Question

How many grams of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) are contained in \(500 . \mathrm{g}\) of a \(5.5 \%\) by mass \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) solution?

Short answer

The mass of Na2CO3 in the 500g solution, which has a 5.5% concentration by mass, can be calculated using the rearranged mass percentage formula: Mass of Na2CO3 = (5.5 × 500) / 100. After performing the calculation, we find that there are \( 27.5g \) of Na2CO3 in the given solution.

Step-by-step solution

Understand the mass percentage formula

The mass percentage formula relates the mass of the solute, the mass of the solution, and the mass percentage of the solution, as follows: Mass percentage = (mass of solute / mass of solution) × 100

Rearrange the formula to find the mass of the solute

We want to find the mass of Na2CO3 in the solution, so we need to rearrange the formula to solve for the mass of the solute: Mass of solute = (mass percentage × mass of solution) / 100

Plug in the given values

We are given a mass percentage of 5.5% and a mass of the solution of 500g. Plug these values into the rearranged formula: Mass of Na2CO3 = (5.5 × 500) / 100

Calculate the mass of Na2CO3

Perform the calculation to find the mass of Na2CO3 in the solution: Mass of Na2CO3 = (5.5 × 500) / 100 = 27.5g So, there are 27.5 grams of Na2CO3 in the 500g solution.

Key concepts

Solution Concentration

Solution concentration is an important aspect of chemistry that tells us how much solute is present in a given amount of solution. It's essentially a way of expressing the ratio of solute to solution, which is critical in many chemical applications. To express concentration by mass percentage, we use the formula:

• Mass percentage = \( \frac{\text{mass of solute}}{\text{mass of solution}} \times 100\)

This formula allows us to express the concentration as a percentage, making it easy to understand the proportion of the solute in the total solution. In the current problem, the concentration is given as 5.5%, meaning that for every 100 grams of solution, 5.5 grams is sodium carbonate (\(\mathrm{Na}_{2} \mathrm{CO}_{3}\)).
Understanding solution concentration helps in various fields such as biology, medicine, and engineering, where it is crucial to have accurate measurements of chemical components.

Mass of Solute

The mass of a solute is how much of a substance is dissolved within a solution. Calculating this helps in determining the exact amount of the solute needed for reactions or processes. In the exercise provided, we need the mass of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) in a given solution.
To find the mass of the solute, we rearrange the mass percentage formula, aiming to solve for the mass of solute:

• Mass of solute = \( \frac{\text{mass percentage} \times \text{mass of solution}}{100} \)

By plugging in the values from the problem, with a 5.5% mass percentage and a 500g mass of solution:

• Mass of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) = \( \frac{5.5 \times 500}{100} = 27.5 \text{g} \)

Thus, there are 27.5 grams of sodium carbonate in the solution. This straightforward approach makes it easy to determine the solute's mass in any given condition.

Chemical Calculations

Chemical calculations involve using mathematical methods to determine the quantities and properties of substances. They are fundamental in chemistry for stoichiometry, preparing solutions, and analyzing reactions.
To perform these calculations successfully, one must understand various principles like conversions, mass-mass relations, and the use of formulas.
In our exercise, we applied chemical calculations through the mass percentage formula to find the solute's mass:

• Identify given data: mass percentage and total mass of solution.
• Convert the percentage into a workable formula to discover the unknown, in this case, the mass of the solute.
• Perform the arithmetic operation to derive the answer.

These calculations not only help in the current context but also provide a foundation for more complex chemical problems. Mastering these basics allows you to expand your understanding of chemistry as a whole, making it easier to tackle future problems effectively.