Problem 71 - Chemistry Matter and Change

Question

Calculate the molarity of a solution that contain 15.7 g of CaCO \(_{3}\) dissolved in 275 \(\mathrm{mL}\) of water.

Step-by-Step Solution

Answer

The molarity of the solution is approximately 0.571 M.

Step 1: Calculate the Molar Mass of CaCO3

To calculate the molarity, we first need to determine the molar mass of calcium carbonate (\(\text{CaCO}_3\)). We find this by adding the atomic masses of all the atoms in the formula: - Calcium (Ca) has an atomic mass of approximately 40.08 g/mol. - Carbon (C) has an atomic mass of approximately 12.01 g/mol. - Oxygen (O) has an atomic mass of approximately 16.00 g/mol.So, the molar mass of \(\text{CaCO}_3 = 40.08 + 12.01 + (16.00 \times 3) = 100.09 \text{ g/mol}.\)

Step 2: Convert Mass to Moles

Using the molar mass, convert the 15.7 g of \(\text{CaCO}_3\) to moles by dividing by its molar mass.\[ \text{moles of CaCO}_3 = \frac{15.7 \text{ g}}{100.09 \text{ g/mol}} \approx 0.157 \text{ moles} \]

Step 3: Convert Volume to Liters

The molarity formula requires the volume to be in liters. Convert 275 mL to liters by dividing by 1000.\[ 275 \text{ mL} = 0.275 \text{ L} \]

Step 4: Calculate Molarity

Now calculate the molarity \( M \) using the formula \( M = \frac{\text{moles of solute}}{\text{liters of solution}} \).\[ M = \frac{0.157 \text{ moles}}{0.275 \text{ L}} \approx 0.571 \text{ M} \]

Key Concepts

Molar Mass

When you hear "molar mass," think of it as the weight of one mole of a substance, given in grams per mole. It's like the DNA of a substance, letting you know how much a mole of it weighs. To find it, add up the atomic masses of each atom in the substance's chemical formula. For calcium carbonate (CaCO3), you simply add the atomic masses:

• Calcium (Ca): approximately 40.08 g/mol
• Carbon (C): approximately 12.01 g/mol
• Oxygen (O): each approximately 16.00 g/mol, and there are three oxygens in the formula

Thus, the molar mass of calcium carbonate is calculated as:\[40.08 + 12.01 + (16.00 \times 3) = 100.09 \text{ g/mol}\]This tells us that one mole of calcium carbonate weighs 100.09 grams.

Calcium Carbonate

Calcium carbonate, or CaCO3, is a common chemical compound you might know from chalk, limestone, and even in some antacids. It's a salt used in a variety of applications, both in industrial and everyday uses. In its pure form, it appears as a fine white powder.
One fascinating aspect about calcium carbonate is its ability to neutralize acid, which is why it's in products like Tums. Its role in chemistry often involves reactions where its carbonates and calcium ions are crucial.
In solutions, especially when calculating molarity, understanding its molar mass is key, as it tells us how much of this compound we're working with in chemical terms.

Conversion of Units

Converting units might sound tedious, but it's a must for proper scientific calculations. When dealing with solutions, having units in the proper format can change how calculations turn out. Take milliliters and liters for instance, often in chemistry, you'll need to convert between them.
Here's how it works for our solution problem:

• Volume in milliliters was given as 275 mL.
• To convert to liters, you divide by 1000 (since 1000 mL equals 1 L).
• Therefore, 275 mL becomes 0.275 L.

This conversion is crucial in calculating molarity, which requires the volume to be in liters.

Chemical Solution

A chemical solution is a homogenous mixture composed of two or more substances. In our example, we're focusing on a solution where calcium carbonate (CaCO3) is dissolved in water.
The main factor you're interested in for chemical solutions is usually the concentration, or how much solute is present per unit of solvent.
Molarity (M) is a measure of concentration that expresses the number of moles of solute per liter of solution.
To find it, use the formula:\[M = \frac{\text{moles of solute}}{\text{liters of solution}}\]In a lab setting, knowing how to prepare a solution of a specific molarity is essential for experiments, giving you control over the reaction conditions.