Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 5: Problem 128

Cyclopropane, a gas that when mixed with oxygen is used as a general anesthetic, is composed of \(85.7 \% \mathrm{C}\) and \(14.3 \% \mathrm{H}\) by mass. If the density of cyclopropane is \(1.88 \mathrm{~g} / \mathrm{L}\) at STP, what is the molecular formula of cyclopropane?

Short Answer

Expert verified
The molecular formula of cyclopropane is \(\mathrm{C}_3\mathrm{H}_6\).

Step by step solution

01

Calculate moles of Carbon and Hydrogen

Assume that we have \(100 \mathrm{~g}\) of Cyclopropane. Moles of Carbon (\(\mathrm{C}\)): \(\% \mathrm{C} = 85.7 \%\) means it has \(85.7 \mathrm{~g}\) of Carbon. The atomic mass of Carbon is \(12.01 \mathrm{~g}/\mathrm{mol}\), so the moles of Carbon are: \[\frac{85.7 \mathrm{~g}}{12.01 \mathrm{~g}/\mathrm{mol}} = 7.14 \mathrm{~mol} \] Moles of Hydrogen (\(\mathrm{H}\)): \(\% \mathrm{H} = 14.3 \%\) means it has \(14.3 \mathrm{~g}\) of Hydrogen. The atomic mass of Hydrogen is \(1.01 \mathrm{~g}/\mathrm{mol}\), so the moles of Hydrogen are: \[\frac{14.3 \mathrm{~g}}{1.01 \mathrm{~g}/\mathrm{mol}} = 14.16 \mathrm{~mol} \]
02

Find the empirical formula

To find the empirical formula, we need to find the simplest mole ratio between Carbon and Hydrogen. Divide each mole value by their smallest common value. Mole ratio: \[\frac{7.14}{7.14} \mathrm{C} : \frac{14.16}{7.14} \mathrm{H} = 1 \mathrm{C} : 1.99 \approx 2 \mathrm{H} \] Thus, the empirical formula is \(\mathrm{CH}_2\).
03

Calculate the molar mass

Now we'll use the given density of cyclopropane and the Ideal Gas Law. The Ideal Gas Law is \[PV = nRT\] where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is temperature in kelvins. At STP, P = 1 atm and T = 273 K. The gas constant R is 0.0821 L.atm/mol.K. Let's assume 1 L (volume) of cyclopropane. Mass = Density × Volume = \(1.88 \mathrm{~g}\) Number of moles, \(n = \frac{Mass}{Molar~mass}\) From the Ideal Gas Law: \[Molar~mass = \frac{Mass × R × T}{P × V} \] Substitute the values and calculate the molar mass: \[Molar~mass = \frac{1.88 \mathrm{~g} × 0.0821 \mathrm{L.atm/mol.K} × 273 \mathrm{K}}{1\mathrm{atm} × 1\mathrm{L}} = 42.08 \mathrm{~g/mol} \]
04

Determine the molecular formula

Molecular formula = Empirical formula × n First, calculate the molar mass of the empirical formula \(\mathrm{CH}_2\): Molar mass of the empirical formula = \((1\times 12.01) + (2\times 1.01) = 14.03 \mathrm{~g/mol}\) To find n, divide the molar mass of the molecular formula by the molar mass of the empirical formula: \[n = \frac{42.08 \mathrm{~g/mol}}{14.03 \mathrm{~g/mol}} \approx 3 \] Therefore, the molecular formula of Cyclopropane is: \(\mathrm{CH}_2 \times 3 = \mathrm{C}_3\mathrm{H}_6\)

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.

Empirical and Molecular Formulas
Understanding the difference between empirical and molecular formulas is fundamental in chemistry. An empirical formula represents the simplest whole-number ratio of elements in a compound. For example, ethylene has an empirical formula of CH2, indicating for every carbon atom, there are two hydrogen atoms.

The molecular formula, on the other hand, conveys the actual number of atoms of each element in a molecule. It could be the same as the empirical formula or a multiple of it. In the case of cyclopropane, its empirical formula is CH2 while its molecular formula is C3H6. The molecular formula tells us that cyclopropane contains three times the number of atoms indicated by its empirical formula.
Mole Concept
The mole concept is a fundamental aspect of chemical quantification, serving as a bridge between the microscopic world of atoms and molecules and the macroscopic world we observe. One mole contains Avogadro's number of entities, which is approximately 6.022 x 1023.

