Question

If \(5 \mathrm{~g}\) solvent, \(0.2 \mathrm{~g}\) solute \(\mathrm{A},\) and \(0.3 \mathrm{~g}\) solute \(\mathrm{B}\) are mixed to form a solution, calculate the weight percent of \(\mathrm{A} .\)

Short answer

The weight percent of solute A is approximately 3.64%.

Step-by-step solution

Define Weight Percent Formula

Weight percent of a solute in a solution is given by the formula \[ \text{Weight percent of solute} = \left( \frac{\text{mass of solute}}{\text{total mass of solution}} \right) \times 100. \]

Calculate Total Mass of Solution

First, calculate the total mass of the solution by adding the masses of the solute and solvent. \[ \text{Total mass of solution} = \text{mass of solvent} + \text{mass of solute } A + \text{mass of solute } B = 5 \text{ g} + 0.2 \text{ g} + 0.3 \text{ g} = 5.5 \text{ g}. \]

Apply the Formula for Weight Percent

Now use the weight percent formula to calculate the weight percent of solute A. Substitute the known values: \[ \text{Weight percent of A} = \left( \frac{0.2 \text{ g}}{5.5 \text{ g}} \right) \times 100. \]

Perform the Calculation

Calculate the expression to find the weight percent: \[ \text{Weight percent of A} = \left( \frac{0.2}{5.5} \right) \times 100 \approx 3.64\%. \]

Key concepts

Solution Composition

In chemistry, understanding the composition of a solution is key to grasping more complex concepts. A solution is a homogenous mixture composed of two primary parts: the solvent and the solute. The solvent is typically the substance present in the greatest quantity and dissolves the solute, which is the substance added into the solvent. For example, when sugar (solute) dissolves in water (solvent), it forms a sugar-water solution.
When working with solutions, it often becomes necessary to know exactly how much of a particular component, or solute, is present. This can be quantitatively expressed in weight percent or other concentration units such as molarity, molality, etc. Each has its unique application or advantage depending on the context of the problem.
In the provided exercise, two solutes (A and B) are mixed into a solvent, and determining the composition, particularly the weight percent, of solute A is crucial for understanding the overall makeup of the solution. This kind of information is fundamental in fields ranging from industrial chemistry to pharmacology.

Chemistry Problem Solving

Solving chemistry problems often involves a step-by-step approach to ensure accuracy, especially in calculations involving solution composition. The method starts with understanding what information is given and what needs to be determined.

• Identify the known values - these include the masses of the solute(s) and the solvent.
• Understand the required calculation - in this case, the weight percent of solute A.
• Apply relevant formulas - the weight percent formula is crucial here.

Breaking down the problem into these manageable steps clarifies the process and minimizes the chance of errors. For instance, correctly calculating the total mass of the solution is necessary before using the weight percent formula.
This gentle approach not only makes the problem more accessible and manageable but also builds crucial problem-solving skills applicable in a wide array of scientific inquiries.

Mass Calculation in Solutions

Calculating mass within solutions involves both precise mathematical manipulations and a solid understanding of the components involved. First, ascertain the total mass of the solution. This is simply the sum of the solvent and the solutes involved. In this exercise, the combined mass of the solvent and two solutes gives us the total mass of the solution.
Once the total mass is known, calculating the weight percent of any component follows a straightforward use of the formula:\[\text{Weight percent of solute} = \left( \frac{\text{mass of solute}}{\text{total mass of solution}} \right) \times 100\]
This fraction shows the proportion of the solute compared to the entire solution and multiplying by 100 converts this fraction into a percentage, a more intuitive measure. This calculation tells us how much of the solution is made up by the specific solute in question. As seen in the solved exercise, precision in these calculations ensures the accuracy of the result, crucial particularly in situations like creating accurate pharmaceutical formulations or industrial mixtures.