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\mathrm{~A} 6.01-\mathrm{g}\( aqueous solution of isopropyl alcohol contains \)3.67 \mathrm{~g}$ of isopropyl alcohol. What is the mass percentage of isopropyl alcohol in the solution?

Short Answer

Expert verified
The mass percentage of isopropyl alcohol in the solution is approximately 61.07%.

Step by step solution

01

Understand the Mass Percentage Formula

The mass percentage of a component in a solution is given by the formula \( \text{Mass percentage} = \left( \frac{\text{mass of the component}}{\text{total mass of the solution}} \right) \times 100 \% \). We need to identify the mass of the isopropyl alcohol and the total mass of the solution.
02

Identify Masses

Here, the mass of isopropyl alcohol is given as 3.67 g and the total mass of the solution is 6.01 g. These values will be used in the formula for calculating the mass percentage.
03

Plug Values into the Formula

Substitute the values into the mass percentage formula: \( \text{Mass percentage} = \left( \frac{3.67}{6.01} \right) \times 100 \% \).
04

Calculate the Mass Percentage

Perform the calculation: \( \frac{3.67}{6.01} \approx 0.6107 \). Then, multiply by 100 to convert into a percentage: \( 0.6107 \times 100 \approx 61.07 \% \).
05

Review and Conclude

Ensure all calculations are correct and consistent with the problem statement. The mass percentage of isopropyl alcohol in the solution is approximately 61.07%.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Isopropyl Alcohol
Isopropyl alcohol, also known as 2-propanol, is a clear, flammable liquid commonly used as a cleaning agent and disinfectant. It features prominently in many households and laboratories because of its versatility in both cleaning and chemical synthesis.

Isopropyl alcohol is made up of carbon, hydrogen, and oxygen atoms arranged in a specific structure. Its formula is \( \text{C}_3\text{H}_8\text{O} \) indicating it has three carbon atoms, eight hydrogen atoms, and one oxygen atom.

Those looking to blend this compound into a solution will find that it mixes well with water. This property makes it highly useful in creating solutions for cleaning that do not leave residues. Its ability to quickly dissolve oils is another reason it's favored in the lab and home.
  • It's often used in medical settings as an antiseptic.
  • Should be handled with care due to flammability.
  • Proper ventilation is necessary when using in closed environments.
Understanding the role of isopropyl alcohol in solution chemistry is key to grasping more complex chemical concepts.
Solution Chemistry
Solution chemistry focuses on the process of dissolving substances to form solutions, a uniform mixture of two or more components. The key terms are solute and solvent.

- **Solute**: The substance being dissolved (e.g., isopropyl alcohol here). - **Solvent**: The substance doing the dissolving, typically in larger amount (e.g., water in this case). In this problem, isopropyl alcohol is the solute and is mixed with water to form an aqueous solution. Aqueous solutions are simply solutions where water is the solvent.
The key concepts when looking into solution chemistry include:
  • Concentration: How much solute is within a solution, which can be defined in various terms like molarity, molality, and mass percentage.
  • Solubility: How well the solute can dissolve in the solvent.
  • Homogeneity: Solutions are completely mixed at a molecular level, giving a uniform appearance and composition throughout.
Being familiar with these concepts will help you understand how different substances interact at the molecular level to form solutions.
Mass Percentage Formula
The mass percentage formula is a convenient way to express concentration of a substance in a mixture. It calculates how much of one component is found within the entire solution, expressed as a percentage. The formula is:
\[\text{Mass percentage} = \left( \frac{\text{mass of the component}}{\text{total mass of the solution}} \right) \times 100 \%\]
This formula is useful in both chemistry and everyday applications, like figuring out how potent a cleaning solution might be.

To apply this formula, simply follow these steps:
  • Identify the mass of the component you're interested in; in this case, isopropyl alcohol was 3.67 g.
  • Determine the total mass of the solution, which was 6.01 g in this exercise.
  • Apply the numbers to the mass percentage formula to calculate the result.
This helps to understand not only what percentage of the solution is active but also to ensure that the solution's composition matches intended uses. Calculating mass percentage allows chemists to create mixtures that are precisely tailored for their purposes.

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Most popular questions from this chapter

An exciting, and often loud, chemical demonstration involves the simple reaction of hydrogen gas and oxygen gas to produce water vapor: $$ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g) $$ The reaction is carried out in soap bubbles or balloons that are filled with the reactant gases. We get the reaction to proceed by igniting the bubbles or balloons. The more \(\mathrm{H}_{2} \mathrm{O}\) that is formed during the reaction, the bigger the bang. Explain the following observations. A bubble containing just \(\mathrm{H}_{2}\) makes a quiet "fffft" sound when ignited. When a bubble containing equal amounts of \(\mathrm{H}_{2}\) and \(\mathrm{O}_{2}\) is ignited, a sizable bang results. When a bubble containing a ratio of 2 to 1 in the amounts of \(\mathrm{H}_{2}\) and \(\mathrm{O}_{2}\) is ignited, the loudest bang results. When a bubble containing just \(\mathrm{O}_{2}\) is ignited, virtually no sound is made.

1 .92 \mathrm{~g} \mathrm{M}^{2+}\( ion reacts with \)0.158 \mathrm{~mol} \mathrm{X}^{-}\( ion to pro- duce a compound, \)\mathrm{MX}_{2}\(, which is \)86.8 \% \mathrm{X}\( by mass. What are the identities of \)\mathrm{M}^{2+}\( and \)\mathrm{X}^{-?}$ ?

Consider the equation $$ 2 \mathrm{KOH}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{K}_{2} \mathrm{SO}_{4}+2 \mathrm{H}_{2} \mathrm{O} $$ If \(25 \mathrm{~g} \mathrm{H}_{2} \mathrm{SO}_{4}\) is reacted with \(7.7 \mathrm{~g} \mathrm{KOH}\), how many grams of \(\mathrm{K}_{2} \mathrm{SO}_{4}\) are produced? For part a of this problem, identify the limiting reactant and calculate the mass of excess reactant that remains after the reaction is completed. Calculate the theoretical yield of the reaction. How many grams of material would you expect to obtain if the reaction has a \(67.1 \%\) yield?

An oxide of tungsten (symbol \(\mathrm{W}\) ) is a bright yellow solid. If \(5.34 \mathrm{~g}\) of the compound contains \(4.23 \mathrm{~g}\) of tungsten, what is its empirical formula?

Aspirin (acetylsalicylic acid) is prepared by heating salicylic acid, \(\mathrm{C}_{7} \mathrm{H}_{6} \mathrm{O}_{3},\) with acetic anhydride, \(\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O}_{3} .\) The other product is acetic acid, \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}\) $$ \mathrm{C}_{7} \mathrm{H}_{6} \mathrm{O}_{3}+\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O}_{3} \longrightarrow \mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}+\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2} $$ What is the theoretical yield (in grams) of aspirin, \(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4},\) when \(2.00 \mathrm{~g}\) of salicylic acid is heated with \(4.00 \mathrm{~g}\) of acetic anhydride? If the actual yield of aspirin is \(1.86 \mathrm{~g}\), what is the percentage yield?

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