Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 24: Problem 47

An organic compound is found to contain 37.5 percent carbon, 3.2 percent hydrogen, and 59.3 percent fluorine by mass. The following pressure and volume data were obtained for \(1.00 \mathrm{~g}\) of this substance at \(90^{\circ} \mathrm{C}\) $$ \begin{array}{cc} \hline P(\mathrm{~atm}) & V(\mathrm{~L}) \\ \hline 2.00 & 0.332 \\ 1.50 & 0.409 \\ 1.00 & 0.564 \\ 0.50 & 1.028 \\ \hline \end{array} $$ The molecule is known to have no dipole moment. (a) What is the empirical formula of this substance? (b) Does this substance behave as an ideal gas? (c) What is its molecular formula? (d) Draw the Lewis structure of this molecule and describe its geometry. (e) What is the systematic name of this compound?

Short Answer

Expert verified
The substance has the empirical formula CFH, does not behave as an ideal gas, has the molecular formula C3F3H3, has a planar geometry with a 120° bond angle, and is named 1,1,1-trifluoroethane.

Step by step solution

01

Calculate the mole ratio

From the percent composition, calculate the number of moles of each element. Use the atomic masses of each element (C = 12.01 g/mol, H = 1.01 g/mol, F = 18.99 g/mol). The number of moles is given by mass (g) divided by atomic mass (g/mol). For Carbon: \((37.5 / 12.01) = 3.12\) mol, Hydrogen: \((3.2 / 1.01) = 3.17\) mol, Fluorine: \((59.3 / 18.99) = 3.12\) mol. The ratio of C:H:F is approximately 1:1:1.
02

Empirical Formula

The empirical formula is the simplest whole number ratio of moles of each element in the compound. Here, it is CFH.
03

Ideal Gas Behavior

According to ideal gas behavior, at constant temperature, the volume of a gas is inversely proportional to its pressure (Boyle’s Law). If the given data is plotted, it should yield a straight line. The data points in this case do not lie on a straight line, suggesting that the gas does not behave ideally.
04

Molecular Formula

If this substance behaves non-ideally, it means that its particles are interacting, and hence, likely to be heavier than predicted by the empirical formula. To confirm this, we can calculate the molecular weight of the gas using the ideal gas law \(PV = nRT\), where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. Converting all the given units to their SI equivalents, and rearranging the formula to solve for n gives \(n = PV / RT = (2.00 atm \times 0.332 L) / (0.0821 L.atm/mol.K \times 90 + 273 K) \approx 0.028 mol. Since there is 1.00 g of substance, this corresponds to a molecular weight of \(1.00 g / 0.028 mol = 35.7 g/mol.\) This is approximately three times the molecular weight of the empirical formula (CHF = 12.01 + 1.01 + 18.99 = 33.01 g/mol), hence the molecular formula of the substance is (CFH)3 or C3F3H3.
05

Lewis Structure and Geometry

The Lewis structure will involve drawing three CHF units linked together. Each Carbon (C) atom will be connected to one Hydrogen (H) and one Fluorine (F) atom, and the three Carbon atoms will be linked in a chain. The molecule will take a planar geometry with a 120° bond angle due to sp2 hybridization.
06

Systematic Name

The systematic name of this compound according to the IUPAC nomenclature for organic compounds would be 1,1,1-trifluoroethane.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Percent Composition
Percent composition describes the percentage by mass of each element in a compound. It is a fundamental aspect of chemistry that allows chemists to determine the empirical formula of a compound, which is the simplest whole number ratio of atoms in the molecule. In a solved exercise, if a compound consists of 37.5 percent carbon, 3.2 percent hydrogen, and 59.3 percent fluorine, one would first convert these percentages to mass values assuming a certain mass of the compound, often 100 g for simplicity.

From these mass values, the number of moles for each element can be calculated by dividing the mass of the element by its molar mass. The ratios of the moles of each element to each other yield the empirical formula. For instance, the empirical formula for a compound with an equal molar ratio of carbon, hydrogen, and fluorine would be CHF.
Ideal Gas Law
The ideal gas law is represented by the equation \(PV = nRT\), where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature in Kelvin. This law combines several gas laws, including Boyle's, Charles's, and Avogadro's, into one formula that can predict the behavior of an ideal gas under various conditions.

