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Acetylsalicylic acid is the active ingredient in aspirin. It took 35.17 mL of 0.5065 M sodium hydroxide to react completely with 3.210 g of acetylsalicylic acid. Acetylsalicylic acid has one acidic hydrogen. What is the molar mass of acetylsalicylic acid?

Short Answer

Expert verified
The molar mass of acetylsalicylic acid is approximately 180.24 g/mol.

Step by step solution

01

Calculate moles of sodium hydroxide

To calculate the moles of sodium hydroxide, use the equation: moles of NaOH = molarity of NaOH * volume of NaOH (in L) moles of NaOH = 0.5065 M * 35.17 mL * (1 L / 1000 mL) = 0.01781 moles
02

Determine moles of acetylsalicylic acid

As the reaction follows a 1:1 stoichiometry, the moles of acetylsalicylic acid will be equal to the moles of sodium hydroxide. moles of acetylsalicylic acid = moles of NaOH = 0.01781 moles
03

Calculate the molar mass of acetylsalicylic acid

Now, we will find the molar mass of acetylsalicylic acid by dividing the mass (in grams) by the moles of acetylsalicylic acid. molar mass of acetylsalicylic acid = mass of acetylsalicylic acid / moles of acetylsalicylic acid molar mass of acetylsalicylic acid = 3.210 g / 0.01781 moles = 180.24 g/mol The molar mass of acetylsalicylic acid is approximately 180.24 g/mol.

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Key Concepts

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

Acetylsalicylic Acid
Acetylsalicylic acid is the chemical name for what most of us know as aspirin. This compound is widely used as a pain reliever and anti-inflammatory drug. It is composed of carbon, hydrogen, and oxygen atoms, with the chemical formula \(C_9H_8O_4\).

One of its key features is the presence of a carboxylic acid group, which contains one acidic hydrogen. This property makes it an acid, meaning it can donate a hydrogen ion to a base in a chemical reaction. Understanding this is crucial for chemical experiments like titrations, where acetylsalicylic acid can react with a base to determine its concentration or purity.

Being familiar with the structure and composition of acetylsalicylic acid is important for anyone studying chemistry, as it ties into concepts such as functional groups and molecular interactions.
Stoichiometry
Stoichiometry is a branch of chemistry that focuses on the quantitative relationships between reactants and products in a chemical reaction. In our exercise, stoichiometry helps determine how many moles of acetylsalicylic acid react with sodium hydroxide.

The balanced chemical equation for the reaction between acetylsalicylic acid and sodium hydroxide is essential. In this particular case, the reaction follows a 1:1 stoichiometry. This means that one mole of acetylsalicylic acid reacts with exactly one mole of sodium hydroxide.

Using stoichiometry allows chemists to calculate the amounts of reactants needed and products formed in a reaction. It's a fundamental tool for predicting the outcomes of reactions and ensuring that experiments are accurately conducted and analyzed.
Neutralization Reaction
A neutralization reaction occurs when an acid and a base react to form water and a salt. In our specific example, acetylsalicylic acid is the acid, and sodium hydroxide (\(NaOH\)) is the base. When they react, sodium acetylsalicylate (the salt) and water are produced.

This type of reaction is very common in chemistry, particularly in acid-base titration experiments, where the purpose is to determine the concentration of an unknown acid or base. The reaction continues until the acid has completely converted into the salt, which is indicated by a change in the pH of the solution.

Understanding neutralization reactions helps in predicting how substances will interact, and it is frequently applied in fields such as medicine, environmental science, and materials engineering.
Sodium Hydroxide
Sodium hydroxide, with the chemical formula \(NaOH\), is a strong base commonly used in various chemical reactions. In our exercise, it reacts with acetylsalicylic acid during a neutralization reaction.

It's known for its caustic nature and ability to easily dissociate in water to form sodium ions and hydroxide ions. This dissociation allows \(NaOH\) to effectively neutralize acids by accepting hydrogen ions to form water.

Handling sodium hydroxide requires caution due to its corrosive properties; it can cause damage to living tissues. Despite this, it is widely used in chemical manufacturing, soap making, and water treatment processes due to its effective reactivity and affordability.
Acid-Base Titration
Acid-base titration is a technique used to determine the unknown concentration of an acid or base solution by reacting it with a solution of known concentration.

In our scenario, a titration allows us to find the concentration of acetylsalicylic acid by titrating with sodium hydroxide, a base of known concentration. In the process, we measure how much base is needed to completely neutralize the acid.

Titrations are performed using indicators, substances that change color at a particular pH level, to signify the endpoint of the reaction. Knowing the exact volume of titrant used to reach this endpoint is vital for calculating the concentration of the unknown solution. This method is an essential process in analytical chemistry, providing precise measurements and insights into chemical properties and reactions.

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Most popular questions from this chapter

The units of parts per million (ppm) and parts per billion (ppb) are commonly used by environmental chemists. In general, 1 ppm means 1 part of solute for every \(10^{6}\) parts of solution. Mathematically, by mass: $$\mathrm{ppm}=\frac{\mu \mathrm{g} \text { solute }}{\mathrm{g} \text { solution }}=\frac{\mathrm{mg} \text { solute }}{\mathrm{kg} \text { solution }}$$ In the case of very dilute aqueous solutions, a concentration of 1.0 \(\mathrm{ppm}\) is equal to 1.0\(\mu \mathrm{g}\) of solute per 1.0 \(\mathrm{mL}\) , which equals 1.0 \(\mathrm{g}\) solution. Parts per billion is defined in a similar fashion. Calculate the molarity of each of the following aqueous solutions. a. 5.0 \(\mathrm{ppb} \mathrm{Hg}\) in \(\mathrm{H}_{2} \mathrm{O}\) b. 1.0 \(\mathrm{ppb} \mathrm{CHCl}_{3}\) in \(\mathrm{H}_{2} \mathrm{O}\) c. 10.0 \(\mathrm{ppm}\) As in \(\mathrm{H}_{2} \mathrm{O}\) d. 0.10 \(\mathrm{ppm} \mathrm{DDT}\left(\mathrm{C}_{14} \mathrm{H}_{9} \mathrm{C} 1_{5}\right)\) in \(\mathrm{H}_{2} \mathrm{O}\)

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