When calculating moles from a given mass, we use the formula:
\[ \text{Moles} = \frac{\text{given mass (g)}}{\text{molar mass (g/mol)}} \]
As illustrated in solving the cyclopropane problem, the number of moles of carbon and hydrogen was determined using their respective atomic masses, which led to finding the empirical formula.
Ideal Gas Law
The ideal gas law is a crucial equation in chemistry that relates the pressure (P), volume (V), temperature (T), and number of moles (n) of a gas through the equation \( PV = nRT \), where R is the gas constant. This law allows for the determination of one of these variables if the others are known. Using the ideal gas law, one can deduce properties such as the density and molar mass of gases under certain conditions, as was done with cyclopropane at standard temperature and pressure (STP).

For our cyclopropane example, the molar mass calculation at STP provided key information needed to obtain the molecular formula from the empirical formula.
Stoichiometry
Stoichiometry refers to the quantitative aspect of chemical formulas and reactions. It's the calculation involving the masses, volumes, and number of moles in a chemical process. Central to stoichiometry is the law of conservation of mass, which states that in a chemical reaction, matter is neither created nor destroyed, only transformed.

By employing stoichiometric calculations, we paired the empirical formula of cyclopropane with its molar mass to uncover its actual molecular makeup. This quantitative agreement allowed us to affirm that one mole of cyclopropane gas has a mass of 42.08 grams, providing a concrete example of stoichiometry in action.

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

Ethene is converted to ethane by the reaction $$ \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) \stackrel{\text { Catalyst }}{\longrightarrow} \mathrm{C}_{2} \mathrm{H}_{6}(g) $$ \(\mathrm{C}_{2} \mathrm{H}_{4}\) flows into a catalytic reactor at \(25.0 \mathrm{~atm}\) and \(300 .{ }^{\circ} \mathrm{C}\) with a flow rate of \(1000 . \mathrm{L} / \mathrm{min}\). Hydrogen at \(25.0 \mathrm{~atm}\) and \(300 .{ }^{\circ} \mathrm{C}\) flows into the reactor at a flow rate of \(1500 . \mathrm{L} / \mathrm{min}\). If \(15.0 \mathrm{~kg}\) \(\mathrm{C}_{2} \mathrm{H}_{6}\) is collected per minute, what is the percent yield of the reaction?

Consider an equimolar mixture (equal number of moles) of two diatomic gases \(\left(\mathrm{A}_{2}\right.\) and \(\mathrm{B}_{2}\) ) in a container fitted with a piston. The gases react to form one product (which is also a gas) with the formula \(\mathrm{A}_{x} \mathrm{~B}_{y}\). The density of the sample after the reaction is complete (and the temperature returns to its original state) is \(1.50\) times greater than the density of the reactant mixture. a. Specify the formula of the product, and explain if more than one answer is possible based on the given data. b. Can you determine the molecular formula of the product with the information given or only the empirical formula?

Consider separate \(1.0\) -L gaseous samples of \(\mathrm{H}_{2}, \mathrm{Xe}, \mathrm{Cl}_{2}\), and \(\mathrm{O}_{2}\) all at STP. a. Rank the gases in order of increasing average kinetic energy. b. Rank the gases in order of increasing average velocity. c. How can separate \(1.0\) -L samples of \(\mathrm{O}_{2}\) and \(\mathrm{H}_{2}\) each have the same average velocity?

Suppose two 200.0-L tanks are to be filled separately with the gases helium and hydrogen. What mass of each gas is needed to produce a pressure of \(2.70 \mathrm{~atm}\) in its respective tank at \(24^{\circ} \mathrm{C} ?\)

A steel cylinder contains \(150.0\) moles of argon gas at a temperature of \(25^{\circ} \mathrm{C}\) and a pressure of \(8.93 \mathrm{MPa}\). After some argon has been used, the pressure is \(2.00 \mathrm{MPa}\) at a temperature of \(19^{\circ} \mathrm{C}\). What mass of argon remains in the cylinder?
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Chemical Reactions

Read Explanation

Inorganic Chemistry

Read Explanation

Kinetics

Read Explanation

Physical Chemistry

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