In the exercise example, by using the ideal gas law, we can determine if the substance behaves like an ideal gas. Discrepancies in behavioral expectations from the ideal gas law may suggest real-gas effects, such as molecule interactions, which indicate that the substance does not adhere strictly to the law. For real applications, such results prompt further exploration into the nature of the substance, taking into account factors like intermolecular forces.
Molecular Geometry
Molecular geometry refers to the three-dimensional arrangement of atoms within a molecule. The geometry affects the molecule's chemical reactivity, polarity, color, phase, magnetism, and biological activity. A molecule with no dipole moment, like the one found in the given problem, is likely to be symmetrical.

Common geometries include linear, bent, trigonal planar, tetrahedral, and others, which are determined based on the valence-shell electron-pair repulsion (VSEPR) theory. Geometry can predict properties like the bond angles within a molecule as well, which are key factors in understanding molecular shape. In the case of our CHF compound, the Lewis structure and lack of dipole moment suggest a planar geometry with equal bond angles of 120°.
Lewis Structure
A Lewis structure is a diagram that shows the bonding between atoms of a molecule and the lone pairs of electrons that may exist in the molecule. It provides a visual representation of the valence electrons, allowing chemists to predict many properties of the molecule, such as shape, polarity, and reactivity.

For the compound CFH, the Lewis structure would involve a central carbon atom with a single bond to both hydrogen and fluorine. If the substance was made up of three CFH units, as suggested by its molecular formula obtained through the ideal gas law, these could be displayed as a chain with carbons bonded to each other and each having one hydrogen and one fluorine atom attached.
IUPAC Nomenclature
The International Union of Pure and Applied Chemistry (IUPAC) nomenclature is a standardized naming system for chemical substances. This system allows chemists from all over the world to communicate clearly about molecular structures and their components without ambiguity. The IUPAC name for a compound is based on the type and number of each atom present in the substance.

In our example, the systematic name for a compound composed of three CFH units would reflect the presence of fluorine atoms and the carbon backbone chain. The correct IUPAC name, 1,1,1-trifluoroethane, encompasses the presence of three fluorine atoms (trifluoro) and a two-carbon alkane base (ethane). Understanding IUPAC nomenclature helps chemists to both name new compounds and to deduce the structure of a compound from its name.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Fat and oil are names for the same class of compounds, called triglycerides, which contain three ester groups where \(\mathrm{R}, \mathrm{R}^{\prime},\) and \(\mathrm{R}^{\prime \prime}\) represent long hydrocarbon chains. (a) Suggest a reaction that leads to the formation of a triglyceride molecule, starting with glycerol and carboxylic acids (see Section 11.3 for structure of glycerol). (b) In the old days, soaps were made by hydrolyzing animal fat with lye (a sodium hydroxide solution). Write an equation for this reaction. (c) The difference between fats and oils is that at room temperature, the former are solids and the latter are liquids. Fats are usually produced by animals, whereas oils are commonly found in plants. The melting points of these substances are determined by the number of \(\mathrm{C}=\mathrm{C}\) bonds (or the extent of unsaturation) present-the larger the number of \(\mathrm{C}=\mathrm{C}\) bonds, the lower the melting point and the more likely that the substance is a liquid. Explain. (d) One way to convert liquid oil to solid fat is to hydrogenate the oil, a process by which some or all of the \(\mathrm{C}=\mathrm{C}\) bonds are converted to \(\mathrm{C}-\mathrm{C}\) bonds. This procedure prolongs shelf life of the oil by removing the more reactive \(\mathrm{C}=\mathrm{C}\) group and facilitates packaging. How would you carry out such a process (that is, what reagents and catalyst would you employ)? (e) The degree of unsaturation of oil can be determined by reacting the oil with iodine, which reacts with the \(\mathrm{C}=\mathrm{C}\) bond as follows:

What is the difference between aliphatic and aromatic hydrocarbons?

Benzene and cyclohexane molecules both contain six-membered rings. Benzene is a planar molecule, and cyclohexane is nonplanar. Explain.

Complete the following equation and identify the products: $$ \mathrm{HCOOH}+\mathrm{CH}_{3} \mathrm{OH} \longrightarrow $$

Name the classes to which the following compounds belong: (a) \(\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{OH}\) (b) \(\mathrm{CH}_{3} \mathrm{OC}_{2} \mathrm{H}_{5}\) (c) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{CHO}\) (d) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\) (e) \(\mathrm{CH}_{3} \mathrm{NH}_{2}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemistry Branches

Read Explanation

Chemical Reactions

Read Explanation

Making Measurements

Read Explanation

The Earths Atmosphere

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free