Chemistry: The Molecular Nature of Matter and Change
Chemistry: The Molecular Nature of Matter and Change
8th Edition
ISBN: 9781259631757
Author: Martin Silberberg Dr., Patricia Amateis Professor
Publisher: McGraw-Hill Education
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 3, Problem 3.14P

(a)

Interpretation Introduction

Interpretation:

The mass of 0.68 mol of KMnO4 is to be calculated.

Concept introduction:

The molar mass is defined as the mass of one mol of a substance in grams. The molar mass of a compound is calculated by adding the molar mass of each element multiplied by its number of atoms present in the chemical formula.

The mass of a compound in grams can be calculated from the given moles and the molar mass of that compound.

The expression to calculate the mass of a chemical substance in grams is as follows:

Mass (g)=(amount(mol))(no. of grams1 mol)

(a)

Expert Solution
Check Mark

Answer to Problem 3.14P

The mass of 0.68 mol of KMnO4 is 1.1×102 g.

Explanation of Solution

The expression to calculate the molar mass of KMnO4 is as follows:

Molar mass of KMnO4=(1)(M of K)+(1)(M of Mn)+(4)(M of O)

Substitute 39.10 g/mol for M of K, 54.94 g/mol for M of Mn, and 16.00 g/mol for M of O in the above equation as follows:

Molar mass of KMnO4=(1)(39.10 g/mol)+(1)(54.94 g/mol)+(4)(16.00 g/mol)=94.04 g/mol+64 g/mol=158.04 g/mol

The molar mass of KMnO4 is 158.04 g/mol.

Calculate the mass of 0.68 mol of KMnO4 as follows:

Mass ofKMnO4(g)=(0.68 mol KMnO4)(158.04 g KMnO41 mol KMnO4)=1.1×102 g KMnO4

Conclusion

Therefore, the mass of 0.68 mol of KMnO4 is 1.1×102 g.

(b)

Interpretation Introduction

Interpretation:

The moles of oxygen atoms in 8.18 g of Ba(NO3)2 is to be calculated.

Concept introduction:

One mole is defined as the amount of substance that contains the same number of entities such as molecules, ions, atoms as the number of atoms in 12 g of C-12. This number is called Avogadro’s number. The value of Avogadro’s number is 6.022×1023 (entitites/mol).

The amount of an element can be calculated from the given amount of compound, the moles of that element present in one mole of the compound and the Avogadro’ number. The expression to calculate the moles of atoms of an element in a compound is as follows:

Amount of element(mol)=(amount ofcompound(mol))(moles of element inmolecular formula1 mol compound)

(b)

Expert Solution
Check Mark

Answer to Problem 3.14P

In 8.18 g of Ba(NO3)2, 0.188 mol O atoms are present.

Explanation of Solution

The expression to calculate the molar mass of Ba(NO3)2 is as follows:

Molar mass of Ba(NO3)2=(1)(M of Ba)+(2)(M of N)+(6)(M of O)

Substitute 137.3 g/mol for M of Ba, 14.01 g/mol for M of N, and 16.00 g/mol for M of O in the above equation as follows:

Molar mass of Ba(NO3)2=(1)(137.3 g/mol)+(2)(14.01 g/mol)+(6)(16.00 g/mol)=165.32 g/mol+96 g/mol=261.3 g/mol

The molar mass of Ba(NO3)2 is 261.3 g/mol.

Calculate the amount of 8.18 g of Ba(NO3)2 as follows:

Amount of Ba(NO3)2(mol)=(8.18 g Ba(NO3)2)(1 mol Ba(NO3)2261.3 g Ba(NO3)2)=0.031305 mol Ba(NO3)2

Therefore, 0.031305 mol of Ba(NO3)2 are present in 8.18 g of Ba(NO3)2.

One mole of Ba(NO3)2 contains 6 mol of oxygen atoms. Calculate the moles of O atoms in 0.031305 mol of Ba(NO3)2 as follows:

Moles of O atoms=(0.031305 mol Ba(NO3)2)(6 mol O atoms1 mol Ba(NO3)2)=0.18783 mol O atoms0.188 mol O atoms

Conclusion

In 8.18 g of Ba(NO3)2, 0.188 mol O atoms are present.

