and molar mass - which is \(0.477g / 44.01g/mol = 0.0108 mol\). Since the molar ratio of CO2 to any of the minerals is 1:1, we also have 0.0108 mol of the mineral.

Step 3: Calculating the expected mass of the mineral

Calculate the expected masses of the minerals based on their molar masses. The molar masses of the minerals are \(100.09 g/mol\) for CaCO3, \(84.31 g/mol\) for MgCO3, and \(184.40 g/mol\) for CaCO3·MgCO3. Multiplying the moles of mineral by these amounts provides the theoretical masses for each mineral, which are \(1.08 g\) for CaCO3, \(0.91 g\) for MgCO3, and \(2.00 g\) for CaCO3·MgCO3.

Step 4: Compare to the original sample mass

The original sample has a mass of 1.000g, so the mineral that decomposes to form a mass closest to this original mass is the correct answer. We can see that the mass of CaCO3 is the closest to the original sample mass.

Key Concepts

Molar Mass Calculation

Understanding molar mass is critical for solving stoichiometry problems, especially when analyzing chemical reactions. Molar mass, the mass of one mole of a substance, is expressed in grams per mole (g/mol). It's equivalent to the sum of the atomic masses of all atoms in a molecule. For instance, carbon dioxide (CO₂) comprises one carbon atom and two oxygen atoms. To calculate its molar mass:

• Identify the atomic mass of Carbon: approximately 12.01 g/mol.
• Identify the atomic mass of Oxygen: approximately 16.00 g/mol.
• Since there are two oxygen atoms in CO₂, multiply the atomic mass of oxygen by two and add it to the atomic mass of carbon: \[12.01 \text{ g/mol} + 2 \times 16.00 \text{ g/mol} = 44.01 \text{ g/mol}\].

This calculation lets students relate the quantity of a substance in a reaction to its mass, a pivotal skill for determining the composition of products and reactants in chemical equations.

Chemical Decomposition

Chemical decomposition is a reaction where a single compound breaks down into two or more simpler substances when heated, often resulting in a metal oxide and a gas. When considering the decomposition of minerals such as calcite (CaCO₃), magnesite (MgCO₃), or dolomite (CaCO₃·MgCO₃), it's essential to recognize that heating triggers the release of carbon dioxide (CO₂) and leaves behind the respective metal oxide.

To determine the original mineral from the mass of CO₂ produced, we compare the ratios of the decomposed gas to the remaining compound - which is typically a 1:1 molar ratio. This fundamental concept of stoichiometry uses balanced equations and the conservation of mass to deduce the identity of an unknown substance by analyzing its decomposition products.

Mineral Analysis

Mineral analysis involves identifying and quantifying the composition of a mineral sample. In our example, the mass of CO₂ released upon heating reveals clues about the mineral's identity. By comparing the theoretical mass of CO₂ to that observed, students can determine which mineral they started with.

In our exercise, we calculate the expected mass of each possible mineral based on its molar mass and the moles implied by the CO₂ produced. The key is to find which mineral's theoretical mass matches the original sample mass (1.000 g). This process not only aids in identifying minerals but also reinforces the practical application of molar mass calculations and stoichiometric principles in real-world scenarios like geological studies and material sciences.