/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 129 A herbicide contains \(2,4-\math... [FREE SOLUTION] | 91影视

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A herbicide contains \(2,4-\mathrm{D}\) \((2,4-\text { dichlorophenoxyacetic acid }), \mathrm{C}_{8} \mathrm{H}_{6} \mathrm{Cl}_{2} \mathrm{O}_{3} . \mathrm{A}\) \(1.236-\mathrm{g}\) sample of the herbicide was decomposed to liberate the chlorine as \(\mathrm{Cl}^{-}\) ion. This was precipitated as AgCl, with a mass of 0.1840 g. What is the mass percent of \(2,4-\mathrm{D}\) in the sample?

Short Answer

Expert verified
The mass percent of 2,4-D in the sample is approximately 11.44%.

Step by step solution

01

Determine moles of AgCl

First, find the moles of AgCl. The molar mass of AgCl is the sum of the atomic masses of Ag (107.87 g/mol) and Cl (35.45 g/mol), which equals 143.32 g/mol. So, the moles of AgCl precipitated is given by: \[ \text{Moles of } \text{AgCl} = \frac{0.1840 \text{ g}}{143.32 \text{ g/mol}} \approx 0.00128 \text{ moles} \]
02

Calculate moles of Cl鈦 ions

Each mole of AgCl contains one mole of Cl鈦 ions, so the moles of Cl鈦 ion is equal to the moles of AgCl. Therefore, there are approximately 0.00128 moles of Cl鈦 ions.
03

Calculate moles of C鈧圚鈧咰l鈧侽鈧

In 2,4-D, each molecule contains two Cl atoms. Therefore, the number of moles of 2,4-D is half the number of moles of Cl鈦 ions. \[ \text{Moles of } C_8H_6Cl_2O_3 = \frac{0.00128}{2} \approx 0.00064 \text{ moles} \]
04

Calculate mass of C鈧圚鈧咰l鈧侽鈧

Now, calculate the mass of C鈧圚鈧咰l鈧侽鈧 using its molar mass. The molar mass of C鈧圚鈧咰l鈧侽鈧 is: \[ (8 \times 12.01) + (6 \times 1.01) + (2 \times 35.45) + (3 \times 16.00) = 221.0 \text{ g/mol} \]The mass is then calculated as:\[ \text{Mass of } C_8H_6Cl_2O_3 = 0.00064 \text{ moles} \times 221.0 \text{ g/mol} = 0.1414 \text{ g} \]
05

Calculate mass percent of C鈧圚鈧咰l鈧侽鈧

The mass percent of C鈧圚鈧咰l鈧侽鈧 in the sample is calculated using the mass of C鈧圚鈧咰l鈧侽鈧 and the mass of the sample:\[ \text{Mass percent of } C_8H_6Cl_2O_3 = \left( \frac{0.1414 \text{ g}}{1.236 \text{ g}} \right) \times 100 \approx 11.44\% \]

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Key Concepts

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

2,4-Dichlorophenoxyacetic Acid
2,4-Dichlorophenoxyacetic acid, commonly abbreviated as 2,4-D, is a type of herbicide used to control broadleaf weeds. Its chemical formula is \( \text{C}_8\text{H}_6\text{Cl}_2\text{O}_3 \). Each molecule contains two chlorine atoms, making it an essential player in the chemistry of herbicide production and its application.
2,4-D works by mimicking a plant hormone that disrupts growing plants, making it a selective herbicide; it targets specific plants while leaving others unharmed. This property makes 2,4-D extremely useful in agriculture and lawn management.
For students studying chemistry, understanding the compound's makeup and behavior is important because it demonstrates the practical application of organic chemistry in everyday agricultural practices. Knowing the chemical composition helps predict reactions and interactions with other substances, such as in precipitation reactions.
Molar Mass
Molar mass is a critical concept in chemistry used to convert between grams and moles. It's the mass of one mole of a substance and is expressed in grams per mole (g/mol). For instance, the molar mass of AgCl (silver chloride) is the sum of silver (107.87 g/mol) and chlorine (35.45 g/mol), equaling 143.32 g/mol.
When calculating the molar mass of a compound, sum the atomic masses of all atoms in its formula. For 2,4-Dichlorophenoxyacetic acid, the calculation is:
  • Carbon (C): \(8 \times 12.01\)
  • Hydrogen (H): \(6 \times 1.01\)
  • Chlorine (Cl): \(2 \times 35.45\)
  • Oxygen (O): \(3 \times 16.00\)
Adding these gives a total molar mass of 221.0 g/mol for 2,4-D.
Understanding molar mass is crucial for stoichiometric calculations because it allows chemists to convert between the mass of a substance and the amount in moles, which is helpful in predicting the yield of chemical reactions.
Stoichiometry
Stoichiometry is the area of chemistry that deals with the relationships between the quantities of reactants and products in a chemical reaction. It uses balanced chemical equations to calculate these quantities.
An essential stoichiometric tool is the mole ratio, derived from coefficients in a balanced equation. For example, in the decomposition of 2,4-D where chlorine is liberated as chloride ions and then precipitated as AgCl, the number of moles of Cl鈦 ions corresponds to moles of precipitated AgCl because the ratio between them in the reaction is 1:1.
In problems involving mass percent calculation, understanding stoichiometry is fundamental for determining how much of a compound is in a mixture. By calculating the moles of a target compound such as 2,4-D, students can find its mass in a given sample and determine the percentage composition, providing insights into the sample's purity and composition.
Precipitation Reaction
A precipitation reaction occurs when two soluble salts in solution react to form one or more insoluble products. These insoluble products are known as precipitates. In our case, when Cl鈦 ions from the decomposition of 2,4-D are treated with Ag鈦 ions, silver chloride (AgCl) precipitates out of the solution.
This type of reaction is useful in various applications, such as removing unwanted ions from solutions or isolating specific compounds from a mixture.
  • To predict whether a precipitation reaction will occur, refer to the solubility rules that estimate the solubility of various ionic compounds.
  • In a laboratory setting, observing a precipitation reaction involves seeing a solid form in a previously clear solution.
Understanding precipitation reactions is essential for effectively analyzing sample compounds, as in determining the mass percent of chlorine in a compound through the formation of insoluble AgCl.

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