Problem 1 - Chemical Principles in the Laboratory

Question

A \(0.4943-\mathrm{g}\) sample of \(\mathrm{CaCO}_{3}\) is dissolved in \(12 \mathrm{M} \mathrm{HCl}\) and the resulting solution is diluted to \(250.0 \mathrm{mL}\) in a volumetric flask. a. How many moles of \(\mathrm{CaCO}_{3}\) are used (formula mass \(=100.1 \mathrm{g} / \mathrm{mol}\) )?_____\(moles\) b. What is the molarity of the \(\mathrm{Ca}^{2+}\) in the \(250.0 \mathrm{mL}\) of solution?____\(M\) c. How many moles of \(\mathrm{Ca}^{2+}\) are in a 25.00 -mL aliquot of the solution in \(1(\mathrm{b}) ?\) _____\(moles\)

Step-by-Step Solution

Answer

a) 0.004937 moles; b) 0.01975 M; c) 0.00049375 moles.

Step 1: Determine Moles of CaCO3

To find the number of moles of CaCO3, use the formula: \[moles = \frac{mass}{molar\,mass}\]Here, the mass of CaCO3 is 0.4943 g and the molar mass is 100.1 g/mol. Substitute these values:\[moles = \frac{0.4943\,g}{100.1\,g/mol} \approx 0.004937\,moles\]

Step 2: Calculate Molarity of Ca2+

Since each molecule of CaCO3 releases one Ca2+ ion, the moles of Ca2+ will be the same as the moles of CaCO3. To find the molarity of Ca2+ in the solution, divide the number of moles by the volume of the solution in liters:\[molarity = \frac{moles}{volume} = \frac{0.004937\,moles}{0.2500\,L} \approx 0.01975\,M\]

Step 3: Find Moles of Ca2+ in a 25.00 mL Aliquot

First, convert the volume from mL to L: 25.00 mL = 0.02500 L. Use the molarity of Ca2+ from step 2 to find the moles in the aliquot:\[moles = molarity \times volume = 0.01975\,M \times 0.02500\,L \approx 0.00049375\,moles\]

Key Concepts

Moles Calculation

Calculating moles is an essential skill in chemistry, providing a bridge between the microscopic world of atoms and the more tangible quantities we can measure. A mole (\(\text{mol}\)) is a unit that represents \(6.022 \times 10^{23}\) entities, similar to how a dozen represents 12. This concept is based on Avogadro's number and is useful for converting between mass and the number of atoms, ions, or molecules.

To calculate moles, you can use the formula:

• Moles = \(\frac{\text{Mass}}{\text{Molar Mass}}\)

The molar mass is the mass of one mole of a substance, typically measured in grams per mole (g/mol). It combines the atomic masses of all the elements in a compound, which you can find on the periodic table.

For example, with calcium carbonate (\(\text{CaCO}_3\)), the molar mass is known to be \(100.1 \, \text{g/mol}\). By dividing the sample mass (\(0.4943 \, \text{g}\)) by the molar mass, you find that there are approximately \(0.004937\) moles of \(\text{CaCO}_3\). Understanding this conversion is fundamental in chemical calculations as it connects the mass scene in the laboratory with the chemical equations used in reaction stoichiometry.

Molarity

Molarity (\(M\)) is a way to express the concentration of a solution. It represents the number of moles of a solute dissolved in one liter of solution. Understanding molarity is crucial because it helps you know how much solute is present in a given volume of a solution.

The formula for molarity is:

• Molarity = \(\frac{\text{Moles of Solute}}{\text{Volume of Solution in Liters}}\)

When you have the moles of calcium ions (\(\text{Ca}^{2+}\)) from calcium carbonate (\(\text{CaCO}_3\)) dissolved, you can calculate the molarity by dividing the moles by the volume of the solution. Here, we dissolve the \(\text{CaCO}_3\) in enough water to make \(250.0 \, \text{mL}\) of solution, which is equivalent to \(0.2500 \, \text{L}\).

Given that \(0.004937\) moles of \(\text{CaCO}_3\) are dissolved, the molarity calculation is \(\frac{0.004937\, \text{moles}}{0.2500\, \text{L}} \approx 0.01975\, M\). This tells us that each liter of solution contains approximately \(0.01975\) moles of \(\text{Ca}^{2+}\) ions.

Dilution

Dilution involves decreasing the concentration of a solution by increasing its volume with a solvent, usually while keeping the amount of solute constant. When performing dilutions, chemists often think in terms of proportions; the key relationship is that \(\text{M}_1 \times \text{V}_1 = \text{M}_2 \times \text{V}_2\), where \(\text{M}\) stands for molarity and \(\text{V}\) for volume.

In the given exercise, after \(\text{CaCO}_3\) was dissolved, the resulting solution was diluted to \(250.0 \, \text{mL}\). We calculated the molarity to be \(0.01975\, M\), which stays constant until a dilution factor is applied. If you take a \(25.00 \, \text{mL}\) aliquot from this main solution, the proportion remains the same:

• Moles in Aliquot = \(\text{Molarity} \times \text{Volume of Aliquot in Liters}\)

Convert \(25.00 \, \text{mL}\) to \(0.02500 \, \text{L}\), and multiply by the molarity: \(0.01975 \, M \times 0.02500 \, L \approx 0.00049375 \, ext{moles}\).

This calculation reveals that the dilution does not change the number of moles of solute, only its concentration in a given volume.