Calculating Mass Fraction Of $CaCl_2$ In A Mixture A Step-by-Step Solution

In the realm of chemistry, quantitative analysis plays a pivotal role in deciphering the composition of various mixtures. This article delves into a fascinating problem involving a mixture of calcium chloride ($CaCl_2$) and sodium chloride (NaCl), where we aim to determine the mass fraction of $CaCl_2$ within the mixture. This exploration will not only demonstrate the practical application of stoichiometry and chemical reactions but also highlight the meticulous approach required to solve such analytical chemistry problems.

Problem Statement: Dissecting the Mixture's Secrets

Imagine a scenario where we have a sample of a mixture containing $CaCl_2$ and NaCl, with a total weight of 2.22 grams. Our mission is to unravel the proportion of $CaCl_2$ present in this mixture. To achieve this, we embark on a chemical journey involving precipitation and conversion reactions. The process begins by treating the mixture to precipitate all the calcium ions ($Ca^{2+}$) as calcium carbonate ($CaCO_3$). This precipitate is then subjected to intense heat, leading to its quantitative conversion into calcium oxide (CaO), weighing 0.84 grams. With these experimental observations at our disposal, our analytical prowess is put to the test – to calculate the mass fraction of $CaCl_2$ in the original mixture. This problem is a classic example of how chemical transformations, combined with precise measurements, can reveal the hidden composition of matter.

The Chemical Transformation: From $CaCl_2$ to CaO

To effectively tackle this problem, a deep understanding of the chemical reactions involved is paramount. The journey from $CaCl_2$ to CaO is a two-step process, each with its own chemical equation and stoichiometric implications. Let's dissect these steps to understand the transformations occurring at the molecular level.

Step 1: Precipitation of Calcium Carbonate ($CaCO_3$)

The first step involves the precipitation of calcium ions ($Ca^{2+}$) from the mixture as calcium carbonate ($CaCO_3$). This is achieved by introducing a suitable precipitating agent, such as sodium carbonate ($Na_2CO_3$), into the solution. The reaction proceeds as follows:

$CaCl_2(aq) + Na_2CO_3(aq) ightarrow CaCO_3(s) + 2NaCl(aq)$

In this reaction, calcium ions from $CaCl_2$ react with carbonate ions from $Na_2CO_3$ to form solid $CaCO_3$, which precipitates out of the solution. The sodium ions and chloride ions remain in the solution. The key here is that all the calcium present in the original mixture as $CaCl_2$ is now quantitatively converted into $CaCO_3$. This quantitative conversion is crucial for our analysis, as it allows us to relate the amount of $CaCO_3$ formed to the initial amount of $CaCl_2$.

Step 2: Thermal Decomposition of Calcium Carbonate ($CaCO_3$) to Calcium Oxide (CaO)

The second step involves subjecting the precipitated $CaCO_3$ to intense heat. This process, known as thermal decomposition, breaks down $CaCO_3$ into calcium oxide (CaO) and carbon dioxide ($CO_2$). The chemical equation for this reaction is:

$CaCO_3(s) ightarrow CaO(s) + CO_2(g)$

Upon heating, $CaCO_3$ decomposes, releasing $CO_2$ gas and leaving behind solid CaO. The crucial aspect of this reaction is that it proceeds quantitatively, meaning that all the $CaCO_3$ is converted into CaO. This quantitative conversion is vital because it allows us to directly relate the mass of CaO obtained to the initial amount of $CaCO_3$ and, ultimately, to the amount of $CaCl_2$ in the original mixture. The problem states that 0.84 grams of CaO is obtained, providing us with a crucial piece of information for our calculations.

Stoichiometry: The Language of Chemical Calculations

Stoichiometry is the cornerstone of quantitative chemical analysis, providing the necessary tools to relate the amounts of reactants and products in a chemical reaction. In this problem, stoichiometry will bridge the gap between the mass of CaO obtained and the mass of $CaCl_2$ present in the original mixture. Let's delve into the stoichiometric relationships that govern our calculations.

Molar Masses: The Conversion Factors

The first step in any stoichiometric calculation is to determine the molar masses of the compounds involved. The molar mass is the mass of one mole of a substance and serves as a crucial conversion factor between mass and moles. For our problem, we need the molar masses of $CaCl_2$ and CaO. These can be calculated by summing the atomic masses of the constituent elements, which can be found on the periodic table.

• Molar mass of $CaCl_2$: 40.08 (Ca) + 2 * 35.45 (Cl) = 110.98 g/mol
• Molar mass of CaO: 40.08 (Ca) + 16.00 (O) = 56.08 g/mol

These molar masses tell us that one mole of $CaCl_2$ weighs 110.98 grams, and one mole of CaO weighs 56.08 grams. These values will be essential for converting between grams and moles in our calculations.

