Chemistry 2012 Student Edition (hard Cover) Grade 11
Chemistry 2012 Student Edition (hard Cover) Grade 11
12th Edition
ISBN: 9780132525763
Author: Prentice Hall
Publisher: Prentice Hall
Question
Book Icon
Chapter 14, Problem 115A

(a)

Interpretation Introduction

Interpretation: To calculate the molar mass of the given substance.

Concept Introduction: Molar mass is defined as the smallest unit of the compound. It is the 112th mass of the Carbon-12. Molar mass is defined as the sum of total mass in grams of all the atoms that make up a mole of the molecule.

(a)

Expert Solution
Check Mark

Answer to Problem 115A

158 g/mol.

Explanation of Solution

The given molecular compound is Ca(CH3CO2)2

There is only one calcium atom in the given compound. The molar mass of calcium atom is 40 g/mol.

There are six hydrogen atoms in the given compound. The molar mass of one hydrogen atom is 1 g/mol. Therefore, the molar mass of six hydrogen atoms will be 6×1gmol=6gmol .

There are four carbon atoms in the given compound. The molar mass of one carbon atom is 12 g/mol. Therefore, the molar mass of four carbon atoms will be 12×4gmol=48gmol

There are four oxygen atoms in the given compound. The molar mass of one oxygen atom is 16 g/mol. Therefore, the molar mass of four oxygen atoms will be 16×4gmol=64gmol .

Therefore, the molar mass of the compound will be 158 g/mol.

(b)

Interpretation Introduction

Interpretation: To calculate the molar mass of the given substance.

Concept Introduction: Molar mass is defined as the smallest unit of the compound. It is the 112th mass of the Carbon-12. Molar mass is defined as the sum of total mass in grams of all the atoms that make up a mole of the molecule.

(b)

Expert Solution
Check Mark

Answer to Problem 115A

98 g/mol

Explanation of Solution

There are three hydrogen atoms in the given compound. The molar mass of one hydrogen atom is 1 g/mol. Therefore, the molar mass of three hydrogen atoms will be 3×1gmol=3gmol

There is only one phosphorous atom in the given compound. The molar mass of Phosphorous atom is 31 g/mol.

There are four oxygen atoms in the given compound. The molar mass of one oxygen atom is 16 g/mol. Therefore, the molar mass of four oxygen atoms will be 16×4gmol=64gmol .

The overall molar mass of the compound will be 98 g/mol

(c)

Interpretation Introduction

Interpretation: To calculate the molar mass of the given substance.

Concept Introduction: Molar mass is defined as the smallest unit of the compound. It is the 112th mass of the Carbon-12. Molar mass is defined as the sum of total mass in grams of all the atoms that make up a mole of the molecule.

(c)

Expert Solution
Check Mark

Answer to Problem 115A

342 g/mol.

Explanation of Solution

There are twelve carbon atoms in the given compound. The molar mass of one carbon atom is 12 g/mol. Therefore, the molar mass of twelve carbon atoms will be 12×12gmol=144gmol

There are 22 hydrogen atoms in the given compound. The molar mass of one hydrogen atom is 1 g/mol. Therefore, the molar mass of 22 hydrogen atoms will be 22×1gmol=22gmol

There are eleven oxygen atoms in the given compound. The molar mass of one oxygen atom is 16 g/mol. Therefore, the molar mass of eleven oxygen atoms will be 16×11gmol=176gmol

The overall molar mass of the compound will be 342 g/mol.

(d)

Interpretation Introduction

Interpretation: To calculate the molar mass of the given substance.

Concept Introduction: Molar mass is defined as the smallest unit of the compound. It is the 112th mass of the Carbon-12. Molar mass is defined as the sum of total mass in grams of all the atoms that make up a mole of the molecule.

(d)

Expert Solution
Check Mark

Answer to Problem 115A

331 g/mol

Explanation of Solution

There is one lead atom in the given compound. The molar mass of the lead atom is 207 g/mol.

There are two nitrogen atoms in the given compound. The molar mass of one nitrogen atom is 14 g/mol. Therefore, the molar mass of the two nitrogen atoms will be 14×2gmol=28gmol

There are six oxygen atoms in the given compound. The molar mass of one oxygen atom is 16 g/mol. Therefore, the molar mass of six oxygen atoms will be 16×6gmol=96gmol .

