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Ch.8 - Basic Concepts of Chemical Bonding
Chapter 8, Problem 103

The compound chloral hydrate, known in detective stories as knockout drops, is composed of 14.52% C, 1.83% H, 64.30% Cl, and 13.35% O by mass, and has a molar mass of 165.4 g/mol. (c) Draw the Lewis structure of the molecule, assuming that the Cl atoms bond to a single C atom and that there are a C–C bond and two C–O bonds in the compound.

Verified step by step guidance
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Step 1: Determine the empirical formula of chloral hydrate using the given mass percentages. Assume a 100 g sample, which gives 14.52 g C, 1.83 g H, 64.30 g Cl, and 13.35 g O. Convert these masses to moles by dividing by their respective atomic masses: C (12.01 g/mol), H (1.008 g/mol), Cl (35.45 g/mol), and O (16.00 g/mol).
Step 2: Calculate the mole ratio of each element by dividing the number of moles of each element by the smallest number of moles calculated in Step 1. This will give you the simplest whole number ratio of atoms in the compound, which is the empirical formula.
Step 3: Determine the molecular formula by comparing the molar mass of the empirical formula to the given molar mass of the compound (165.4 g/mol). Multiply the subscripts in the empirical formula by a whole number to match the molar mass of the compound.
Step 4: Using the molecular formula, draw the Lewis structure. Start by arranging the atoms according to the given bonding information: Cl atoms bond to a single C atom, and there are a C–C bond and two C–O bonds. Place the remaining H and O atoms to satisfy the valency of each atom.
Step 5: Complete the Lewis structure by adding lone pairs to satisfy the octet rule for each atom, especially the oxygen and chlorine atoms. Ensure that the total number of valence electrons used in the structure equals the sum of the valence electrons from all atoms in the molecular formula.

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Molecular Composition and Percent Composition

Understanding the molecular composition involves knowing how to interpret the percentage by mass of each element in a compound. In this case, chloral hydrate's composition indicates the relative amounts of carbon, hydrogen, chlorine, and oxygen. This information is crucial for determining the empirical formula and ultimately the molecular structure.
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Molar Mass and Molecular Formula

The molar mass of a compound is the mass of one mole of its entities, typically expressed in grams per mole. For chloral hydrate, the given molar mass of 165.4 g/mol helps in calculating the molecular formula from the empirical formula derived from the percent composition. This relationship is essential for accurately drawing the Lewis structure.
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Lewis Structures and Bonding

Lewis structures are diagrams that represent the bonding between atoms in a molecule and the lone pairs of electrons. They are essential for visualizing how atoms are connected and the arrangement of electrons. In this case, the instruction to assume specific bonds (C–C and C–O) guides the drawing of the Lewis structure for chloral hydrate, illustrating its molecular geometry.
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Related Practice
Textbook Question

Two compounds are isomers if they have the same chemical formula but different arrangements of atoms. Use Table 8.3 to estimate H for each of the following gas-phase isomerization reactions and indicate which isomer has the lower enthalpy. (d) Methyl isocyanide → Acetonitrile

Textbook Question

The electron affinity of oxygen is -141 kJ/mol, corresponding to the reaction O(g) + e- → O-(g). The lattice energy of K2O(s) is 2238 kJ/mol. Use these data along with data in Appendix C and Figure 7.11 to calculate the 'second electron affinity' of oxygen, corresponding to the reaction O-(g) + e- → O2-(g)

Textbook Question

Under special conditions, sulfur reacts with anhydrous liquid ammonia to form a binary compound of sulfur and nitrogen. The compound is found to consist of 69.6% S and 30.4% N. Measurements of its molecular mass yield a value of 184.3 g/mol. The compound occasionally detonates on being struck or when heated rapidly. The sulfur and nitrogen atoms of the molecule are joined in a ring. All the bonds in the ring are of the same length. (a) Calculate the empirical and molecular formulas for the substance.

Textbook Question

A common form of elemental phosphorus is the tetrahedral P4 molecule, where all four phosphorus atoms are equivalent: 

(b) How many P-P bonds are there in the molecule?

Textbook Question

A common form of elemental phosphorus is the tetrahedral P4 molecule, where all four phosphorus atoms are equivalent: 

Draw a Lewis structure for a linear P4 molecule that satisfies the octet rule. Does this molecule have resonance structures?