(c)

Interpretation Introduction

Interpretation:

The number of oxygen atoms in 7.3×103 g of CaSO42H2O is to be calculated.

Concept introduction:

One mole is defined as the amount of substance that contains the same number of entities such as molecules, ions, atoms as the number of atoms in 12 g of C-12. This number is called Avogadro’s number. The value of Avogadro’s number is 6.022×1023 (entitites/mol).

The number of an atom of an element can be calculated from the given amount of compound, the moles of that element present in one mole of the compound and the Avogadro’ number.

The expression to calculate the amount of a chemical substance in moles is as follows:

Amount(mol)=(Mass (g))(1 molno. of grams)

The expression to calculate the amount of atoms of an element in a compound is as follows:

Amount of element(mol)=(amount ofcompound(mol))(moles of element inmolecular formula1 mol compound)

The expression to calculate the number of atoms of an element is as follows:

Number of atoms=(amount(mol))(6.022×1023 atoms1 mol)

(c)

Expert Solution
Check Mark

Answer to Problem 3.14P

The number of oxygen atoms in 7.3×103 g of CaSO42H2O is 1.5×1020 O atoms.

Explanation of Solution

The expression to calculate the molar mass of CaSO42H2O is as follows:

Molar mass of CaSO42H2O=(1)(M of Ca)+(1)(M of S)+(4)(M of H)+(6)(M of O)

Substitute 40.08 g/mol for M of Mg, 1.008 g/mol for M of H, 16.00 g/mol for M of O, and 32.06 g/mol for M of S in the above equation as follows:

Molar mass of CaSO42H2O=[(1)(40.08 g/mol)+(1)(32.06 g/mol)+(4)(1.008 g/mol)+(6)(16.00 g/mol)]=40.08 g/mol+32.06 g/mol+4.032 g/mol+96.00 g/mol=172.17 g/mol

Therefore, the molar mass of CaSO42H2O is 172.17 g/mol.

Calculate the amount of 7.3×103 g of CaSO42H2O as follows:

Amount of CaSO42H2O(mol)=(7.3×103 g)(1 mol CaSO42H2O172.17 g CaSO42H2O)=4.239995×105 mol CaSO42H2O

Therefore, 4.239995×105 mol of CaSO42H2O are present in 7.3×103 g of CaSO42H2O.

One mole of CaSO42H2O contains 6 mol of oxygen atoms. Calculate the moles of O atoms in 4.239995×105 mol of CaSO42H2O as follows:

Moles of O atoms=(4.239995×105 mol CaSO42H2O)(6 mol O atoms1 mol CaSO42H2O)=2.543997×105 mol O atoms

Therefore, 2.543997×105 mol of oxygen atoms are present in 4.239995×103 mol of CaSO42H2O.

Calculate the number of oxygen atoms in 2.543997×105 mol of oxygen atoms as follows:

Number of O atoms=(2.543997×105 mol O atoms)(6.022×1023 O atoms1 mol O atom)=1.5320×1020 O atoms1.5×1020 O atoms

Conclusion

In 7.3×103 g of CaSO42H2O , 1.5×1020 O atoms are present.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Chapter 3 Solutions

Chemistry: The Molecular Nature of Matter and Change

Ch. 3.1 - Prob. 3.6AFPCh. 3.1 - For many years, compounds known as...Ch. 3.1 - Use the information in Follow-up Problem 3.6A to...Ch. 3.1 - Prob. 3.7BFPCh. 3.2 - Prob. 3.8AFPCh. 3.2 - A sample of an unknown compound contains 6.80 mol...Ch. 3.2 - A sample of an unknown compound is found to...Ch. 3.2 - Prob. 3.9BFPCh. 3.2 - Prob. 3.10AFPCh. 3.2 - Prob. 3.10BFPCh. 3.2 - A dry-cleaning solvent (ℳ = 146.99 g/mol) that...Ch. 3.2 - Prob. 3.11BFPCh. 3.3 - Prob. 3.12AFPCh. 3.3 - Prob. 3.12BFPCh. 3.3 - Prob. 3.13AFPCh. 3.3 - Prob. 3.13BFPCh. 3.4 - Prob. 3.14AFPCh. 3.4 - The tarnish that forms on objects made of silver...Ch. 3.4 - Prob. 3.15AFPCh. 3.4 - In the reaction that removes silver tarnish (see...Ch. 3.4 - Prob. 3.16AFPCh. 3.4 - Prob. 3.16BFPCh. 3.4 - Prob. 3.17AFPCh. 3.4 - Prob. 3.17BFPCh. 3.4 - Prob. 3.18AFPCh. 3.4 - Prob. 3.18BFPCh. 3.4 - In the reaction in Follow-up Problem 3.18A, how...Ch. 3.4 - Prob. 3.19BFPCh. 3.4 - Prob. 3.20AFPCh. 3.4 - Prob. 3.20BFPCh. 3.4 - Marble (calcium carbonate) reacts with...Ch. 3.4 - Prob. 3.21BFPCh. 3 - Prob. 3.1PCh. 3 - Prob. 3.2PCh. 3 - Why might the expression “1 mol of chlorine” be...Ch. 3 - Prob. 3.4PCh. 3 - Prob. 3.5PCh. 3 - Prob. 3.6PCh. 3 - Prob. 3.7PCh. 3 - Prob. 3.8PCh. 3 - Calculate the molar mass of each of the...Ch. 3 - Prob. 3.10PCh. 3 - Prob. 3.11PCh. 3 - Calculate each of the following quantities: Mass...Ch. 3 - Calculate each of the following quantities: Amount...Ch. 3 - Prob. 3.14PCh. 3 - Prob. 3.15PCh. 3 - Prob. 3.16PCh. 3 - Prob. 3.17PCh. 3 - Prob. 3.18PCh. 3 - Prob. 3.19PCh. 3 - Calculate each of the following: Mass % of H in...Ch. 3 - Calculate each of the following: Mass % of I in...Ch. 3 - Calculate each of the following: Mass fraction of...Ch. 3 - Calculate each of the following: Mass fraction of...Ch. 3 - Oxygen is required for the metabolic combustion of...Ch. 3 - Cisplatin (right), or Platinol, is used in the...Ch. 3 - Allyl sulfide (below) gives garlic its...Ch. 3 - Iron reacts slowly with oxygen and water to form a...Ch. 3 - Prob. 3.28PCh. 3 - Prob. 3.29PCh. 3 - The mineral galena is composed of lead(II) sulfide...Ch. 3 - Prob. 3.31PCh. 3 - Prob. 3.32PCh. 3 - List three ways compositional data may be given in...Ch. 3 - Prob. 3.34PCh. 3 - Prob. 3.35PCh. 3 - Prob. 3.36PCh. 3 - Prob. 3.37PCh. 3 - Prob. 3.38PCh. 3 - Prob. 3.39PCh. 3 - What is the molecular formula of each...Ch. 3 - Prob. 3.41PCh. 3 - Prob. 3.42PCh. 3 - Find the empirical formula of each of the...Ch. 3 - An oxide of nitrogen contains 30.45 mass % N. (a)...Ch. 3 - Prob. 3.45PCh. 3 - A sample of 0.600 mol of a metal M reacts...Ch. 3 - Prob. 3.47PCh. 3 - Prob. 3.48PCh. 3 - Prob. 3.49PCh. 3 - Prob. 3.50PCh. 3 - Prob. 3.51PCh. 3 - Prob. 3.52PCh. 3 - Prob. 3.53PCh. 3 - Prob. 3.54PCh. 3 - Prob. 3.55PCh. 3 - Prob. 3.56PCh. 3 - Prob. 3.57PCh. 3 - Prob. 