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Consider the following compounds: I. DDT II. gammexane III. carbon tetrachloride IV. chlorobenzene The correct sequence of these compounds in the increasing order of percentage of chlorine in them is (a) I, II, III, IV (b) IV, I, II, III (c) IV, II, I, III (d) III, I, II, IV

Short Answer

Expert verified
The correct sequence is (b) IV, I, II, III.

Step by step solution

01

Determine the Chemical Formula of Each Compound

Identify the chemical formula for each compound: - DDT (I) is C₁₄H₉Cl₅. - Gammexane (II) is C₆H₆Cl₆, also known as hexachlorocyclohexane. - Carbon tetrachloride (III) is CCl₄. - Chlorobenzene (IV) is C₆H₅Cl.
02

Calculate the Molar Mass of Each Compound

Calculate the total molar mass for each compound by adding up the mass of each element involved: - DDT: Carbon (14x12) + Hydrogen (9x1) + Chlorine (5x35.5) = 354.5 g/mol. - Gammexane: Carbon (6x12) + Hydrogen (6x1) + Chlorine (6x35.5) = 290.5 g/mol. - Carbon tetrachloride: Carbon (1x12) + Chlorine (4x35.5) = 153.5 g/mol. - Chlorobenzene: Carbon (6x12) + Hydrogen (5x1) + Chlorine (1x35.5) = 112.5 g/mol.
03

Calculate the Mass of Chlorine in Each Compound

For each compound, calculate the total mass of chlorine: - DDT: 5x35.5 = 177.5 g of chlorine. - Gammexane: 6x35.5 = 213 g of chlorine. - Carbon tetrachloride: 4x35.5 = 142 g of chlorine. - Chlorobenzene: 1x35.5 = 35.5 g of chlorine.
04

Calculate the Percentage of Chlorine

Calculate the percentage of chlorine in each compound using \[ \text{Percentage of Chlorine} = \left( \frac{\text{mass of chlorine}}{\text{molar mass of compound}} \right) \times 100 \]- DDT: \( \left( \frac{177.5}{354.5} \right) \times 100 \approx 50.07\% \)- Gammexane: \( \left( \frac{213}{290.5} \right) \times 100 \approx 73.31\% \)- Carbon tetrachloride: \( \left( \frac{142}{153.5} \right) \times 100 \approx 92.52\% \)- Chlorobenzene: \( \left( \frac{35.5}{112.5} \right) \times 100 \approx 31.56\% \)
05

Arrange Compounds by Increasing Chlorine Percentage

Compare the chlorine percentages calculated above and arrange them in ascending order: - Chlorobenzene: 31.56% - DDT: 50.07% - Gammexane: 73.31% - Carbon tetrachloride: 92.52% Thus, the correct sequence is IV, I, II, III.

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Key Concepts

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

Chemical Formula
Chemical formulas are essential tools in chemistry that provide detailed information about the types and numbers of atoms in a chemical compound. They act like a cheat sheet for understanding the composition of a substance. Take DDT as an example; its chemical formula is \(\text{C}_{14}\text{H}_9\text{Cl}_5\). Here's what that means:
  • Carbon: There are 14 atoms of carbon present.
  • Hydrogen: The compound contains 9 hydrogen atoms.
  • Chlorine: It includes 5 chlorine atoms.

For another compound, gammexane, written as \(\text{C}_6\text{H}_6\text{Cl}_6\), the formula tells us there are 6 carbon, 6 hydrogen, and 6 chlorine atoms. Each element in the formula tells us its role within the molecule, allowing for predictions about how the compound behaves during chemical reactions.
Molar Mass Calculation
The molar mass is the total mass of all atoms present in a mole of a substance. It's a handy calculation for determining how much of a compound you have, and it's measured in grams per mole (g/mol). Let’s see how we can calculate this for DDT:
  • Carbon: There are 14 carbon atoms, each with an atomic mass of 12, summing up to \(14 \times 12 = 168\).
  • Hydrogen: With 9 hydrogen atoms, each weighing only 1, the contribution is \(9 \times 1 = 9\).
  • Chlorine: For the five chlorine atoms, each with an atomic mass of 35.5, we calculate \(5 \times 35.5 = 177.5\).
Add these up: \(168 + 9 + 177.5 = 354.5\, \text{g/mol}\). That’s the molar mass of DDT. Understanding and calculating molar mass is crucial for any conversation surrounding chemical reactions, quantities, and stoichiometry.
Chlorine Percentage
The percentage of chlorine in a compound indicates how much of the substance's weight can be attributed to chlorine atoms. This measurement is especially vital for determining the chemical's effectiveness or functionality in various applications. The formula for calculating chlorine percentage is:\[\text{Percentage of Chlorine} = \left( \frac{\text{Mass of Chlorine}}{\text{Molar Mass of Compound}} \right) \times 100\]
For chlorobenzene, this computation is:
  • Mass of chlorine: 35.5 g
  • Molar mass of chlorobenzene: 112.5 g/mol

