Question

Calcium carbonate, \(\mathrm{CaCO}_{3},\) can be obtained in a very pure state. Standard solutions of calcium ion are usually prepared by dissolving calcium carbonate in acid. What mass of \(\mathrm{CaCO}_{3}\) should be taken to prepare \(500 .\) mL of \(0.0200 M\) calcium ion solution?

Short answer

To prepare 500 mL of 0.0200 M calcium ion solution, 1.001 g of calcium carbonate (\(CaCO_3\)) should be taken.

Step-by-step solution

Find the molar mass of CaCO3

Calculate the molar mass of \(CaCO_3\) by summing the atomic masses of the individual elements in the compound. Atomic masses: - Calcium (Ca): 40.08 g/mol - Carbon (C): 12.01 g/mol - Oxygen (O): 16.00 g/mol So, the molar mass of \(CaCO_3\) is: Molar mass = 40.08 + 12.01 +(3 × 16.00) = 100.09 g/mol

Use the molarity formula to find moles of CaCO3

Next, we'll use the molarity formula to find the required moles of \(CaCO_3\). Recall that: Molarity (M) = moles of solute/liters of solution Rearranging the formula to solve for moles of solute, we have: Moles of solute = molarity × liters of solution We are given the molarity (0.0200 M) and the volume of the solution (500 mL) which must be converted to liters: Volume = 500 mL × (1 L/1000 mL) = 0.500 L Now, plug these values into the formula: Moles of CaCO3 = 0.0200 M × 0.500 L = 0.0100 moles

Convert moles of CaCO3 to mass

Finally, convert moles of \(CaCO_3\) to mass using the molar mass: Mass = moles × molar mass Substitute the moles and molar mass calculated earlier in this formula: Mass of CaCO3 = 0.0100 moles × 100.09 g/mol = 1.001 g So, 1.001 g of calcium carbonate (CaCO3) should be taken to prepare 500 mL of 0.0200 M calcium ion solution.

Key concepts

Molar Mass Calculation

When calculating the molar mass of a compound, you need to add the atomic masses of all the atoms in the formula. For calcium carbonate, CaCO3, you will sum the masses of calcium, carbon, and oxygen. Here’s how:

• Calcium (Ca): Atomic mass of 40.08 g/mol
• Carbon (C): Atomic mass of 12.01 g/mol
• Oxygen (O): Atomic mass of 16.00 g/mol (but since there are three oxygens, you multiply this by 3)

Combine these values to obtain the molar mass of CaCO3:\[\text{Molar mass of } \text{CaCO}_3 = 40.08 + 12.01 + (3 \, \times \, 16.00) = 100.09 \, \text{g/mol}\]Understanding molar mass is crucial as it serves as a bridge between the molecular scale and the practical scale, converting moles into grams. Always double-check your atomic masses using a reliable periodic table.

Molarity Formula

Molarity is a way to express the concentration of a solution. It's defined as the number of moles of solute per liter of solution. This is how we use that formula:\[\text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}}\]To find the number of moles of solute, you rearrange the formula:\[\text{Moles of solute} = \text{Molarity} \, \times \, \text{Liters of solution}\]For instance, if you have a solution with a molarity of 0.0200 M and a volume of 500 mL, you must convert the volume to liters (500 mL = 0.500 L). Substituting in the values, you get:\[\text{Moles of } \text{CaCO}_3 = 0.0200 \, \text{M} \, \times \, 0.500 \, \text{L} = 0.0100 \, \text{moles}\]This formula is essential for preparing standard solutions and ensuring accurate concentrations in any chemical experiment.

Stoichiometry

Stoichiometry is a method used in chemistry to relate quantities of reactants and products. It involves mole ratios from balanced chemical equations.For solutions, stoichiometry helps to convert between moles and grams using the molar mass. Using the moles calculated from molarity, you can find out how much of a solid must be weighed out to achieve a specific solution concentration.In this exercise, knowing that you need 0.0100 moles of CaCO3, you will use the molar mass to find the mass:\[\text{Mass of } \text{CaCO}_3 = \text{Moles} \, \times \, \text{Molar mass}\]For our calculation:\[0.0100 \, \text{moles} \, \times \, 100.09 \, \text{g/mol} = 1.001 \, \text{g}\]Stoichiometry allows for converting between moles, mass, and volume, ensuring detailed and accurate chemical analysis.

Solution Preparation

The final chemistry task is to prepare the solution, after calculating the required mass of CaCO3. Begin by measuring 1.001 g of pure CaCO3. Follow these steps for precise solution preparation:

• Measure: Use a balance to weigh exactly 1.001 g of CaCO3.
• Dissolve: Carefully add the CaCO3 to a container with some dilute acid to ensure it reacts completely.
• Fill: Once dissolved, pour the solution into a 500 mL volumetric flask.
• Top Up: Add distilled water until the total volume reaches 500 mL.

This ensures you now have a 0.0200 M calcium ion solution ready for use. Consistency and precision in preparation are vital to achieving exact concentrations needed for your experiments.