The overall molar mass of the given compound will be 331 g/mol

Chapter 14 Solutions

Chemistry 2012 Student Edition (hard Cover) Grade 11

Ch. 14.2 - Prob. 11SPCh. 14.2 - Prob. 12SPCh. 14.2 - Prob. 13SPCh. 14.2 - Prob. 14SPCh. 14.2 - Prob. 15SPCh. 14.2 - Prob. 16SPCh. 14.2 - Prob. 17LCCh. 14.2 - Prob. 18LCCh. 14.2 - Prob. 19LCCh. 14.2 - Prob. 20LCCh. 14.2 - Prob. 21LCCh. 14.2 - Prob. 22LCCh. 14.2 - Prob. 23LCCh. 14.2 - Prob. 24LCCh. 14.2 - Prob. 25LCCh. 14.3 - Prob. 26SPCh. 14.3 - Prob. 27SPCh. 14.3 - Prob. 28SPCh. 14.3 - Prob. 29SPCh. 14.3 - Prob. 30LCCh. 14.3 - Prob. 31LCCh. 14.3 - Prob. 32LCCh. 14.3 - Prob. 33LCCh. 14.3 - Prob. 34LCCh. 14.3 - Prob. 35LCCh. 14.3 - Prob. 36LCCh. 14.4 - Prob. 37SPCh. 14.4 - Prob. 38SPCh. 14.4 - Prob. 39SPCh. 14.4 - Prob. 40LCCh. 14.4 - Prob. 41LCCh. 14.4 - Prob. 42LCCh. 14.4 - Prob. 43LCCh. 14.4 - Prob. 44LCCh. 14.4 - Prob. 45LCCh. 14.4 - Prob. 46LCCh. 14 - Prob. 47ACh. 14 - Prob. 48ACh. 14 - Prob. 49ACh. 14 - Prob. 50ACh. 14 - Prob. 51ACh. 14 - Prob. 52ACh. 14 - Prob. 53ACh. 14 - Prob. 54ACh. 14 - Prob. 55ACh. 14 - Prob. 56ACh. 14 - Prob. 57ACh. 14 - Prob. 58ACh. 14 - Prob. 59ACh. 14 - Prob. 60ACh. 14 - Prob. 61ACh. 14 - Prob. 62ACh. 14 - Prob. 63ACh. 14 - Prob. 64ACh. 14 - Prob. 65ACh. 14 - Prob. 66ACh. 14 - Prob. 67ACh. 14 - Prob. 68ACh. 14 - Prob. 69ACh. 14 - Prob. 70ACh. 14 - Prob. 71ACh. 14 - Prob. 72ACh. 14 - Prob. 73ACh. 14 - Prob. 74ACh. 14 - Prob. 75ACh. 14 - Prob. 76ACh. 14 - Prob. 77ACh. 14 - Prob. 78ACh. 14 - Prob. 79ACh. 14 - Prob. 80ACh. 14 - Prob. 81ACh. 14 - Prob. 82ACh. 14 - Prob. 83ACh. 14 - Prob. 84ACh. 14 - Prob. 85ACh. 14 - Prob. 86ACh. 14 - Prob. 87ACh. 14 - Prob. 88ACh. 14 - Prob. 89ACh. 14 - Prob. 90ACh. 14 - Prob. 91ACh. 14 - Prob. 92ACh. 14 - Prob. 93ACh. 14 - Prob. 94ACh. 14 - Prob. 95ACh. 14 - Prob. 96ACh. 14 - Prob. 97ACh. 14 - Prob. 98ACh. 14 - Prob. 99ACh. 14 - Prob. 100ACh. 14 - Prob. 101ACh. 14 - Prob. 102ACh. 14 - Prob. 106ACh. 14 - Prob. 107ACh. 14 - Prob. 108ACh. 14 - Prob. 109ACh. 14 - Prob. 110ACh. 14 - Prob. 111ACh. 14 - Prob. 112ACh. 14 - Prob. 113ACh. 14 - Prob. 114ACh. 14 - Prob. 115ACh. 14 - Prob. 116ACh. 14 - Prob. 117ACh. 14 - Prob. 118ACh. 14 - Prob. 119ACh. 14 - Prob. 120ACh. 14 - Prob. 121ACh. 14 - Prob. 122ACh. 14 - Prob. 123ACh. 14 - Prob. 1STPCh. 14 - Prob. 2STPCh. 14 - Prob. 3STPCh. 14 - Prob. 4STPCh. 14 - Prob. 5STPCh. 14 - Prob. 6STPCh. 14 - Prob. 7STPCh. 14 - Prob. 8STPCh. 14 - Prob. 9STPCh. 14 - Prob. 10STPCh. 14 - Prob. 11STPCh. 14 - Prob. 12STP
Knowledge Booster
Background pattern image
Recommended textbooks for you
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781259911156
Author:Raymond Chang Dr., Jason Overby Professor
Publisher:McGraw-Hill Education
Text book image
Principles of Instrumental Analysis
Chemistry
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9780078021558
Author:Janice Gorzynski Smith Dr.
Publisher:McGraw-Hill Education
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY