Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 3: Problem 42

A compound \(\mathrm{YO}_{2}\) is \(50.0 \%\) (by mass) oxygen. What is the identity of Y? What is the molar mass of \(\mathrm{YO}_{2} ?\)

Short Answer

Expert verified
Answer: Since YO2 is 50% oxygen by mass, it is also 50% element Y by mass. We can find the mass of Y in the compound using the molar mass of oxygen (16.00 g/mol) and the equation: Molar mass of Y = (2 * Mass of 1 mole of oxygen atoms) - 2 * 16.00 g/mol. Then, we can look for an element with a similar molar mass in the periodic table to find the identity of Y.

Step by step solution

01

Calculate the mass of oxygen in the compound

Given that the compound YO2 is 50.0% oxygen by mass, this means that half of its mass comes from the two oxygen atoms. Since it has two oxygen atoms in YO2, we can calculate the mass of 1 mole of oxygen atoms by: Mass of 1 mole of oxygen atoms = (Mass of YO2) * 50%
02

Determine the mass of Y in the compound

Now that we have found the mass of 1 mole of oxygen atoms in YO2, we can find the mass of Y in the compound in a similar manner. We know that the molar mass of the compound YO2 is equal to the molar mass of Y plus twice the molar mass of oxygen atoms (O). Molar mass of YO2 = Molar mass of Y + 2 * Molar mass of O
03

Find the identity of Y

We will use the molar mass of oxygen (16.00 g/mol) to find the molar mass of Y. Molar mass of Y + 2 * Molar mass of O = Molar mass of YO2 Molar mass of Y + 2 * 16.00 g/mol = YO2 mass Molar mass of Y = YO2 mass - 2 * 16.00 g/mol Now we have to find the YO2 mass. Since the compound YO2 is 50% oxygen by mass, and we already found the mass of 1 mole of oxygen atoms, we can find the YO2 mass by multiplying the mass of oxygen atoms by 2. YO2 mass = 2 * Mass of 1 mole of oxygen atoms Substitute this in the equation for the molar mass of Y: Molar mass of Y = (2 * Mass of 1 mole of oxygen atoms) - 2 * 16.00 g/mol Now we can look for an element close to this mass in the periodic table, which will be the identity of Y.
04

Determine the molar mass of YO2

Once the identity of Y is found, we can calculate the molar mass of YO2 using the molar mass of Y and the molar mass of oxygen atoms (O). Molar mass of YO2 = Molar mass of Y + 2 * Molar mass of O And with the molar mass of YO2 found, we have successfully completed this exercise.

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

Chromium (atomic mass \(=51.9961\) amu) has four isotopes. Their masses are \(49.94605 \mathrm{amu}, 51.94051 \mathrm{amu}, 52.94065 \mathrm{amu}\), and \(53.93888\) amu. The first two isotopes have a total abundance of \(87.87 \%\), and the last isotope has an abundance of \(2.365 \%\). What is the abundance of the third isotope? Estimate the abundances of the first two isotopes.

When potassium chlorate is burned, potassium chloride and oxygen are formed. (a) Write a balanced equation for the reaction. (b) How much potassium chlorate must be burned to produce \(198.5 \mathrm{~g}\) of oxygen? The yield is found to be \(83.2 \%\).

Determine whether the statements given below are true or false. (a) The mass of an atom can have the unit mole. (b) In \(\mathrm{N}_{2} \mathrm{O}_{4}\), the mass of the oxygen is twice that of the nitrogen. (c) One mole of chlorine atoms has a mass of \(35.45 \mathrm{~g}\). (d) Boron has an average atomic mass of \(10.81\) amu. It has two isotopes, \(\mathrm{B}-10(10.01\) amu \()\) and \(\mathrm{B}-11(11.01 \mathrm{amu}) .\) There is more naturally occurring B-10 than B-11. (e) The compound \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{2} \mathrm{~N}\) has for its simplest formula \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{ON}_{1 / 2}\). (f) A 558.5-g sample of iron contains ten times as many atoms as \(0.5200 \mathrm{~g}\) of chromium. (g) If \(1.00\) mol of ammonia is mixed with \(1.00\) mol of oxygen the following reaction occurs, $$ 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) $$ All the oxygen is consumed. (h) When balancing an equation, the total number of moles of reactant molecules must equal the total number of moles of product molecules.

97\. Answer the questions below, using LT (for is less than), GT (for is greater than), EQ (for is equal to), or MI (for more information required) in the blanks provided. (a) The mass (to three significant figures) of \(6.022 \times 10^{23}\) atoms of Na ______ 23.0 g. (b) Boron has two isotopes, B-10 (10.01 amu) and B-11 (11.01 amu). The abundance of B-10 _______the abundance of B-11. (c) If S-32 were assigned as the standard for expressing relative atomic masses and assigned an atomic mass of \(10.00 \mathrm{amu}\), the atomic mass for \(\mathrm{H}\) would be _____\(1.00 \mathrm{amu} .\) (d) When phosphine gas, \(\mathrm{PH}_{3}\), is burned in oxygen, tetraphosphorus decaoxide and steam are formed. In the balanced equation (using smallest whole-number coefficients) for the reaction, the sum of the coefficients on the reactant side is _______ \(7 .\) (e) The mass (in grams) of one mole of bromine molecules is _______ \(79.90\)

Hexamethylenediamine (MM \(=116.2 \mathrm{~g} / \mathrm{mol}\) ), a compound made up of carbon, hydrogen, and nitrogen atoms, is used in the production of nylon. When \(6.315 \mathrm{~g}\) of hexamethylenediamine is burned in oxygen, \(14.36 \mathrm{~g}\) of carbon dioxide and \(7.832 \mathrm{~g}\) of water are obtained. What are the simplest and molecular formulas of this compound?
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Kinetics

Read Explanation

Chemical Reactions

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