Plug the values into the formula: \[\left( \frac{35.5}{112.5} \right) \times 100 \approx 31.56\%\]This result shows that 31.56% of the total weight of chlorobenzene is due to chlorine. Understanding these percentages helps in ranking compounds and comparing them by their chlorine contents, which can affect their application in industries like agriculture and chemistry.
Organic Chemistry
Organic chemistry involves the study of carbon-containing compounds, primarily those with carbon-carbon bonds. DDT, gammexane, carbon tetrachloride, and chlorobenzene are classic examples of organic compounds, each with distinct arrangements of carbon, hydrogen, and chlorine atoms. Let's break down some key concepts:
  • Carbon Backbone: Organic molecules are built upon carbon chains that can become very complex or remain quite straightforward, like in carbon tetrachloride, \(\text{CCl}_4\), where the single carbon atom is bonded to four chlorine atoms.
  • Functional Groups: These are specific groupings of atoms within molecules that have distinct chemical properties. In each of these compounds, the chlorine atoms significantly impact the chemical behavior of the molecule.
  • Reactivity and Usage: Organic compounds vary widely in terms of reactivity. Some, like DDT, were initially utilized as pesticides, while others, such as carbon tetrachloride, serve as solvents in industrial applications.
The interplay of carbon, hydrogen, and chlorine in these compounds not only dictates their chemical and physical properties but also determines their utility and environmental impact. Understanding these basics is fundamental to mastering organic chemistry.

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Most popular questions from this chapter

Identify \(Z\) in the following series: \(\mathrm{CH}_{2}=\mathrm{CH}_{2} \stackrel{\mathrm{HBr}}{\longrightarrow} \mathrm{X} \stackrel{\text { Hydrolysis }}{\longrightarrow} \mathrm{Y} \stackrel{\mathrm{I}_{2}, \text { Excess }}{\longrightarrow} \mathrm{Z}\) (a) \(\mathrm{CHI}_{3}\) (b) \(\mathrm{CH}_{3} \mathrm{CHO}\) (c) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}\) (d) \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{OH}\)

Match the following: List I (Reactants) 1\. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br}+\) alc. \(\mathrm{KOH}\) 2\. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}(\mathrm{Br}) \mathrm{CH}_{3}+\) alc. \(\mathrm{KOH}\) 3\. \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}+\mathrm{alc} . \mathrm{KOH}\) List II (Products) A. cis-but-2-ene B. trans-but- 2 -ene C. but-1-ene D. 2 -methylprop-1-ene The correct matching is \(\begin{array}{lll}1 & 2 & 3\end{array}\) \(\begin{array}{lll}1 & 2 & 3\end{array}\) (a) \(\mathrm{C} \quad \mathrm{B} \mathrm{D}\) (b) \(\mathrm{C} \mathrm{A} \mathrm{D}\) (c) \(\mathrm{C}\) D A (d) \(\mathrm{B} \mathrm{A} \mathrm{C}\)

A compound \(\mathrm{P}\left(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{Cl}_{2}\right)\) on reaction with an alkali either gives a compound \(\mathrm{Q}\left(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}\right)\) or \(\mathrm{R}\left(\mathrm{C}_{3} \mathrm{H}_{4}\right)\). On oxidation, Q gives a compound \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{2} . \mathrm{R}\), on reacting with dilute \(\mathrm{H}_{2} \mathrm{SO}_{4}\) containing \(\mathrm{Hg}^{2+}\) ion, gives a compound \(\mathrm{S}\left(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}\right)\), which reacts with bromine and alkali to give sodium salt of \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}\). Then \(\mathrm{P}\) is (a) \(\mathrm{CH}_{2} \mathrm{ClCH}_{2} \mathrm{CH}_{2} \mathrm{Cl}\) (b) \(\mathrm{CH}_{3} \mathrm{CCl}_{2} \mathrm{CH}_{3}\) (c) \(\mathrm{CH}_{3} \mathrm{CHClCH}_{2} \mathrm{Cl}\) (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHCl}_{2}\)

CCC(C)Cl (a) meso-form (b) racemic mixture (c) d-form (d) l-form

The compound shown below is allowed to react with \(\mathrm{H}_{2} \mathrm{O} .\) Predict the number of optically active isomers formed in the reaction. CC(Br)C(Br)CCO Optically active isomers

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