Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 3: Problem 27

Without doing detailed calculations, arrange the following in order of increasing \(\% \mathrm{Cr},\) by mass, and explain your reasoning: \(\mathrm{CrO}, \mathrm{Cr}_{2} \mathrm{O}_{3}, \mathrm{CrO}_{2}, \mathrm{CrO}_{3}\)

Short Answer

Expert verified
The compounds in increasing order by Chromium mass percentage are: CrO, Cr2O3, CrO2, CrO3.

Step by step solution

01

Understand the Chemistry

First, we need to understand that the higher the number of Oxygen atoms in the compound, the lesser will be the percentage of Chromium. This is due to the addition of the mass of the Oxygen atoms which will cause the percentage by mass of Chromium to decrease.
02

Get the Cr and O Atomic Mass

Using the periodic table, Chromium (Cr) has an atomic weight of approximately 52 while Oxygen (O) has an approximate atomic weight of 16. This information will help to understand how the presence of oxygen impacts the mass percentage of Chromium.
03

Arrange Compounds

Since we're not required to do the detailed calculations, we can now arrange the compounds in increasing order of Chromium by mass percentage: CrO3, CrO2, Cr2O3, CrO. In this order, the oxygen atoms increase from one to three causing the Chromium percentage to decrease.

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!

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

The compound \(\mathrm{C}_{7} \mathrm{H}_{7} \mathrm{NO}_{2}\) contains (a) 17 atoms per mole; (b) equal percents by mass of \(\mathrm{C}\) and \(\mathrm{H} ;\) (c) about twice the percent by mass of \(\mathrm{O}\) as of \(\mathrm{N} ;\) (d) about twice the percent by mass of \(\mathrm{N}\) as of \(\mathrm{H}\).

Liquid ethyl mercaptan, \(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{S},\) has a density of \(0.84 \mathrm{g} / \mathrm{mL} .\) Assuming that the combustion of this compound produces only \(\mathrm{CO}_{2}, \mathrm{H}_{2} \mathrm{O},\) and \(\mathrm{SO}_{2},\) what masses of each of these three products would be produced in the combustion of \(3.15 \mathrm{mL}\) of ethyl mercaptan?

Determine the mass percent of each of the elements in the fungicide copper(II) oleate, $\mathrm{Cu}\left(\mathrm{C}_{18} \mathrm{H}_{33} \mathrm{O}_{2}\right)_{2}$

Assign suitable names to the compounds (a) ICl; (b) \(\mathrm{ClF}_{3} ;\) (c) \(\mathrm{SF}_{4} ;\) (d) \(\mathrm{Br} \mathrm{F}_{5} ;\) (e) \(\mathrm{N}_{2} \mathrm{O}_{4} ;\) (f) \(\mathrm{S}_{4} \mathrm{N}_{4}\).

A piece of gold (Au) foil measuring \(0.25 \mathrm{mm} \times\) \(15 \mathrm{mm} \times 15 \mathrm{mm}\) is treated with fluorine gas. The treatment converts all the gold in the foil to \(1.400 \mathrm{g}\) of a gold fluoride. What is the formula and name of the fluoride? The density of gold is \(19.3 \mathrm{g} / \mathrm{cm}^{3}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

Inorganic Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Kinetics

Read Explanation

Chemical Analysis

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