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Chapter 21: Problem 148

Consider a sample of a hydrocarbon at 0.959 atm and \(298 \mathrm{K}\). Upon combusting the entire sample in oxygen, you collect a mixture of gaseous carbon dioxide and water vapor at 1.51 atm and 375 K. This mixture has a density of 1.391 g/L and occupies a volume four times as large as that of the pure hydrocarbon. Determine the molecular formula of the hydrocarbon and name it.

Short Answer

Expert verified
The molecular formula of the hydrocarbon is \(C_3H_8\), and its name is propane.

Step by step solution

01

Find the moles of carbon dioxide and water vapor produced

We are given the density, pressure, and temperature of the carbon dioxide and water vapor mixture after the combustion of the hydrocarbon. We can use the ideal gas law and the density formula to determine the moles of the mixture. Since we know that the density \(\rho = \frac{m}{V}\), we can rewrite the ideal gas law, \(PV = nRT\), with density as follows: \(PV = \frac{m}{\rho}RT\) Now, solve for the mass, m, of the combustion products: \(m = \rho V\) The given mixture has a density of \(\rho = 1.391 \,\text{g/L}\) and occupies a volume four times as large as that of the pure hydrocarbon. Therefore, the volume of the combustion products \(V = 4V_h\), where \(V_h\) is the volume of the hydrocarbon sample. By plugging in the density, volume, pressure, and temperature, we can calculate the moles and mass of the combustion products.
02

Determine moles and mass of carbon dioxide produced

From the combustion of the hydrocarbon, we produce carbon dioxide and water vapor. Let's assume that we have produced x moles of carbon dioxide and y moles of water vapor. Since we know the mass of carbon dioxide and water vapor produced, we can use their molar masses to determine the number of moles of each substance: \(m_{CO2} = x \cdot M_{CO2}\) \(m_{H2O} = y \cdot M_{H2O}\) Here, \(M_{CO2}\) and \(M_{H2O}\) are the molar masses of carbon dioxide and water vapor, respectively.
03

Determine the molecular formula of the hydrocarbon

Knowing the moles and mass of carbon dioxide and water vapor produced, we can express the moles of carbon and hydrogen in the hydrocarbon using stoichiometry: Let the molecular formula of the hydrocarbon be \(C_nH_m\). Then, the balanced combustion equation will be: \(C_nH_m+ \frac{n+m}{4} O_2 \rightarrow n CO_2 + \frac{m}{2} H_2O\) For each mole of the hydrocarbon produced, n moles of carbon dioxide and \(m/2\) moles of water vapor are produced. Therefore, we can write the moles of carbon and hydrogen in the hydrocarbon as: \(n_h = \frac{x}{n}\) \(m_h = \frac{2y}{m}\) From the given initial conditions of the hydrocarbon, we have its pressure \(P_h = 0.959 \,\text{atm}\) and temperature \(T_h = 298 \,\mathrm{K}\). Now, using the ideal gas law for the hydrocarbon, we can write: \(P_hV_h = n_hRT_h\) By knowing the values of \(P_h\), \(T_h\), \(R\), \(x\), \(y\), \(n\), and \(m\), we can solve for the volume of the hydrocarbon sample, \(V_h\). Finally, we can determine the molecular formula of the hydrocarbon by finding the integer values of n and m that satisfy the stoichiometry, moles, and mass equations. This will give us the molecular formula of the hydrocarbon and its name.

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Key Concepts

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

Molecular Formula Determination
Molecular formula determination refers to the process of identifying the actual number of atoms of each element in a molecule. This involves using experimental data from methods such as combustion analysis, mass spectrometry, or percentages of elements by mass. In our combustion analysis example, the molecular formula of a hydrocarbon is deduced from the moles of carbon dioxide and water produced during combustion.

By knowing the number of moles of carbon and hydrogen atoms present in the CO2 and H2O, respectively, we can infer the stoichiometry of the original hydrocarbon. If one mole of the hydrocarbon yields 'n' moles of CO2 and 'm/2' moles of H2O, then the empirical formula of the hydrocarbon can be represented as CHm/n. However, to determine the molecular formula, we must consider the molar mass of the hydrocarbon, which could be an integer multiple of the empirical formula's molar mass. Therefore, accurate mass measurements are crucial.

Calculating Moles from Combustion Products

By using the ideal gas law and the mass of the mixture obtained from the density, we can calculate the total moles of the gases after combustion. This helps us backtrack to the stoichiometry of the hydrocarbon. Subsequently, the molecular formula is found by determining the simplest whole number ratio of carbon to hydrogen atoms in the hydrocarbon, then adjusting the empirical formula based on the molecular weight.
Ideal Gas Law
The ideal gas law is a fundamental equation relating the pressure (P), volume (V), temperature (T), and number of moles (n) of an ideal gas, where R is the gas constant. The equation is expressed as:
\[PV = nRT\].

Understanding the ideal gas law is essential in solving many problems in chemistry, including combustion analysis. When a substance combusts to form gases, if the conditions approximate an ideal gas scenario, the ideal gas law can help determine the number of moles of gas produced based on the known pressure, volume, and temperature conditions post-combustion. As demonstrated in the example, by rearranging the ideal gas law formula to include density (\(\rho\)) and mass (m), we can find the mass and volume of the combustion products:
\[m = \rho V\].

Applying the Ideal Gas Law in Calculations

The law is used multiple times: first to find the volume of the pure hydrocarbon and then to find the mass of the combustion products. This iterative application can directly lead to the mole calculation, and thus, contribute significantly to determining the compound's molecular formula.
Stoichiometry
Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It is based on the conservation of mass and the mole concept. For a balanced chemical equation, the mole ratio of any two substances can be obtained, which is then used to calculate required masses, volumes, or numbers of moles of various components.

