/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 69 Several important compounds cont... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 69

Several important compounds contain only nitrogen and oxygen. Place the following compounds in order of increasing mass percent of nitrogen. a. NO, a gas formed by the reaction of \(\mathrm{N}_{2}\) with \(\mathrm{O}_{2}\) in internal combustion engines b. \(\mathrm{NO}_{2}\), a brown gas mainly responsible for the brownish color of photochemical smog c. \(\mathrm{N}_{2} \mathrm{O}_{4}\), a colorless liquid used as fuel in space shuttles d. \(\mathrm{N}_{2} \mathrm{O}\), a colorless gas sometimes used as an anesthetic by dentists (known as laughing gas)

Short Answer

Expert verified
The order of the compounds based on increasing mass percent of nitrogen is: NO鈧� (30.45%), N鈧侽鈧� (30.45%), NO (46.68%), and N鈧侽 (63.62%).

Step by step solution

01

Determine the Molar Mass of Nitrogen and Oxygen

The first step is to determine the molar mass of nitrogen (N) and oxygen (O). Using the periodic table, we get the following values: Molar mass of Nitrogen (N) = 14.01 g/mol Molar mass of Oxygen (O) = 16.00 g/mol
02

Calculate the Mass Percentage of Nitrogen in NO

Now we will calculate the mass percentage of nitrogen in NO. The molar mass of NO = molar mass of N + molar mass of O = 14.01 + 16.00 = 30.01 g/mol Mass percentage of Nitrogen in NO = \(\frac{14.01}{30.01}\) 脳 100 = 46.68%
03

Calculate the Mass Percentage of Nitrogen in NO鈧�

Next, we will determine the mass percentage of Nitrogen in NO鈧�. The molar mass of NO鈧� = molar mass of N + 2 脳 molar mass of O = 14.01 + 2 脳 16.00 = 46.01 g/mol Mass percentage of Nitrogen in NO鈧� = \(\frac{14.01}{46.01}\) 脳 100 = 30.45%
04

Calculate the Mass Percentage of Nitrogen in N鈧侽鈧�

Now, let's find the mass percentage of Nitrogen in N鈧侽鈧�. The molar mass of N鈧侽鈧� = 2 脳 molar mass of N + 4 脳 molar mass of O = 2 脳 14.01 + 4 脳 16.00 = 92.02 g/mol Mass percentage of Nitrogen in N鈧侽鈧� = \(\frac{2 \times 14.01}{92.02}\) 脳 100 = 30.45%
05

Calculate the Mass Percentage of Nitrogen in N鈧侽

Finally, we will calculate the mass percentage of Nitrogen in N鈧侽. The molar mass of N鈧侽 = 2 脳 molar mass of N + molar mass of O = 2 脳 14.01 + 16.00 = 44.02 g/mol Mass percentage of Nitrogen in N鈧侽 = \(\frac{2 \times 14.01}{44.02}\) 脳 100 = 63.62%
06

Arrange the Compounds in Order of Increasing Mass Percent of Nitrogen

Now we will arrange the compounds in order of increasing mass percentage of nitrogen: 1. NO鈧�: 30.45% 2. N鈧侽鈧�: 30.45% 3. NO: 46.68% 4. N鈧侽: 63.62% So, the order is NO鈧�, N鈧侽鈧�, NO, and N鈧侽.

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Understanding Mass Percent Composition
Mass percent composition helps us understand how much of a specific element is present in a compound compared to the entire mass of the compound. To calculate it, we need the molar mass of each element in the compound. First, add up the molar mass of all elements to find the total molar mass of the compound. Next, take the molar mass of the element of interest and divide it by the total molar mass. Multiply by 100 to convert to a percentage.

This calculation tells us the proportion of a specific element within the compound, which is crucial for understanding the nature and function of the compound.
Exploring Nitrogen Compounds
Nitrogen compounds play a significant role in various chemical processes and industries. Compounds like NO and NO鈧� are common in environmental chemistry, especially in air pollution.

N鈧侽鈧� is used in aerospace engineering, highlighting how nitrogen's versatile chemical properties can be utilized. On the other hand, N鈧侽, commonly known as laughing gas, is used in medicine.

Understanding these compounds' composition can provide insights into their chemical behavior and applications.
Decoding Chemical Formulas
Chemical formulas are shorthand notations representing the elements in a compound. For instance, NO means one nitrogen atom is bonded to one oxygen atom. In NO鈧�, one nitrogen atom is bonded to two oxygen atoms. Similarly, N鈧侽 shows two nitrogen atoms bonded with one oxygen atom, while N鈧侽鈧� displays two nitrogen atoms with four oxygen atoms.

