Problem 55 - Chemistry The Central Science

Question

Commercial aqueous nitric acid has a density of 1.42 \(\mathrm{g} / \mathrm{mL}\) and is 16 \(\mathrm{M} .\) Calculate the percent \(\mathrm{HNO}_{3}\) by mass in the solution.

Step-by-Step Solution

Answer

The percent HNO3 by mass in the commercial aqueous nitric acid solution is approximately 71%.

Step 1: Calculate the mass of the solution

We are given the density of the commercial aqueous nitric acid as 1.42 g/mL. To find the mass of 1 litre (1000 mL) of the solution, we can use the formula: Mass = Density × Volume Mass = 1.42 g/mL × 1000 mL Mass = 1420 g So, the mass of 1 litre of the commercial aqueous nitric acid solution is 1420 g.

Step 2: Calculate the moles of HNO3 in 1 litre of the solution

We are given the molarity (M) of the solution as 16 M. By definition, molarity is the number of moles of solute per litre of solution. Thus, in 1 litre of the solution, we have: Moles of HNO3 = Molarity × Volume Moles of HNO3 = 16 mol/L × 1 L Moles of HNO3 = 16 mol So, there are 16 moles of HNO3 in 1 litre of the solution.

Step 3: Convert the moles of HNO3 into mass of HNO3

Now we need to find the mass of HNO3, given the number of moles. We can use the molar mass of HNO3 to do this conversion. The molar mass of HNO3 is: HNO3: (1 × 1.01) + (1 × 14.01) + (3 × 16.00) = 1.01 + 14.01 + 48.00 = 63.02 g/mol Now, we can calculate the mass of HNO3 using the moles and molar mass: Mass of HNO3 = Moles × Molar mass Mass of HNO3 = 16 mol × 63.02 g/mol Mass of HNO3 = 1008.32 g So, the mass of HNO3 in 1 litre of the solution is 1008.32 g.

Step 4: Calculate the percent HNO3 by mass in the solution

Finally, we can find the percent HNO3 by mass in the solution using the formula: Percent HNO3 = (Mass of HNO3 / Mass of the solution) × 100 Percent HNO3 = (1008.32 g / 1420 g) × 100 Percent HNO3 ≈ 71% Thus, the percent HNO3 by mass in the commercial aqueous nitric acid solution is approximately 71%.

Key Concepts

Density Calculation

Density is a core concept in chemistry. It tells us how much mass is contained in a specific volume of a substance. It’s expressed as mass per unit volume. In this exercise, we are dealing with the density of aqueous nitric acid, which is 1.42 g/mL. This means that every milliliter of the solution contains 1.42 grams of mass.

To calculate the mass of a solution, you multiply the density by the volume. For example, if we have 1000 mL (or 1 liter) of the solution, you multiply it by the density to get the total mass: 1.42 g/mL × 1000 mL = 1420 g. This calculation is vital because it helps us understand how much the solution weighs, which is important for further calculations. Remember:

• Density is always mass/volume
• Make sure your volume is in the correct units, typically milliliters or liters in these calculations

Molarity

Molarity (M) is the measure of the concentration of a solute in a solution. It’s expressed in moles of solute per liter of solution. In our problem, the solution has a molarity of 16 M. This simply means there are 16 moles of nitric acid (HNO3) in every liter of the solution.

To find the number of moles, you'd use the relationship:

• Moles = Molarity × Volume
• In our case: 16 mol/L × 1 L = 16 moles

This concept is particularly important because it helps us understand how concentrated a chemical solution is. It also serves as a bridge to calculate the actual amount of a substance, in moles, which we often need in reactions and further analyses.

Percent Composition

Percent composition is a way to express the concentration of a particular substance in a solution. It’s often described as the percent by mass of one component compared to the total mass. In this exercise, we want to find out how much of the mass of the entire solution is made up of nitric acid.

To find it, you take the mass of the specific component (HNO3 in this case), which is calculated by converting moles to mass using the molar mass, and divide it by the total mass of the solution. Then multiply by 100 to convert it into a percentage:

• Mass of HNO3 = 16 moles × 63.02 g/mol = 1008.32 g
• Percent HNO3 = \( \frac{1008.32 \text{ g}}{1420 \text{ g}} \times 100 \)
• The final result is approximately 71%

Understanding percent composition is crucial. It provides insight into the purity and concentration of solutions, helping in both industrial applications and simple lab experiments.