Problem 1 - Chemical Principles in the Laboratory

Question

A 0.4243 -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 \(=100.1\) )? b. What is the molarity of the \(\mathrm{Ca}^{2+}\) in the \(250 \mathrm{mL}\) of solution? c. How many moles of \(\mathrm{Ca}^{2+}\) are in a \(25.00-\mathrm{mL}\) aliquot of the solution in \(1 \mathrm{b} ?\)

Step-by-Step Solution

Answer

a. 0.00424 mol; b. 0.01696 M; c. 0.000424 mol

Step 1: Calculate Moles of CaCO3

To find the number of moles of \( \mathrm{CaCO}_{3} \), use the formula: \[\text{Moles of } \mathrm{CaCO}_{3} = \frac{\text{mass of } \mathrm{CaCO}_{3}}{\text{molar mass of } \mathrm{CaCO}_{3}}\]Substituting in the given values:\[\text{Moles of } \mathrm{CaCO}_{3} = \frac{0.4243 \, \text{g}}{100.1 \, \text{g/mol}} \approx 0.00424 \, \text{mol}\]

Step 2: Determine Molarity of Ca2+ in Solution

Since \(\mathrm{CaCO}_{3} \) fully dissociates into \(\mathrm{Ca}^{2+} \) ions when dissolved, the moles of \(\mathrm{CaCO}_{3} \) equal the moles of \(\mathrm{Ca}^{2+} \).The total volume of the solution is 250 mL, which is 0.250 L.The molarity \( M \) is calculated as:\[M = \frac{\text{moles of } \mathrm{Ca}^{2+}}{\text{volume of solution in L}} = \frac{0.00424 \, \text{mol}}{0.250 \, \text{L}} \approx 0.01696 \, \text{M}\]

Step 3: Calculate Moles of Ca2+ in 25.00 mL Aliquot

In a 25.00 mL aliquot, which is 0.0250 L, calculate the moles of \(\mathrm{Ca}^{2+} \) using the molarity:\[\text{Moles of } \mathrm{Ca}^{2+} = M \times \text{Volume in L} = 0.01696 \, \text{M} \times 0.0250 \, \text{L} = 0.000424 \, \text{mol}\]

Key Concepts

Stoichiometry

Stoichiometry is all about calculating the amounts of substances in chemical reactions. Through balanced equations, it helps us relate quantities of reactants and products. In this example, we start with a chemical compound, calcium carbonate (\(\text{CaCO}_3\)), and track its dissociation in hydrochloric acid (HCl). To master stoichiometry:

• Understand the mole concept, which is a basic unit in chemistry to express amounts of substances.
• Use mole ratios from balanced chemical equations to connect different substances.

In the problem, \(\text{CaCO}_3\) dissociates to form calcium ions (\(\text{Ca}^{2+}\)). Stoichiometry helps in calculating how much calcium carbonate is required to yield a certain amount of calcium ion.

Volumetric Flask

A volumetric flask is a key piece of laboratory glassware used to prepare precise solutions. It has a long neck with a single, precise measurement mark that shows the total volume of the liquid it contains. To use a volumetric flask:

• Dissolve the solute completely before diluting to the mark.
• Accurately fill the flask with the solvent just to the calibration mark.
• Mix the solution thoroughly by inverting the flask several times with the stopper on.

In the original exercise, the solution was diluted to 250 mL using a volumetric flask, ensuring precise concentration for accurate molarity calculations.

Moles of Calcium Carbonate (\(\text{CaCO}_3\))

To calculate the moles of \(\text{CaCO}_3\), we use its mass and molar mass. A mole in chemistry is a way of expressing a fixed quantity, about \(6.022 \times 10^{23}\) particles, atoms, or molecules.Here's a step-by-step solution:

• Determine the mass of \(\text{CaCO}_3\), which is given as 0.4243 grams.
• Find the molar mass of \(\text{CaCO}_3\), which is provided as 100.1 g/mol.
• Use the formula: \(\text{Moles of } \text{CaCO}_3 = \frac{\text{mass of } \text{CaCO}_3}{\text{molar mass of } \text{CaCO}_3}\).
• Plug in the values: \(\frac{0.4243 \, \text{g}}{100.1 \, \text{g/mol}} \approx 0.00424 \, \text{mol}\).

This calculation reveals how many moles of \(\text{CaCO}_3\) are used in the reaction.

Molar Mass

Molar mass is essential for converting between grams and moles, facilitating stoichiometric calculations. It represents the mass of one mole of a substance and is expressed in grams per mole (g/mol).For \(\text{CaCO}_3\):

• The molar mass is calculated by summing the atomic masses of its constituent elements: calcium (Ca), carbon (C), and oxygen (O).
• Calcium has an atomic mass of about 40.08 g/mol, carbon 12.01 g/mol, and oxygen 16.00 g/mol.
• Thus, the molar mass of \(\text{CaCO}_3\) is \(40.08 + 12.01 + (3 \times 16.00) = 100.1\) g/mol.

With the molar mass known, we can easily determine the amount of substance in moles from its mass, an important step in any stoichiometric or solution preparation problem.