Mole Ratios: Connecting Reactants and Products

The balanced chemical equations for the reactions provide the mole ratios between reactants and products. In our case, we have two reactions to consider:

1. $CaCl_2(aq) + Na_2CO_3(aq) ightarrow CaCO_3(s) + 2NaCl(aq)$
2. $CaCO_3(s) ightarrow CaO(s) + CO_2(g)$

From the first equation, we see that one mole of $CaCl_2$ produces one mole of $CaCO_3$. From the second equation, one mole of $CaCO_3$ decomposes to form one mole of CaO. This 1:1 mole ratio between $CaCl_2$, $CaCO_3$, and CaO is the key to connecting the mass of CaO obtained to the initial amount of $CaCl_2$. It tells us that the number of moles of CaO formed is equal to the number of moles of $CaCO_3$ that decomposed, which, in turn, is equal to the number of moles of $CaCl_2$ initially present in the mixture. This simple yet profound relationship allows us to trace the calcium atoms from their initial form in $CaCl_2$ to their final form in CaO.

Calculation: Unraveling the Mass Fraction

Now, armed with the chemical equations, molar masses, and stoichiometric relationships, we can embark on the calculation to determine the mass fraction of $CaCl_2$ in the mixture. This involves a step-by-step process, converting the mass of CaO to moles, then using the mole ratios to find the moles of $CaCl_2$, and finally converting moles of $CaCl_2$ to grams to calculate the mass fraction.

Step 1: Converting Mass of CaO to Moles

We are given that 0.84 grams of CaO is obtained. To convert this mass to moles, we use the molar mass of CaO (56.08 g/mol):

Moles of CaO = Mass of CaO / Molar mass of CaO Moles of CaO = 0.84 g / 56.08 g/mol Moles of CaO ≈ 0.015 mol

This calculation tells us that 0.84 grams of CaO corresponds to approximately 0.015 moles of CaO. This value is a crucial intermediate in our calculation, linking the experimental observation to the molar quantity of the product.

Step 2: Determining Moles of $CaCl_2$

Using the 1:1 mole ratio between CaO and $CaCl_2$, we know that the number of moles of $CaCl_2$ initially present in the mixture is equal to the number of moles of CaO formed:

Moles of $CaCl_2$ = Moles of CaO Moles of $CaCl_2$ ≈ 0.015 mol

This simple yet powerful deduction is a direct consequence of the stoichiometry of the reactions. It allows us to infer the amount of $CaCl_2$ present in the original mixture based on the amount of CaO obtained in the experiment.

Step 3: Converting Moles of $CaCl_2$ to Grams

Now that we know the number of moles of $CaCl_2$, we can convert this to grams using the molar mass of $CaCl_2$ (110.98 g/mol):

Mass of $CaCl_2$ = Moles of $CaCl_2$ * Molar mass of $CaCl_2$ Mass of $CaCl_2$ ≈ 0.015 mol * 110.98 g/mol Mass of $CaCl_2$ ≈ 1.66 g

This calculation reveals that approximately 1.66 grams of $CaCl_2$ were present in the original mixture. This is a significant piece of information, bringing us closer to our ultimate goal of determining the mass fraction of $CaCl_2$.

Step 4: Calculating Mass Fraction of $CaCl_2$

The mass fraction of a component in a mixture is the mass of that component divided by the total mass of the mixture. We know the mass of $CaCl_2$ (1.66 g) and the total mass of the mixture (2.22 g), so we can calculate the mass fraction:

Mass fraction of $CaCl_2$ = Mass of $CaCl_2$ / Total mass of mixture Mass fraction of $CaCl_2$ ≈ 1.66 g / 2.22 g Mass fraction of $CaCl_2$ ≈ 0.748

To express this as a percentage, we multiply by 100:

Mass fraction of $CaCl_2$ ≈ 0.748 * 100% Mass fraction of $CaCl_2$ ≈ 74.8%

Conclusion: Unveiling the Mixture's Composition

Through a meticulous step-by-step analysis, we have successfully determined the mass fraction of $CaCl_2$ in the mixture. Our calculations reveal that approximately 74.8% of the mixture, by mass, is composed of $CaCl_2$. This result showcases the power of quantitative chemical analysis in unraveling the composition of complex mixtures. By combining experimental observations with stoichiometric principles, we can gain valuable insights into the nature of matter.

This problem serves as a testament to the importance of understanding chemical reactions, stoichiometry, and molar masses in solving analytical chemistry problems. The ability to convert between mass and moles, apply mole ratios, and perform accurate calculations is crucial for any aspiring chemist or scientist. The journey from a seemingly simple mixture to the precise determination of its composition highlights the elegance and power of chemistry as a quantitative science.