In the context of our hydrocarbon combustion example, stoichiometry enables us to relate the moles of carbon dioxide and water (the combustion products) back to the moles of the hydrocarbon (the reactant). By using the balanced chemical equation for the combustion reaction, we can establish the ratio in which the reactants combine and the products form, allowing us to piece together the hydrocarbon's molecular formula.

Converting Moles of Products to Reactant Composition

For example, in the balanced equation, every 'n' moles of CO2 corresponds to n moles of carbon in the hydrocarbon. Similarly, every 'm/2' moles of H2O corresponds to m moles of hydrogen in the hydrocarbon. Stoichiometry helps to find the exact n and m values that reflect the molecular formula of the hydrocarbon. Hence, the determination of the hydrocarbon's identity hinges on accurate stoichiometric calculations.

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Most popular questions from this chapter

Integrative Problems. These problems require the integration of multiple concepts to find the solutions. An organometallic compound is one containing at least one metal-carbon bond. An example of an organometallic species is \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right) \mathrm{MBr},\) which contains a metal-cthyl bond. How does the hybridization of the starred carbon atom change, if at all, in going from reactants to products? c. What is the systematic name of the product? (Hint: In this shorthand notation, all the \(\mathrm{C}-\mathrm{H}\) bonds have been eliminated and the lines represent \(\mathrm{C}-\mathrm{C}\) bonds, unless shown differently. As is typical of most organic compounds, cach carbon atom has four bonds to it and the oxygen atoms have only two bonds.) a. If \(\mathrm{M}^{2+}\) has the electron configuration \([\mathrm{Ar}] 3 d^{10},\) what is the percent by mass of \(\mathrm{M}\) in \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right) \mathrm{MBr} ?\) b. A reaction involving \(\left(\mathrm{CH}, \mathrm{CH}_{2}\right) \mathrm{MBr}\) is the conversion of a ketone to an alcohol as illustrated here:

Marathon Problems. These problems are designed to incorporate several concepts and techniques into one situation. For each of the following, fill in the blank with the correct response. All of these fill-in-the-blank problems pertain to material covered in the sections on alkanes, alkenes and alkynes, aromatic hydrocarbons, and hydrocarbon derivatives. a. The first "organic" compound to be synthesized in the laboratory, rather than being isolated from nature, was _____________which was prepared from____________. b. An organic compound whose carbon-carbon bonds are all single bonds is said to be______. c. The general orientation of the four pairs of electrons around the carbon atoms in alkanes is_______. d. Alkanes in which the carbon atoms form a single unbranched chain are said to be_______alkanes. e. Structural isomerism occurs when two molecules have the same number of each type of atom but exhibit different arrangements of the_______between those atoms. f. The systematic names of all saturated hydrocarbons have the ending_______ added to a root name that indicates the number of carbon atoms in the molecule. g. For a branched hydrocarbon, the root name for the hydrocarbon comes from the number of carbon atoms in the _________continuous chain in the molecule. h. The positions of substituents along the hydrocarbon framework of a molecule are indicated by the ________of the carbon atom to which the substituents are attached. i. The major use of alkanes has been in_______reactions,as a source of heat and light. J. With very reactive agents, such as the halogen elements, alkanes undergo _______reactions, whereby a new atom replaces one or more hydrogen atoms of the alkane. k. Alkenes and alkynes are characterized by their ability to undergo rapid, complete ______ reactions, by which other atoms attach themselves to the carbon atoms of the double or triple bond. 1\. Unsaturated fats may be converted to saturated fats by the process of ________ m. Benzene is the parent member of the group of hydrocarbons called ________ hydrocarbons. n. An atom or group of atoms that imparts new and characteristic properties to an organic molecule is called a ___________ group. 0.4 _________alcohol is one in which there is only one hydrocarbon group attached to the carbon atom holding the hydroxyl group. p. The simplest alcohol, methanol, is prepared industrially by the hydrogenation of ________ q. Ethanol is commonly prepared by the __________of certain sugars by yeast. r. Both aldehydes and ketones contain the _______group but they differ in where this group occurs along the hydrocarbon chain. s. Aldehydes and ketones can be prepared by _______ of the corresponding alcohol. t. Organic acids, which contain the __________ group, are typically weak acids. u. The typically sweet-smelling compounds called ____________ result from the condensation reaction of an organic acid with an___________.

Cumene is the starting material for the industrial production of acetone and phenol. The structure of cumene is Give the systematic name for cumene.

A confused student was doing an isomer problem and listed the following six names as different structural isomers of \(\mathbf{C}_{7} \mathbf{H}_{16}.\) a. \(1-\) sec-butylpropane d. 1 -ethyl- 1 -methylbutane b. 4 -methylhexane e. 3 -methylhexane c. 2 -ethylpentane f. 4 -ethylpentane How many different structural isomers are actually present in these six names?

If one hydrogen in a hydrocarbon is replaced by a halogen atom, the number of isomers that exist for the substituted compound depends on the number of types of hydrogen in the original hydrocarbon. Thus there is only one form of chloroethane (all hydrogens in ethane are equivalent), but there are two isomers of propane that arise from the substitution of a methyl hydrogen or a methylene hydrogen. How many isomers can be obtained when one hydrogen in each of the compounds named below is replaced by a chlorine atom? a. \(n\) -pentane c. 2,4 -dimethylpentane b. 2 -methylbutane d. methylcyclobutane
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