These formulas help us quickly identify the atomic composition of a compound and are crucial for calculating properties like molar mass and mass percent composition.
Getting to Know Periodic Table Elements
The periodic table is an essential tool in chemistry that organizes elements based on their atomic number and properties. Understanding where nitrogen (N) and oxygen (O) are located helps us understand their properties.

Nitrogen, with an atomic number of 7, is a non-metal that forms strong bonds with oxygen. Oxygen, atomic number 8, is also a non-metal and highly reactive.

The periodic table gives us the molar masses needed for our calculations, making it crucial for determining the mass percent composition of compounds.

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

A compound contains only carbon, hydrogen, nitrogen, and oxygen. Combustion of \(0.157 \mathrm{~g}\) of the compound produced \(0.213 \mathrm{~g}\) \(\mathrm{CO}\), and \(0.0310 \mathrm{~g} \mathrm{H}_{2} \mathrm{O} .\) In another experiment, it is found that \(0.103 \mathrm{~g}\) of the compound produces \(0.0230 \mathrm{~g} \mathrm{NH}_{3} .\) What is the empirical formula of the compound? Hint: Combustion involves reacting with excess \(\mathrm{O}_{2}\). Assume that all the carbon ends up in \(\mathrm{CO}_{2}\) and all the hydrogen ends up in \(\mathrm{H}_{2} \mathrm{O}\). Also assume that all the nitrogen ends up in the \(\mathrm{NH}_{3}\) in the second experiment.

Consider the following data for three binary compounds of hydrogen and nitrogen: $$ \begin{array}{lcc} & \% \mathrm{H} \text { (by Mass) } & \text { \% N (by Mass) } \\ \hline \text { I } & 17.75 & 82.25 \\ \text { II } & 12.58 & 87.42 \\ \text { III } & 2.34 & 97.66 \end{array} $$ When \(1.00 \mathrm{~L}\) of each gaseous compound is decomposed to its elements, the following volumes of \(\mathrm{H}_{2}(g)\) and \(\mathrm{N}_{2}(g)\) are obtained: $$ \begin{array}{lcc} & \mathrm{H}_{2} \text { (L) } & \mathrm{N}_{2} \text { (L) } \\ \hline \text { I } & 1.50 & 0.50 \\ \text { II } & 2.00 & 1.00 \\ \text { III } & 0.50 & 1.50 \end{array} $$ Use these data to determine the molecular formulas of compounds I, II, and III and to determine the relative values for the atomic masses of hydrogen and nitrogen.

Tetrodotoxin is a toxic chemical found in fugu pufferfish, a popular but rare delicacy in Japan. This compound has a \(\mathrm{LD}_{50}\) (the amount of substance that is lethal to \(50 . \%\) of a population sample) of \(10 . \mu \mathrm{g}\) per \(\mathrm{kg}\) of body mass. Tetrodotoxin is \(41.38 \%\) carbon by mass, \(13.16 \%\) nitrogen by mass, and \(5.37 \%\) hydrogen by mass, with the remaining amount consisting of oxygen. What is the empirical formula of tetrodotoxin? If three molecules of tetrodotoxin have a mass of \(1.59 \times 10^{-21} \mathrm{~g}\), what is the molecular formula of tetrodotoxin? What number of molecules of tetrodotoxin would be the LD \(_{50}\) dosage for a person weighing \(165 \mathrm{lb}\) ?

Consider the following unbalanced equation: \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{CaSO}_{4}(s)+\mathrm{H}_{3} \mathrm{PO}_{4}(a q)\) What masses of calcium sulfate and phosphoric acid can be produced from the reaction of \(1.0 \mathrm{~kg}\) calcium phosphate with \(1.0\) \(\mathrm{kg}\) concentrated sulfuric acid \(\left(98 \% \mathrm{H}_{2} \mathrm{SO}_{4}\right.\) by \(\left.\mathrm{mass}\right)\) ?

DDT, an insecticide harmful to fish, birds, and humans, is produced by the following reaction: $$ 2 \mathrm{C}_{6} \mathrm{H}_{3} \mathrm{Cl}+\mathrm{C}_{2} \mathrm{HOCl}_{3} \longrightarrow \mathrm{C}_{14} \mathrm{H}_{4} \mathrm{Cl}_{5}+\mathrm{H}_{2} \mathrm{O} $$ \(\begin{array}{ll}\text { orobenzenc chloral } & \mathrm{D}\end{array}\) In a government lab, \(1142 \mathrm{~g}\) of chlorobenzene is reacted with \(485 \mathrm{~g}\) of chloral. a. What mass of DDT is formed? b. Which reactant is limiting? Which is in excess? c. What mass of the excess reactant is left over? d. If the actual yield of DDT is \(200.0 \mathrm{~g}\), what is the percent yield?
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Inorganic Chemistry

Read Explanation

Kinetics

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.