Question

A mixture is prepared by mixing \(50.0 \mathrm{g}\) of ethanol, 50.0 g of water, and 5.0 g of sugar. What is the mass percent of each component in the mixture? How many grams of the mixture should one take in order to have \(1.5 \mathrm{g}\) of sugar? How many grams of the mixture should one take in order to have \(10.0 \mathrm{g}\) of ethanol?

Short answer

The mass percent of each component is as follows: ethanol: \( 45.45\% \), water: \( 45.45\% \), and sugar: \( 4.55\% \). To obtain \( 1.5 \mathrm{g} \) of sugar from the mixture, one should take \( 32.97 \mathrm{g} \) of the mixture, and to have \( 10.0 \mathrm{g} \) of ethanol, one should take \( 22.0 \mathrm{g} \) of the mixture.

Step-by-step solution

Calculate the total mass of the mixture

First, we need to find the total mass of the mixture. In order to do this, we sum the masses of ethanol, water, and sugar. Total mass = \( 50.0 \mathrm{g} \) of ethanol + \( 50.0 \mathrm{g} \) of water + \( 5.0 \mathrm{g} \) of sugar

Find the mass percent of each component

Next, we will use the following formula to find the mass percent of each component in the mixture: Mass percent = (mass of a component / total mass) × 100 Calculate the mass percent of ethanol, water, and sugar using the formula: Mass percent of ethanol = (mass of ethanol / total mass) × 100 Mass percent of water = (mass of water / total mass) × 100 Mass percent of sugar = (mass of sugar / total mass) × 100

Determine the amount of mixture for the desired mass of sugar

We want to find out how many grams of the mixture are needed to have \( 1.5 \mathrm{g} \) of sugar. Using the mass percent of sugar found in Step 2, we can set up the following equation: mixture mass for desired sugar = (\( 1.5 \mathrm{g} \) of sugar × total mass) / mass of sugar

Determine the amount of mixture for the desired mass of ethanol

Similarly, we want to find out how many grams of the mixture are needed to have \( 10.0 \mathrm{g} \) of ethanol. Using the mass percent of ethanol from Step 2, we can set up the following equation: mixture mass for desired ethanol = (\( 10.0 \mathrm{g} \) of ethanol × total mass) / mass of ethanol

Key concepts

Stoichiometry

Stoichiometry is a branch of chemistry that deals with the quantitative relationships of the reactants and products in a chemical reaction. In the context of a mixture, stoichiometry can help us understand the proportion of each component relative to the others.

For example, in the given exercise, stoichiometric principles are applied to calculate the mass percent of each component in the solution. It's important in stoichiometry to have a balanced equation, but for mixtures, the focus is on the mass balance. The total mass is simply the sum of the individual masses since there is no chemical reaction altering the mass of substances when making a mixture.

Stoichiometry is also used to determine how much of a mixture one would need to obtain a specific amount of a component, which is critical when ensuring precise measurements in chemical formulations, cooking recipes, or even pharmaceutical dosages.

Solutions Chemistry

In solutions chemistry, the focus is on the homogeneous mixtures where one substance (the solute) is dissolved in another (the solvent). Even though the given exercise doesn't involve a solute dissolving, the concept of concentration is still applicable.

When calculating mass percent, we effectively determine the concentration of each component in the mixture. The mass percent tells us how much of each substance is present compared to the total mixture, similar to how we might discuss the concentration of salt in a saltwater solution. Concentration measures such as mass percent are vital in preparing solutions in laboratories and industries to ensure the desired outcome of the chemical processes they are part of.

Understanding mass percent helps students grasp more complex concentration measures like molarity and molality, which are central to solutions chemistry.

Chemical Compounds

Chemical compounds are substances composed of two or more different types of elements, held together by chemical bonds. In this exercise, ethanol and sugar represent chemical compounds, while water serves as a common solvent. The way compounds interact with each other can dictate the properties of a mixture.

Knowing the properties of individual chemical compounds is crucial when predicting how they will behave in a mixture. For instance, sugar (a carbohydrate) will dissolve in water, and ethanol (an alcohol) is also miscible with water. These compounds do not react chemically in the mixture; instead, the focus is on their distribution within the mass of the mixture.

Understanding the unique features of compounds, such as solubility, helps in predicting the outcome of mixing them, which is not only essential for chemistry students but also for professions like pharmacology and environmental science.

Mixture Composition

The composition of a mixture refers to the types and amounts of each substance within it. The mixture in our exercise is made up of ethanol, water, and sugar. The composition is quantified in terms of mass percent, which tells us how concentrated each component is within the total mixture.

The knowledge of mixture composition is important in several fields, including food technology, material science, and pharmacology, where the properties of a mixture can dramatically change with slight variations in composition. This concept teaches us that even though the components retain their individual properties, the overall character of the mixture can be quite different from that of its components.

Exercise Improvement Advice

It's quite beneficial to also explain and visualize the process of how each component is measured and combined. Using diagrams or animations of the process can help solidify understanding. Ensuring students can practically relate to measuring and mixing ingredients can make the abstract concept of 'mixture composition' much more tangible and comprehensible.