3.58PCh. 3 - Prob. 3.59PCh. 3 - Prob. 3.60PCh. 3 - Prob. 3.61PCh. 3 - Prob. 3.62PCh. 3 - Prob. 3.63PCh. 3 - Prob. 3.64PCh. 3 - Prob. 3.65PCh. 3 - Prob. 3.66PCh. 3 - Prob. 3.67PCh. 3 - Prob. 3.68PCh. 3 - Prob. 3.69PCh. 3 - Prob. 3.70PCh. 3 - Prob. 3.71PCh. 3 - Prob. 3.72PCh. 3 - Prob. 3.73PCh. 3 - Prob. 3.74PCh. 3 - Elemental phosphorus occurs as tetratomic...Ch. 3 - Prob. 3.76PCh. 3 - Solid iodine trichloride is prepared in two steps:...Ch. 3 - Prob. 3.78PCh. 3 - Prob. 3.79PCh. 3 - Prob. 3.80PCh. 3 - Prob. 3.81PCh. 3 - Prob. 3.82PCh. 3 - Prob. 3.83PCh. 3 - Prob. 3.84PCh. 3 - Prob. 3.85PCh. 3 - Prob. 3.86PCh. 3 - Prob. 3.87PCh. 3 - Prob. 3.88PCh. 3 - Prob. 3.89PCh. 3 - Prob. 3.90PCh. 3 - Prob. 3.91PCh. 3 - Prob. 3.92PCh. 3 - Prob. 3.93PCh. 3 - Prob. 3.94PCh. 3 - When 20.5 g of methane and 45.0 g of chlorine gas...Ch. 3 - Prob. 3.96PCh. 3 - Prob. 3.97PCh. 3 - Prob. 3.98PCh. 3 - Prob. 3.99PCh. 3 - Prob. 3.100PCh. 3 - Sodium borohydride (NaBH4) is used industrially in...Ch. 3 - Prob. 3.102PCh. 3 - The first sulfur-nitrogen compound was prepared in...Ch. 3 - Prob. 3.104PCh. 3 - Prob. 3.105PCh. 3 - Prob. 3.106PCh. 3 - Serotonin () transmits nerve impulses between...Ch. 3 - In 1961, scientists agreed that the atomic mass...Ch. 3 - Prob. 3.109PCh. 3 - Isobutylene is a hydrocarbon used in the...Ch. 3 - The multistep smelting of ferric oxide to form...Ch. 3 - Prob. 3.112PCh. 3 - Prob. 3.113PCh. 3 - Prob. 3.114PCh. 3 - Prob. 3.115PCh. 3 - Prob. 3.116PCh. 3 - Prob. 3.117PCh. 3 - Prob. 3.118PCh. 3 - Prob. 3.119PCh. 3 - Prob. 3.120PCh. 3 - For the reaction between solid tetraphosphorus...Ch. 3 - Prob. 3.122PCh. 3 - Prob. 3.123PCh. 3 - Prob. 3.124PCh. 3 - Prob. 3.125PCh. 3 - Prob. 3.126PCh. 3 - Prob. 3.127PCh. 3 - Prob. 3.128PCh. 3 - Ferrocene, synthesized in 1951, was the first...Ch. 3 - Prob. 3.130PCh. 3 - Prob. 3.131PCh. 3 - Citric acid (below) is concentrated in citrus...Ch. 3 - Prob. 3.133PCh. 3 - Nitrogen monoxide reacts with elemental oxygen to...Ch. 3 - Prob. 3.135PCh. 3 - Prob. 3.136PCh. 3 - Manganese is a key component of extremely hard...Ch. 3 - The human body excretes nitrogen in the form of...Ch. 3 - Aspirin (acetylsalicylic acid, C9H8O4) is made by...Ch. 3 - Prob. 3.140PCh. 3 - Prob. 3.141PCh. 3 - Prob. 3.142PCh. 3 - When powdered zinc is heated with sulfur, a...Ch. 3 - Cocaine (C17H21O4N) is a natural substance found...Ch. 3 - Prob. 3.145P
Knowledge Booster
Background pattern image
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
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
Bonding (Ionic, Covalent & Metallic) - GCSE Chemistry; Author: Science Shorts;https://www.youtube.com/watch?v=p9MA6Od-zBA;License: Standard YouTube License, CC-BY
Stoichiometry - Chemistry for Massive Creatures: Crash Course Chemistry #6; Author: Crash Course;https://www.youtube.com/watch?v=UL1jmJaUkaQ;License: Standard YouTube License, CC-BY